okay so this is called summation

notation or Sigma notation so it means

we're going to be adding some stuff

together so we're going to take this

we're going to plug in n equals 1 and

then we're going to plug in n equals 2

and then we're going to plug in n equals

3 and we stop when we get to this number

up here so for this first one we plug in

1 for the N for this one we plug in 2

and for this one we plug in 3. 1 squared

is just one so this is all brought down

three times one is three three plus nine

is 12. plus whatever this simplifies to

2 squared is 4 times 3 plus 9. 3 times 4

is 12. 12 plus 9 is 21 plus whatever's

inside here 3 squared is 9 times 3 plus

9. 3 times 9 is 27 plus the 9 is 36. so

now we just want to add up these three

terms two plus one is three plus six is

nine and then one plus two is three plus

three is six so these add up to 69 so

this is equal to 69 and we can put a box

around it nice how exciting

Sigma Sum


hey guys this looks like a fun one it

says solve for x and it's giving us this

shape right here we have a 75° angle

here we have this segment the same

length as this segment the same length

as this segment and there's a right

angle right here and it wants us to find

the value of x if you want to try it on

your own posit right now because I'm

going to solve it in 3 2 1 first let's

focus on this right triangle any

triangle can be inscribed in a circle so

let's inscribe it in a circle and for

every inscribed right triangle the

hypoten is the diameter of the circle so

we know that the center is going to lie

on this hypotenuse and since we know

this is congruent to this we know the

center is going to be right there and

each of these are radius of the circle

and we can draw another radius from the

center of the circle to the vertex of

the right angle now I think this is

everything we need to solve the problem

if you want to pause it now cuz I'm

going to actually solve it in 3 2 1 for

the next step if this angle is X this

angle right here will also be X and

these are called vertical angles and

then we don't know how big this angle is

so let's call it y so now if we focus on

this angle right here if the inscribed

angle is y The Arc that it subtends is

always double the inscribed angle so

this Arc from here to here is 2 y de and

then if we focus on this angle the

central angle is always congruent to the

measure of the arc so if this Arc is 2 y

de this angle will be 2 y de next we can

label this angle right here Z degre and

if we focus on this isoceles triangle in

an isoceles triangle you have two sides

congruent and you also have the base

angles congruent so this angle right

here will also be equal to Z de so now

we're ready to do some algebra we can

focus on these three angles right here

we have 2 y + z + 75 is equal to 180 and

then next we can focus on this triangle

right here the sum of the interior

angles of any triangle is 180° we know

that x + y + 75 is equal to 180 and last

let's focus on this right triangle we

know that x + z + 90 will equal 180 now

let's clean these up a little bit for

the first one we can subtract 75 from

both sides and we can do the same thing

for the second one and then for the

third one we can subtract 90 from both

sides 75 - 75 is 0 75 - 75 is 0 and 90 -

90 is 0 and then 180 - 75 is 105 180 -

75 is 105 and 180 - 90 is 90 right now

we have an x a y and a z and we want to

solve for x we're going to do this with

elimination now let's rearrange things a

little bit for this second equation

let's multiply both sides of the

equation by -2 this -2 will distribute

to both of these giving us -2X and -2 Y

and then on the right hand side 105 * -2

is -210 and let's shift that up here now

we can add these two equations together

we can bring down the -2X pos2 y -2 y

will go away and we can bring down the

positive Z and then on this side 105 -

210 is

105 and now we're down to x and z so we

just got to eliminate the Z we're going

to multiply both sides of this bottom

equation by1 the negative 1 will

distribute to give USX and negative Z

and on the right hand side it will be 90

and then we can bring that up here and

add the two equations together -2X - 1X

will be -3x and5 - 90 is-

1995 now to solve for x let's divide

both sides by 3 on the left hand side we

have X and on the right hand side -195 /

-3 is POS 65 so now we found this

unknown angle X is equal to 65 and this

is the answer to our question let's put

a box around it how exciting here's the

next problem I plan on working on if you

guys want to try it out I'll post the

solution as soon as I solve it how

exciting

Today's Geometry Challenge

I hope you guys like this one!


hey guys welcome to this video we're

going to solve for x

first step is we want to distribute this

4 to both of these terms

4 times x is 4x 4 times 1 is 4

plus and then we want to distribute this

2 to both of these terms

2 times 3x is 6x and 2 times 1 is 2

and it's still equal to 16. now we

combine like terms this

4x and this plus 6x become 10x

and this plus 4 and this plus 2 becomes

6 and it's still equal to 16.

now subtract 6 from both sides this

cancels

we end up with 10x equals 10.

then we divide both sides by 10 and we

get the answer of x

equals 1. comment below on what you'd

like me to solve for next and i'll talk

to you guys soon

bye

Solve for x using Distribution Fast Example #Shorts

Check out my website!
https://andymath.com/
#Shorts


this video is sponsored by brilliant hey

guys this looks like a fun one we're

given that these three angles are equal

and it wants to know what fraction of

the entire rectangle is shaded if you

want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 1 first since it's a rectangle we know

all these angles are right angles this

is a 90° angle up here and since all

these angles are equal each of them has

to be 30° next let's look at this top

triangle if this is 30 and this is 90

this angle has to be 60 and same thing

for this lower triangle if this angle's

30 this one's 90 this one has to be 60°

since both of these are 30 6090

triangles let's look at the notes for

306090 triangles in every 306090

triangle the hypotenuse is always twice

as big as the shortest side and to get

the length of the side opposite the 60°

you multiply the shortest side by the

square < TK of 3 and this is always true

for every single 30 6090 triangle so how

are we going to use these notes if we

don't have any side lengths we can

assign the variable X to this piece

right here so we can take our triangle

out and to find this length we can take

our shortest side and multiply it by

square < TK of three now let's bring the

triangle back up here we have this piece

in terms of X and we have the height of

our entire rectangle in terms of X and

in a rectangle opposite sides are

congruent so this side is also X < tk3

now let's pull this triangle out and

let's find the length of this side in

terms of X we're going to take this xk3

and multiply it by S < TK of 3 so we end

up with x < tk3 * < tk3 < tk3 * < tk3 is

equal to < TK of 9 and theun of 9 is

equal to 3 so this side length is equal

to 3x so let's bring this triangle back

up here now looking at this rectangle we

have both the height and the width in

terms of X since we know the whole width

is 3x and this piece is X that means

this piece is equal to 2X and now we're

ready to answer the question we want to

do the Shaded area divided by the total

area the Shaded area is this triangle

it's going to be 1/2 base time height so

it's 1/2 * the base of 2x * the height

of X < tk3 and then to get the total

area of the rectangle we're going do

base time height the base is equal to 3x

and the height is equal to X < tk3 and

now we can clean this up a little bit

this 1/2 * this 2 is equal to one and

this xk3 and this xk3 are going to

cancel each other out so now we have 1 X

over 3x well these two x's will also

cancel each other out now we have the

answer to our question 1/3 of this

entire rectangle is shaded let's put a

box around it I really like this problem

I thought it was brilliant speaking of

brilliant let me give you an update on

Brilliance I went through this

predicting with probability I love to

use historical data to form probability

distributions about future events it's

such a cool concept make predictions

about weather medicine traffic even

artificial intelligence this is what I

was looking forward to the most the Bas

theorem stuff I think they chose a

really good example for the lesson and

then I love the practice problems for

each of the problems they had a hint and

a full explanation I found it fun to

First not use the formula and just see

if I could logically deduce what would

occur and then I use the formula to

verify it I don't want to go through too

much detail in case you want to go

through it yourself speaking of going

through it yourself if you want to try

out brilliant there's a free 30-day

trial visit brilliant.org animath or

click on the link in the description you

can also get 20% off an annual premium

plan how exciting

What fraction is shaded?

To try everything Brilliant has to offer for free for a full 30 days, visit https://brilliant.org/AndyMath/ . You'll also get 20% off an annual premium subscription.


hey guys this is one of the most

requested videos I've ever had and for

good reason this is such a cool problem

I did not expect it to have the outcome

that it had so the problem gives us a

semi-circle right here and a quarter

Circle right here and then this red line

with the length of five that goes from

the vertex of this rectangle tangent to

the semicircle if you want to try this

on your own pause it right now because

I'm going to solve it in three two one

so first up let's assign a radius to

this semicircle a lot of people wanted

to assign this radius to match but there

really is not evidence that this radius

is the same as this radius so let's call

it something else like X using our

method we're not going to need this

radius to equal this radius and that

right there is one of my favorite things

about this problem we don't care if the

radii are the same so this last distance

right here will also be R because it's a

radius of the same semicircle and this

height will also be X because it's a

radius of this quarter Circle and now we

can come up with a formula for the area

of the rectangle the area of a rectangle

is base times height the base is going

to be equal to X Plus 2R and the height

is going to be equal to X and then we

can distribute this x x times x is going

to give us x squared and x times 2R is

going to give us 2xr and this is a very

nice formula for the area for the next

step let's draw a segment from this

point of tangency to the center of the

semicircle and this distance is also a

radius of the semicircle so it's going

to be R this right here is going to be a

right angle every radius that intersects

a tangent line at the point of tangency

is at right angles so now we have a nice

right triangle so we can do Pythagorean

theorem r squared plus 5 squared equals

the quantity X plus r squared we can

bring down the r squared 5 squared is 25

and X plus r squared means X plus r

times X plus r and after we multiply

that out we get x squared plus 2xr plus

r squared and then we can subtract r

squared from both sides on the left hand

side we have 25 and on the right hand

side we have x squared plus 2xr and now

we're pretty much done we wanted to know

what is the area of the rectangle but we

know that the area is equal to x squared

plus 2xr so we can plug that in right

here but we also know that X squared

plus 2xr is equal to 25 so we can change

this to 25 and that's the answer to our

question let's give it a label of

centimeters squared and put a box around

it even if these radii are not equal to

each other we can find the area of this

rectangle is exactly 25 centimeters

squared how exciting Isn't that cool

Super Popular Problem

https://andymath.com/geometry-challenges/

A lot of people sent me requests to go over this problem. I hope you love it!


hey guys this looks like a fun one it

says how long is AB and it's giving us

that all semicircles have the same

radius and it's kind of hard to see but

these are red arrows pointed to the

spaces between the semicircles if you

want to try it on your own pause it

right now CU I'm going to solve it in 3

2 1 so first let's identify a radius and

since we're given that they all have the

same radius all of these are going to be

the same and then let's look at the top

and the bottom the top is 6 RS plus a 3

plus a 3 and the bottom has four RS plus

a 66 + a 12 + a 66 and we can simplify

both of these 3 + 3 is equal to 6 and 66

+ 12 + 66 is equal to 144 and ultimately

we're trying to find ab ab is going to

be equal to either this top equation or

the bottom equation let's just use the

6r + 6 and then once we solve for R

we'll be able to find ab and the way

we're going to solve for R is this top

has to equal the bottom so that means

that 6r + 6 has to equal 4 r + 144 and

now we have a nice nice equation let's

solve for R we can subtract 4 R from

both sides and subtract six from both

sides on the left hand side the sixes

will cancel each other out and that

leave us with 2 R on the right hand side

the RS will cancel each other out and

that leaves us with 144 - 6 which is 138

next we can divide both sides by two on

the left hand side we have R and on the

right hand side 138 ided 2 is 69 nice

that means that each of these radii are

equal to 69 and now we're ready to solve

for ab in the place of R we can plug in

69 6 * 69 is equal to 414 and 414 + 6 is

420 that is the answer to our question

let's put a box around it the length of

ab in this diagram is equal to 420 how

exciting

Nice Question


says solve for x no calculator the first

step I think I want to set everything

equal to zero so I'm going to subtract

2x from both sides and add 3x squared to

both sides so on the right hand side

it's going to be equal to 0. we have X

cubed plus 3x squared minus 2X x is 0 is

1 so it's going to be minus six from

here we're going to do Factor by

grouping we're going to pull out the

common x squared for these two terms and

a common negative 2 for these two terms

and it's still equal to zero x squared

times what gives us X cubed that's x x

squared times what gives us 3x squared

that's three negative 2 times what gives

us negative 2x that's X and negative 2

times y gives us negative 6 that's

positive 3. now both of these have an X

plus 3 term so I'm going to factor that

out so I have X plus 3 times what gives

me all of this that's going to be x

squared and then X plus 3 times what

gives me all this that's minus 2. and

from here we know either X plus 3 equals

0 or x squared minus two equals 0. we'll

subtract 3 from both sides and we get x

equals negative three and for this one

we add 2 to both sides get x squared

equals two square root both sides and we

get x equals plus or minus root two and

these are the three answers for X how

exciting

Factor by Grouping Challenge

Collab with Geeklyhub!


this video is sponsored by brilliant hey

guys this looks like a fun one there's a

chord cutting the square in half and it

has a length of six and it wants to know

what is the area of the semicircle if

you want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 1 the thing we're solving for is the

area of a semicircle that's going to be

equal to Pi R 2 / 2 and for the first

step I want to hide the semicircle and

focus on the Square let's take both

diagonals of the square and the

intersection of those diagonals would be

the center of the square and you can see

this also goes through the center and

that's true for any line that bisects

the square it will always go through the

center of the square and now let's focus

on the given chord the center of the

square will cut this chord in half so we

get three and three now we can bring

back the semicircle and let's focus on

the semicircle through this midpoint

let's draw a perpendicular line it's

going to be the perpendicular bis sector

of our chord and the perpendicular bis

sector of any chord will always go

through the center of a circle so we can

mark this center right here next I'm

thinking let's go back to the square and

let's take this piece right here and

extend it all the way to the end of the

square since these two are equal to

three these two will also be equal to

three and that is something else that's

universally true as long as this is

cutting the square in half and this is

perpendicular to it these portions will

always be equal to each other I'm mostly

interested in these three down here we

don't really need this one and let's

bring back our semicircle and next I

want to look at the whole circle and

let's extend this down here and this

piece will be equal to the radius that

we're trying to solve for this red

radius is equal to this blue radius

since this blue thing is equal to R and

this piece is equal to 3 that leaves Rus

3 for this little piece right here and

if we focus on this entire green length

right here it's equal to r + 3 so let's

write it as r + 3 and now we can use

another set of notes this one is about

the relationship of intersecting chords

in circles a * B will always equal C * D

and for our Circle here 3 * 3 will equal

the quantity r + 3 * the quantity Rus 3

on the left hand side 3 * 3 is equal to

9 and r + 3 * r - 3 is a difference of 2

squares it simplifies to r^ 2 - 9 and

here's the notes right here and it ends

up being a + b * Aus B is equal to a^ 2

- b^ 2 now we have 9 is equal to r^ 2 -

9 to get the r 2 by itself let's add 9

to both sides 99 + 9 is equal to 18 and

these 9es cancel each other out leaving

us with r 2 then we can square root both

sides of the equation and we get R is

equal to the < TK of 18 and this is

really all we need to find the area of

our semicircle in the place of this R we

can plug in square < TK of 18 this

square root and this Square will cancel

each other out and 18 / 2 is equal to 9

and we now have the area of our

semicircle it's equal to 9 pi unit squar

let's put a box around it I really like

this question I thought it was a

brilliant one Speaking of brilliant I

want to give you guys another update on

brilliant I've been going through

brilliant for a while now and it has

thousands of lessons to choose from and

all of them are interactive you learn by

doing this stuff I was focusing on

programming for while but now I've

returned back to more math stuff and

also data like this exploring data

visually it's interesting I've always

been fascinated by data visualization

being able to spot important trends at a

glance and making quick decisions is so

cool but you have to be careful you have

to make sure you understand the

visualizations otherwise you can make

some bad decisions there's a lot of

things you want to look out for

especially something like a manipulated

Axis or scales and I suspect a lot of

people aren't familiar with things like

log scales I think brilliant has a great

interactive way to learn about those I'm

also looking at predicting with

probability it's been a lot of fun I

haven't looked at it yet but I'm looking

forward to this Bas roll stuff one of

the only videos I ever deleted was about

base roll the video was correct but

people didn't believe it so I ended up

deleting the video maybe after I go

through this I'll come up with better

ways to explain it if you want to try

brilliant there's a free 30-day trial

just visit brilliant.org Andy or click

on the link in the description you can

also get 20% off an annual premium plan

how exciting

This was a hard one!


guys let's find the perimeter of this

compound figure if you want to try this

on your own positive right now because

I'm going to solve it in three two one

it's nice they gave us all these right

angles but I'm going to get them out of

the way so to find the perimeter we want

to add up all the sides so the width

across the figure is equal to 14. that

means if I added the lengths of these

three sides they will also be 14. to

find the total height I'm going to have

to add 12 to something and let's call

this unknown distance X so the height of

our figure is going to be X Plus 12. and

let's add that to our green stuff next

we have the given sides the 8 and the 12

we can add them to our list as well so

we're almost done with our perimeter but

we have one more piece this yellow piece

right here temporarily let's call it

question mark we know that this X plus

the question mark is going to be equal

to 8. and after we subtract X from both

sides we get the question mark is equal

to 8 minus X and we can change that in

our diagram and let's bring this last

side down to our sum after we add up all

of these we will have the perimeter of

our figure so the the green 14 plus 14

is equal to 28 and the blue 8 plus 12 is

equal to 20. let's bring this yellow 8

up here and this 12 plus 8 is also equal

to 20. let's move the 20 down here and

20 plus 20 is equal to 40. we can move

the 40 up here and the minus X up here

plus X and negative X will cancel each

other out and 28 plus 40 is equal to 68.

that is the perimeter of our figure now

we can give it a label of units and put

a box around it so what I really like

about this problem is X can be anything

between 0 and 8 not including 0 and 8. x

could be one or two or three or five six

or seven or any other number between 0

and 8 and it changes the shape of the

figure just a little bit but all of

these figures have the exact same

parameter 68 units how exciting

Perimeter Challenge Problem

I thought this was a cool one. I like how there is 1 correct answer for the perimeter, but an infinite number of possible figures that make that perimeter.


hey guys I think this is kind of a cool

one it's asking for the area of this

rectangle that contains these three unit

circles so the area of a rectangle is

base times height so we just need to

find the base and the height so first

let's do the base let's start with this

circle the radius is equal to one so

this radius would also be equal to one

now for this circle this radius is one

so this radius is also one so the base

of this rectangle is going to be equal

to one two three four centimeters that

was the easy one now we got to do the

height so this is a radius of a circle

that's equal to one and this right here

would also be equal to the radius of one

of these circles and those are pretty

easy to spot but this green portion is

not quite so easy but we can still find

the length of this green part let's

complete the right triangle right here

this side of the triangle is equal to

one centimeter because it is the radius

of this circle and this piece right here

is also one centimeter because it's a

radius and this piece right here is

equal to one centimeter because it's a

radius so now we have a nice right

triangle let's pull it out this side is

equal to one and this side is equal to

two and our unknown side length we'll

call X centimeters since it's a right

triangle we can do the Pythagorean

theorem one squared plus x squared is

equal to 2 squared 1 squared is equal to

one two squared is equal to four then we

can subtract 1 from both sides these

cancel each other out and four minus 1

is equal to 3. we can square root both

sides and we get X is equal to square

root of three so the length of this

green segment is square root of three

centimeters so the length of this side

of the rectangle is going to be equal to

1 plus the square root of three plus one

centimeters and then one plus one is

equal to two so now we have our height 2

plus root 3 centimeters the area of a

rectangle is equal to base times height

the base is equal to four and the height

is equal to two plus root three so our

area is going to be equal to four times

two which is eight plus four times root

three and we got to give it a label of

centimeters squared this is the answer

to the question this is the area of the

rectangle let's put a box around it how

exciting

Area of this Rectangle


hey guys so we want to solve the radius

of this circle first let's find the

center of the circle and this right here

is going to be the radius I first

thought was maybe it equals 1 but it

doesn't cuz if we bring this down that's

not in the middle here cuz the edge of

the circle is not in the middle either

so we can't really do that next let's

look at this diagonal here we can

actually find out what this diagonal is

cuz this is a right angle so if we call

this side a this side B and this side C

we know that a s + b s is equal to c^2 a

right here is equal to 2 the B is the

same as this right here so it's equal to

2 and then C we'll just leave a c^ s 2 s

is 4 + 2^ 2 is 4 is equal to c^ 2 4 + 4

is 8 we square root both sides on the

left hand side we can break < TK of8

into < of 4 * < TK of two and theun of 4

is 2 and on the right hand side it just

becomes C so we can clean this stuff up

up here and let's put 2 < tk2 right here

for this piece of the diagonal let's get

rid of all this stuff unfortunately the

diameter here is not 2 < tk2 because it

doesn't go all the way to the corner but

we can say that the rest of the diagonal

right up here is also 2 < tk2 so we can

calculate the diagonal of the big square

and it's going going to be equal to the

2 < tk2 plus the 2 < tk2 which is just 4

< tk2 and we can put a box around that

let's clean this all up a little bit and

this right here is the radius we'll call

that R this is what we're trying to

solve for so we know this we know this

we just got to find this piece right

here that's not too bad because we can

make a square right here we know each of

these are equal to the radius as well so

we can call both the sides of this

Square R then we can do another diagonal

we'll move this R over to here and if we

make the diagonal equal to C then we

know that R 2 + R2 = c^2 well R2 + r 2

is 2 r s we can square root both sides

the left hand side can be broken out

into < TK of 2 * < TK of R 2 and the

right hand side would just bequ < TK of

c squ and then squ of R squ is just R

andun of c s is just C and we can smush

these together and we have squ > of 2 R

is equal to C up here was our diagonal C

so we can change that into < tk2 R clean

things up a little bit and now what do

we have big diagonal again this time

it's equal to this diagonal here the 2 <

tk2 plus the r plus the < tk2 r and we

can put a box around that but we know

that the big diagonal has to equal the

big diagonal so we're just going to set

the 4 < tk2 for this side of the big

diagonal equal to all of this and now we

only have one variable R and one

equation so once we solve for R we know

the answer we can subtract 2 < tk2 from

both sides the left hand side becomes 2

< tk2 the right hand side that's gone so

it's just this right here and we can

scoot these over and let's Factor an R

out of the right hand side so R * what

gives us R that's one so R * 1 is r and

then R * what gives us < tk2 R that's

going to be < tk2 so if you imagine

Distributing this it would bring us back

to where we were up here so we factored

out the r then on the left hand side

leave that as 2 < tk2 the 1+ < tk2 is

just a number so we can divide both

sides by that number and we end up with

r being equal to 2 < tk2 over 1 + < tk2

and we can plug it into a calculator and

get a decimal approximation of 17157 and

that keeps going it's irrational and now

we have the answer to our question the

radius is approximately

17157 and we can put a box around that

how exciting

Old School Math

I hope you guys love this one!


hey guys this looks like a fun one we

have an equilateral triangle right here

we're given that the area of this

smaller circle is 105 meters squared and

they want us to find the area of the

larger Circle if you want to try this on

your own posit because I'm going to

solve it in three two one so first I'm

thinking since it's an equilateral

triangle we know that each of these

angles is equal to 60 degrees and if we

find the center of these circles and

drop the angle bisector to that point

it's going to cut the 60 degree angle

into two thirty degree angles and next

let's connect this radius from here to

here now let's label it R since it's the

radius of the smaller Circle this side

of the triangle is touching the circle

in one point that means this is part of

the tangent line and at this point of

tangency it meets at right angles to the

radius and now if we focus on this

triangle right here the sum of the

interior angles of any triangle is 180

degrees which leaves us 60 degrees for

the last angle so this right here is a

special right triangle it's a 30 60 90

triangle in every 30 60 90 triangle the

hypotenuse is always twice as long as

the smallest side so if we rotate these

no notes we can match this with the

orientation of this triangle up here so

once again the hypotenuse is twice as

long as the smallest side so for this

one this hypotenuse needs to be twice as

big as this side so this is going to be

2R and we don't need these notes anymore

let's take these two things and copy

them down here so let's focus on the

small circle area let's gray out all

this stuff the area of a circle is pi

times the radius squared well in this

case it's going to be pi r squared so

now we can focus on the largest Circle

so let's create out all of this stuff

there you have the large circle is also

going to be pi times the radius squared

or the radius for this one is 2R let's

get rid of the question mark we can

bring over the pi this 2 is going to get

squared to give us 4 and R is going to

be squared to give us r squared so now

we have pi times 4r squared we can swap

it to be 4 times pi r squared and we

already know that pi r squared is equal

to 105 meters squared so I'm going to

substitute in for this pi r squared 105

meters squared and then 4 times 105 is

420. so the area of this larger circle

is 420 meters squared and that's the

answer to our question we can put a box

around it how exciting

Find the Area of the Large Circle

I thought this was a fun geometry challenge!


with the SAT coming up a lot of people

have asked me about this exact type of

question I found this cool video going

over an example the tricky thing here is

we want to get X alone so when we're

done we should have x equals something

the thing people struggle with is how do

we get X alone when there's two x's so

let's watch the video and see how it's

done hey guys this looks like a fun one

it says solve for x it's 2x over 3 Pi +

the quantity 1 - x/ 2 = 0 if you want to

try it on your own pause it right now

cuz I'm going to solve it in three 3 2 1

first thing I don't really like

fractions so let's get rid of these

denominators and we're going to do that

by multiplying everything by the least

common denominator in this case between

3 pi and 2 6 Pi is the least common

denominator so we're going to multiply

everything by 6 Pi for this first term 6

Pi * 2x is 12 pi x and it's still going

to be divided by 3 pi and then for this

part the 6 Pi is going to distribute to

both of these so we're going to have two

separate fractions divided by two for

the first one 6 piun * 1 is equal to 6

piun and then we're going to subtract

and 6 piun * X is 6 piun x and on the

right hand side 0 * 6 Pi is still 0er

now we can simplify things to get rid of

the denominators for this first term the

two pies will divide each other out and

12 / 3 is equal to 4 for this term 6 / 2

is equal to 3 and for this term 6 / 2 is

equal to 3 now we have 4x + 3 piun - 3

piun x = 0 since we're trying to solve

for x we want all the terms that contain

X on one side and all the terms that do

not contain X on the other side so we

want to move this 3 pi over to the other

side and we can do that by subtracting 3

PI from both sides we can copy this down

and simplify 3 Pi - 3 Pi will go to 0

and 0 - 3 Pi is - 3 Pi now on the left

hand side every term contains an X let's

factor out that X when we Factor an X

Out of 4X we're left with four and then

we can bring down the subtraction and

when we Factor an X out of 3 pi x we're

left with 3 pi and we're still equal to3

Pi now to get X by itself we can divide

both sides by 4 - 3 pi and let's copy

this step down and simplify it on the

left hand side the 4 - 3 Pi on top and

bottom will divide each other out we're

left with X is equal to -3 pi over 4 - 3

Pi this is probably fine as an answer

but you can still kind of simplify it on

top we can factor out a negative 1 so

instead of -3 Pi we have -1 * 3 pi and

on bottom we can also factor out a 1 so

1 * what gives us 4 well that's -4 and

then -1 * what gives us -3 Pi well

that's positive 3 Pi now this might look

even more complicated but these two

negative ones will divide each other out

and now we can get rid of these

parentheses and rearrange it like this

so I don't know which one do you prefer

-3 pi over 4 - 3 Pi or 3 pi over 3 Pi -

4 I think this one looks a little bit

neater let's put a box around it the

question asks us to solve for x so this

is the answer to our question how

exciting

Popular SAT Concept "Solve for x."

I've had several people ask me about this exact type of question, I hope this helps a bunch of people!


hey guys this looks like a fun one it

gives us that x * Y is = 36 and that x^2

+ y^2 is equal 78 and it wants to know

what is x 4 + y 4th if you want to try

this on your own pause it right now cuz

I'm going to solve it in 3 2 1 in order

to figure this out we don't have to

completely solve for x and y there's a

pretty cool shortcut that we can apply

let me show you let's copy down the x^2

+ y^2 = 78 and let's Square both sides

of the equation the squares mean that

we're going to have this times this

equals this time this on the right hand

side 78 * 78 is equal to

684 and on the left hand side you can

multiply this out let's use one of these

squares x^2 * x^2 = x 4 x^2 * y^2 is =

x^2 y^2 and then we have another x^2 *

y^2 so that'll be x^2 y^2 and then we

have y^2 * y^2 which is equal to Y 4 and

these insides are what we would get if

we were to multiply it out next let's

focus on this x^2 y^2 x^2 y^2 is the

same thing as xy^ 2 and then we can copy

down everything else in the place of the

XY we're going to plug in 36 and then 2

* 36 * 36 is equal to

2592 we can subtract 2592 from both

sides on the left hand side these will

cancel each other out so we're left with

X 4th + y 4th and on the right hand side

684 - 2592 is

3492 and that's the answer to our

question let's put a box around it how

exciting

Cool Math Shortcut

I thought this was a neat shortcut, I hope you guys like this one!


hey guys this is a fun one we want to

find the area of this red region if you

want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 one for the first step let's call this

red region X I'm going to find X by

tackling everything else around it we

have a square right here and another

Square here we have this triangle here

we have this region right here I don't

even know what you would call that and

then we have the same type of region

right here and then right here we have a

thing called a segment of a circle and

then up here we have another segment of

a circle so if we can find the area of

all of these blue pieces then we can add

that to X and set it equal to the whole

Square so first let's do these two

squares the sides of these squares are

equal to four and the area of a square

is equal to the side length squared so

we'll plug in four for the side length

and four squar is equal to 16 so both of

these have an area of 16 and now that

we've solved these let's make them

orange and let's put them back and next

let's look at this triangle both of

these sides are equal to four and the

area of a triangle is base * height / 2

we can plug in four for the base and

four

Full Video 👆

This was a popular video, I wanted to share it with more people on shorts. Click on the link to see the whole thing.


hey guys I'm pretty excited about this

one you know the three four five

triangles that we see everywhere three

four five triangles show up all over the

place in the SAT and high school

geometry courses even construction

workers use three four five triangles to

make sure that runes are built square

but there's something else special about

three four five triangles and I

genuinely think this is going to blow

some people's minds if you inscribe a

circle inside of any of those three four

five triangles guess what the area of

this circle is it's exactly Pi or Pi

unit squared do you want to see the math

first let's find the center of this

circle and from the center to this point

of tangency let's draw a radius and

let's do the same thing for this point

of tangency so radius will always be at

right angles to the tangent line and

same thing for this one to this tangent

line and this forms a quadrilateral and

since we have three right angles the

fourth one has to also be a right angle

we have two adjacent sides congruent

that means this whole quadrilateral is a

square so this side will also be R and

this side will also be R this entire

side is three and this Porsche question

is are this last piece is going to be 3

minus r and the same thing here this

base has a length of 4 and this portion

of it is R which means the remaining

part is going to be 4 minus r now let's

gray out everything except for the

circle and these two tangent lines if

you have the intersection of two tangent

lines of a circle the distance from here

to here is equal to the distance from

here to here so this will also be 3

minus r and same thing down here these

two tangent lines are intersecting at

this point which means this is going to

have the same distance as this so this

will also be 4 minus r and let's focus

on the hypotenuse we have this portion

here which is three minus r plus this

portion here which is 4 minus r and

that's all going to be equal to the

whole thing which was 5. now we can

solve for R let's subtract 3 from both

sides and let's subtract 4 from both

sides these threes will cancel each

other out these fours will cancel each

other out and we have negative R minus r

which is negative two R and then to

figure out the right hand side we can

combine the negative 3 minus 4 to get

negative 7 and 5 minus 7 is negative 2.

now to solve for R we can divide both

sides by negative 2 and that gives us

our radius equal to one and this is

definitely important let's put a box

around it so now let's clean up this

triangle let's bring it back to the

three four five triangle and we want to

find the area of the inscribed circle of

this 345 Triangle Well the area of a

circle is pi r squared and then we could

plug in 1 for this R and then 1 squared

is just one and then multiplying by 1

doesn't do anything so we know that the

area of this circle is equal to Pi and

this is true for every three four five

triangle how exciting

3-4-5 Triangles and Pi


hey guys this looks like a fun one it

says find the area of the square now

we're told that each of the rectangles

inside of the square have equal areas if

you're going to try this on your own

pause it right now cuz I'm going to

solve it in 3 2 1 since the areas of all

five rectangles are equal to each other

that means that each of these rectangles

has an area that's 1 of the entire

Square so if I make the square something

like X by X the entire square is going

to be x^2 and then each of these

rectangles would be x^2 / 5 so that

feels like a pretty strong first step

for the next step out of all of these

rectangles the blue one is the only one

that has X entirely as one of its sides

so let's focus on that one we already

know that the height is X let's call the

base B we know that the base of B times

the height of X is going to be equal to

the area x^2 5 and I want to know what

is B in terms of X so let's divide both

sides of the equation by X on the left

hand side these will cancel so we just

have B and on the right hand side you

can think of x^2 as x * X and then one

of these X's will cancel from top and

bottom so that means B is equal to X

over 5 or 1 x so let's change this B to

1/5 X and the same thing for the top it

will also be/ X and now let's bring back

the rest of our Square each side of this

square is equal to X so if this blue

portion is 1/5 X that means that the

orange portion is going to be 4 fths x

and now we have all the parts of this

orange rectangle so let's focus on that

and then once again base time height

will equal the area so we know that the

base of 4S x * the height of 3 will

equal the area of x^2 5 so on the left

hand side let's write this as a single

fraction on top we can do 4 * x * 3

which gives us 12x and on bottom we just

have the 5 and this is still equal to

x^2 over 5 next we can multiply both

sides by 5 overx and we are allowed to

divide by X here because we know X is

not equal to zero on the left hand side

these two fives cancel each other out

and these two x's cancel each other out

all we have left is 12 and on the right

hand side these two fives cancel each

other each other out and x^2 overx is

just X so now we know that 12 is equal

to X going back to our original Square X

was the side lengths of the square so

this square is a 12x 12 square so to

find the area of this square is just

going to be 12 squar and 12 squ is equal

to 144 and let's give it a label of

square units and this is the answer to

our question the given square has an

area of 144 square units let's put a box

around it how exciting if you guys want

to try more of these geometry challenge

problems I have this page on mat.com

there's a link in the description

Multicolor Square Problem

https://andymath.com/geometry-challenges/

For more geometry challenges, check out the above link! Let me know what you think!


that wants to know what is the value of

x and what is the value of each of these

other three angles we can do this the

sum of the interior angles of any

four-sided figure is always 360 degrees

so we can just say this angle plus this

angle plus this angle plus this angle is

equal to 360. and then X Plus 3x plus 2X

is equal to 6X and then all these

numbers are going to combine to give us

negative 2154 and it's still equal to

360. if we add 2154 to both sides we get

6X on the left hand side and 25 14 on

the right hand side divide both sides by

6 and we get x equals 419 and we can

plug the X in up here so now that we

know what x is we can plug that in for

each of these other angles copy down the

values for each of the angles plug in

419 for the X's 3 times 419 is 1257 and

2 times 419 is 838 419 minus 352 is 67.

1257 minus 1189 is 68. and 838 minus 768

is 70. how exciting

Nice Quadrilateral #shorts

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#shorts


now that we've made the bases match we

can drop the base Drop the

Base and we can set the exponents equal

to each other and solve for x since the

Bas is matched the only way these two

sides can be equal to each other is if

the exponents are equal to each other so

in other words we can just drop the

base and that gives us 2x is equal to

138 the only way these two sides of the

equation are going to be equal is if the

exponents are equal so we can just drop

the base Drop the

Base

now since both these have the same base

of two the only way they're going to be

equal is if these exponents are equal or

in other words we can drop the

[Music]

base so we end up with 4x + 1 = 9x - 3

Drop the Base Compilation

I thought it was fun to see how these changed over time.


hey guys this looks like a fun one we're

trying to find X the area of this red

region if you want to try it on your own

pause it right now cuz I'm going to

solve it in 3 2 1 Let's copy it down and

let's focus on this left semicircle

let's cut it in half and split this into

five and five and then let's focus on

this top semicircle and cut it in half

and make this five and five so now we've

split our white region up into these

three pieces this one right here is a

square of 5 * 5 which means the area is

25 and this blue portion right here was

half of the semicircle or in other words

a quarter Circle the area of the quarter

circle is PK r^ 2 / 4 and the radius of

this quarter circle is equal to 5 so we

can plug in five for the r and 5^ s is

equal to 25 so the area of the blue

region is 25 pi over4 and this quarter

Circle right here is the exact same so

it's also 25 pi over4 and now we know

the area of our white region we can just

add these three things together and 25i

4 + 25i 4 4 would be 50 piun 4 we can

simplify this to 25 piun / 2 this is the

area of our white region and if we add X

to this that'll give us the area of the

entire Square which is 10x 10 or 100 now

we just got to solve for x we can

subtract 25 from both sides these two

will cancel each other out and 100 minus

25 is 75 then we can subtract 25 pi over

2 from both sides on the left hand side

these will cancel out and on the right

hand side we'll have 75 - 25 pi over 2

and that's the answer to our question

let's give it a label of square units

and put a box around it and the

approximate value of this is 35.7 3

square units how exciting

Find the Red Area Area


hey guys somebody sent me this request

and I kind of like it it looks like a

fun one and I was a little bit curious

how I would visualize this so I'm going

to pull them out of here so we have the

red bar and the green bar and then we

also have this purple wall these are the

Shadows that they're making and we're

given all of these measurements the

green box is 2 m high with the shadow of

three along the ground this Shadow

measures 6 M and up the wall it measures

4 M now we're trying to solve for the

height of the red bar so let's call that

X we don't need most of this wall let's

smush it down and we can imagine if the

wall was isn't there the shadow would go

about this far but we don't know how far

this is so let's give it a variable y

let's focus on this part and this part

right here we can set up a proportion we

can say that 2 over 3 is equal to 4 over

y because the Shadows are proportional

to the heights of the objects and now to

solve for y we can cross multiply 2 * y

will give us 2 Y and that's going to be

equal to 3 * 4 which is 12 and then we

can divide both sides by two and we get

Y is equal to 6 so we can update this y

to be 6 M we know from here to here the

shadow is 6 M and then from here to here

it's 6 M the whole thing is going to be

12 M so now we no longer need this thing

and we can set up another proportion the

2 over 3 will equal x over 12 and now to

solve for x we're going to cross

multiply 3 * X is 3x and that's going to

be equal to 2 * 12 which is 24 and then

we can divide both sides by 3 and we get

x = 24 over 3 which is 8 x was the

height of our red bar so we can change

this to 8 m and now we're done that's

the answer to our question let's put a

box around it how exciting

Shadow Math

I am enjoying the longer shorts. I hope you guys love this one!


okay so this says solve for x if you

want to try this on your own positive

because I'm going to solve it in three

two one for the first step let's copy it

down and then next let's multiply both

sides by 10. on left hand side this 10

and this 10 will cancel each other out

and on the right hand side this will

just be 10 x to the 21st next let's

subtract 10x the 21st from both sides

let's bring this down and 1X to the 21

minus 10x to the 21 is equal to negative

9 x to the 21 and the right hand side is

now equal to zero from here let's factor

out the greatest common factor it's

going to be x to the 21. now x to the 21

times what gives us x to the 22 that's X

that will bring down the subtract then x

to the 21 times what gives us 9x to the

21. well that's going to be 9. here we

can set each of these equal to zero x to

the 21 equals 0 or x minus 9 equals zero

for this one if we take the 21st root of

both sides we get x equals zero and for

this one if we add 9 to both sides we

get x equals nine and these are the

solutions to the equation how exciting

New Math Challenge Problem #shorts

https://andymath.com/

#shorts


hey guys welcome to this product rule

video use the product rule to find the

derivative of f of X equals x cubed plus

two times x squared minus one so the

product rule is written right here so f

prime of X is gonna be equal to we bring

down the first function X cubed plus two

and we take the derivative of the second

one derivative of this is 2x plus

because of this plus right here bring

down the second x squared minus one

times the derivative of the first which

is 3x squared so that's the first step

some teachers will accept this as an

answer but we can simplify it distribute

the 2x so we get 2x to the fourth plus

4x plus and then distribute this 3x

squared which will give us 3x to the

fourth minus 3x squared and now we can

just combine like terms 2x to the fourth

plus 3x to the fourth is 5x to the

fourth minus 3x squared plus 4x and

that's the answer to the question so if

this makes sense if not I've linked down

below two more product rule practice on

Andy math calm feel free to check it out

please like this video subscribe to my

channel I'll talk to you guys soon bye

[Music]

Product Rule- Derivatives

For more visit http://andymath.com.

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hey guys this looks like a fun one these

A's and B's are placeholders so this

doesn't mean a * a it means some

matching double digit number like 22 or

33 now we're trying to solve for A and B

so that this plus this equals this we

probably could guess and check but I

want to see if we can figure it out

logically since the a plus b gives us a

three in the one spot we know that

either a plus b = 3 or a plus b = 13

let's explore these possibilities if a

plus b equals 3 that means this a plus

this B would be 3 and then in the 10

place the a plus b would also give us a

three in the 10's place in this case

nothing will carry over so in the

hundred's place we have a zero but we

know from up here the hundred's place

needs to have a one so it is not

possible for a plus b to equal 3 so we

know for sure that a plus b is equal to

13 so that means when we do a plus b

we're going to be carrying a one for the

10's place and now let's focus on the

10's column this one represents 10 this

a represents 10 a this B represents 10 B

and then we're going to add all those

together and set it equal to this 1

which represents 100 plus this a which

represents 10 a and now we want to solve

this system of equations let's subtract

10 from both sides and let's subtract 10

a from both sides on the left hand side

this stuff all cancels out so we just

have 10 B and on the right hand side

this stuff all cancels out so we just

have 90 after we divide both sides by 10

we get B is equal to 9 and that's

important so let's put a box around it

and then let's copy down this a plus b =

13 and let's substitute 9 for this B

after we subtract 9 from both sides we

get a is equal to 4 and let's put a box

around that so now we have an answer a

equal 4 and b equal 9 let's check our

work let's change this AA into 44 and

let's change this a down here into a

four and then the BB we can change into

99 and we have 4 + 9 is equal to 3 carry

a 1 1 + 4 + 9 is 14 so this checks out

so a does equal 4 and B does equal 9 how

exciting

Math Logic Warmup

I hope you guys like this one!


I got something that I bet your math

teacher doesn't even know hey guys this

looks like a fun one did you know that

there are two quadratic formulas anytime

you have ax^2 + BX + cal0 there are two

different quadratic formulas to find

what x could equal you might be familiar

with negative B plus orus < TK of b^2 -

4 a c whole thing over 2 a but there's

also 2 C over the quantity b - + < TK of

b s - 4 a c this quadratic formula will

give the exact same answers as this

quadratic formula would you like see an

example let's try out 3x^2 + 7 x + 2 = 0

the ACs are a little tight let's give us

a little more room and let's do this the

leading coefficient is three which means

a is equal to three so this a will

become three this a will become three

and this a will become three the 7x

means that b is equal to 7 so we'll

change this B to 7 this B to 7 this B to

7 and this bet to 7 and last the

constant is equal to two so we can make

this C2 this C2 and this C2 and now

let's simplify these let's bring down

the -7 and we have plus or minus the

square root 7^2 is equal to 49 and then

we're going to subtract 4 * 3 * 2 which

is equal to 24 and on bottom 2 * 3 is

equal to 6 and then over here 2 * 2 is

equal to 4 we could bring down the -7

and we have minus plus the Square t 7 2

is equal to 49 and then we're going to

subtract 4 * 3 * 2 which is equal to 24

and now we can simplify these on bottom

here here we have the six and on top

here we'll have the four on top we can

copy down the -7 plus or minus Square Ro

T and on bottom we can copy down the -7

minus or plus Square t inside of the

square root 49 - 24 is 25 and same thing

49 - 24 is 25 on the left hand side squ

< TK of 25 is 5 and on the right hand

side Square < TK of 25 is 5 and now

let's smush everything together and

we're ready to finish up we have x

equals and the plus or minus right here

means we're going to have two values of

X both values of X will have a -7 over 6

but the first one is going to be a + 5

and the second one's going to be a

minus5 and on the right hand side the

minus plus means there's going to be two

answers the both have a four and a -7

but the first one is going to be ne5 and

the second one will be + 5 so let's

simplify these top two -7 + 5 is equal

to -2 and let's copy over the six and on

the right hand side we have a 4 and -7 -

5 is -12 both both of these can be

simplified and they both simplify to-

1/3 they both gave the exact same answer

let's put a box around it next let's do

the second one -7 - 5 is -2 and that'll

be over 6 and on this side we can bring

over the four and -7 + 5 is -2 both of

these can be simplified and they both

simplify to -2 once again we got the

exact same answer let's put a box around

it so it actually worked out this

quadratic formula gave the same answer

as this quadratic formula one

interesting observation is this was

minus plus while this was plus minus the

minus version of this gave us the same

answer as the plus version of this and

that will always be true here's another

example where there's only one solution

1/3 both equations gave the same thing

and here's an example with irrational

solutions they both gave the same

irrational Solutions and here's another

example with imaginary Solutions how

exciting so I'm thinking there are two

possible follow-up videos the first has

to show an example of why this is useful

and there is a cool example why it's

useful for another follow-up video we

could actually prove that this always

works I showed a couple of examples here

but we can actually prove this if you

guys want to see either one of those

follow-ups let me know in the comments

how exciting

Different Version of Quadratic Formula

I thought this was a fun one. I hope you guys love it!


he has welcome to this problem so this

one may be able to do in your head but

if you want to show some work you could

say okay X is a variable so what is

three two one well that would give us

three so we know X does not equal one

what's 3 to the 2 well that's nine so we

know X doesn't equal to what's three to

the 30 through 23 mines means 3 times 3

times 3 and that equals 27 oh that's

true so the answer is x equals 3 so

really just trying to figure out what is

the variable in here that would give you

27 and this is what's going on so I hope

this makes sense and i'll talk to you

guys soon hi

Solving for Exponents

Exponents Problem #3 UNDER A MINUTE! For more math made easy visit andymath.com.

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hey guys this looks like a fun one we

have three of these nine to the X Powers

so I'm going to write that as 3 times 9

to the X power next I want to get rid of

this three so I'm going to divide both

sides by 3. these two threes cancel

leaving us with just nine to the X power

on the right hand side we have a 139

threes being multiplied by each other on

top and we have a single three on bottom

one of these 139 threes up here is going

to cancel with this 3 down here so we

only have 138 threes being multiplied

which is 3 to the 138 power so this 9 to

the X power we can rewrite the 9 as 3

squared and that'll still be to the X

power this is going to be equal to 3 to

the 2x power you can just multiply the

two powers and then we'll just bring

down the right hand side so we have 3 to

the 2x is equal to 3 to the 138 since

the bases match the only way these two

sides could be equal to each other is if

the exponents are equal to each other

and that gives us 2x is equal to 138

after we divide both sides by 2 we get X

is equal to 69 nice and let's put a box

around that how exciting

Exponent Challenge


hey guys this looks like a fun one we're

trying to find the ratio of the pink

area to the green area let's do this so

first let's try to find the area of

these yin-yang things so let's pull them

out and if we put them together they

should make a circle so it just be half

of that Circle and they don't these make

an ellipse all right scrap that idea

let's try something else so let's draw

the radius right here in fact let's draw

all these radiuses so that we have a

square in the middle right here and

let's focus on that square each of these

sides are two radiuses so we can call it

2 R and yes radiuses is the plural of

radius we could also say radi so the

area of the square is going to be 2 R *

2 R that's the same thing as 4 R 2 let's

look at another copy of the square down

here let's pull apart these quarter

circles bring down this green part let's

rotate all these quarter circles to make

a circle the area of this circle will be

P pi r s the area of the green stuff is

then going to be the area of this Square

minus the area of this circle now we

don't need this circle anymore so we can

get rid of that let's bring this up here

and this is important so let's put a box

around it let's go back up to this

square up here let's cut the square in

half and let's find the length of this

diagonal well this ends up making a 4545

0 triangle so it ends up being < TK of 2

* one of these sides so the diagonal is

equal to 2 R < TK of2 if you don't know

about the 454590 shortcut here's the

Pythagorean theorem right here pause it

if you want to check it let's clean all

this stuff up and put a box around this

now let's find the area of our large

Circle so we know that the diagonal the

square is equal to 2 < tk2 R and now we

don't need the square anymore and then

this right here is a radius and this

right here is a radius the diameter will

be equal to two of the radiuses plus 2 >

2 R and if we want to know what the

radius of the largest circle is we would

just divide all that stuff by 2 which

gives us r + < tk2 r and now to find the

area of this circle we're going to do PI

radius squ where this right here is the

radius and then a binomial squared ends

up being the binomial times itself so we

can foil this out we'll leave the pi out

front and then R * R is R 2 R * < tk2 R

is < tk2 R 2 < tk2 R * R is < tk2 R 2

and < tk2 R * < tk2 R is = to 2 R 2 and

then we can clean this up R 2 + 2 R 2 is

3 R 2 and right here we have 2 < tk2 R 2

so that would be 2 < tk2 R 2 and then we

can distribute the pi so we now have the

area of this large Circle and we can put

a box around that we're trying to find

the pink area divided by the green area

I think that is what this says so the

pink area is going to be this whole

circle minus this green stuff and then

the green area is going to be this green

stuff we can get rid of these

parentheses and distribute the negative

all six of these terms has an R squ so

we can factor out the r squ and then we

can cancel those R SS and then on top

this 3 pi and this 1 Pi can become 4 pi

and now we are done this is the exact

value of the answer and end up being

approximately

20308 how exciting

Awesome Pink Challenge

This is an older one, I hope you guys love it!


hey guys I want to show you something

really cool about the Pythagorean

theorem you're probably familiar with A2

+ B2 = c^2 and here's a cool

visualization that shows it if you add

the areas of these two red squares it's

equal to the area of this blue square

but these don't have to be squares it

works for isos right triangles they

could be equilateral triangles they

could be regular pentagons they could

even be semicircles for each of these

the red areas add up to the blue area

and this can work for any similar figure

we could even look at something like the

state of orgon if we align the state of

organ along this side this side and this

side it'll be true that the sum of these

two red areas will equal the blue area

or other words a plus b is equal to C

how exciting

Pythagorean Variations

I thought this was fun!


hey guys welcome to this arithmetic

sequence problem these are the notes

that we want to use since it's an

arithmetic sequence we're going to be

interested in this one right here so I

want to find the missing value a n

that's what we're trying to solve for a

eight is equal to a sub 1 is 3 plus D is

5 and then n is 8 so it's 8 minus 1 this

is 3 plus 5 times 7 5 times 7 is 35 plus

3 so a 8 is equal to 38 and that is the

answer the question so we just found the

equation we wanted to use plugged into

values and solved

I hope that makes sense please like and

subscribe check out Andy math comm for

more videos and study guides I'll talk

to you guys soon bye

you

Arithmetic Sequences

For more visit http://andymath.com

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hey guys so this is a fun geometry

problem right here we have four circles

and the area of all four of these

circles is 144 pi and they tell us that

the radius of these smaller circles is x

the radius of this medium circle is 3x

and the radius of this larger circle is

5X and I want us to solve for x first

the area of a single circle is pi r

squared so if we're going to add four

circles we're going to be adding up four

pi r squareds we're going to set this

sum equal to 144 Pi because that was

given as the area of the four circles

for this first Circle that'll be the top

one we'll plug in X for R and for the

second Circle let's do this bottom one

so we'll plug in X for the radius for

the third circle let's plug in 3x for

the radius now we're going to plug in 5x

for the last one since everything is

multiplied by pi independently and we

have an equation we can divide both

sides by pi and I'll just get rid of all

of our Pi's so then let's bring down the

144 this would just be x squared this is

just x squared but then 3x squared the 3

is going to get a squared and the x is

going to get a squared so that's going

to be 9x squared and this case is the 5

good squared as well as the X so we get

25 x squared we can bring down the 144

again now we're going to combine like

terms all of these are X squareds so

we're ultimately just going to add the

coefficients 1 plus 1 plus 9 plus 25

gives us 36 x squared so we can divide

both sides by 36 144 divided by 36 is 4

and this right hand side the 36 is

cancel to give us x squared to solve for

x we square root both sides square root

of 4 is 2 and the square root of x

squared is X now we don't have to worry

about plus or minus 2 because you can't

have a negative value in a geometry

problem like this so this is the answer

to the question x equals two

Geometry Challenge Problem

https://andymath.com/area-and-circumference-circle-2/

Collab with Geeklyhub


hey guys so let's simplify this

expression it's kind of fun it's got a

bunch of three five sevens in it let's

see what happens when we simplify it so

let's start with inside the parentheses

357 over 357 is equal to one and then

one plus three plus five plus seven is

equal to 16. 3 times 5 times 7 is 105.

take the square root of 16 to the 105.

anytime we take a square root with an

exponent you can just divide that

exponent by two but 105 is an odd number

I want to try to keep things clean so

there is a way for us to make this even

we can rewrite the 16 inside the

parentheses is 4 squared and then this 2

is going to multiply by the 105 to give

us 4 to the 210 power now this 210 is

even so when I take the square root of

it it's just going to divide this by 2

which gives me 105. I want to take the

7th root I'm just going to divide this

exponent by 7 which gives me 4 to the

power of 15. then when I take the fifth

root I'm going to divide this 15 by 5 to

give me 4 cubed and then when I take the

cube root that's going to cancel with

the cube to give me just 4. and 7 times

4 is 28 28 times 5 is 140 3 times 140 is

420. nice so this simplifies to 420. how

exciting

Expression with lots of roots #shorts

https://andymath.com/

#shorts


hey guys welcome to this problem we're

going to find the probability of red or

a card given this table here first thing

we can do is fill in the totals just a

little bit of addition so now probably a

red or a car there's two ways to do this

we can circle all the red stuff and

all the cars

and

just add these values up

35 + 70 + 25 + 45 these are all the

things that satisfy it and then we

divide our you might 315 because that's

the total number

so add these up

we get a 175 over 315 can reduce it so

our answers 5 nights there's one other

way we can do this and that is

probability of red

plus the probability of car

minus the probability of red and car

so we'll do that real quick to

probability of red is this total

105 over 315

plus

probability of car is this total 105

over 3 15

minus this overlap of red and car which

is 35 / 3 15 after we do the math here

we end up with 175 / 3 15 which is the

same thing we had above so our final

answer is 59 so you can do it either way

identify everything that's true for or

apply this formula hope it makes sense

to talk to soon bye

you

How to do Probability Problems

Probability Problem #2 MAKES SENSE NOW! For more math made easy visit andymath.com.

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okay so we're going to evaluate this

definite integral this is a polynomial

so we're just going to use the power

rule so we have the constant 2 we bring

down the X and this 3 right here we add

1 to it and we also divide by that

number plus and then we're going to do

the same thing for this one we have the

constant three we're going to add 1 to

the 2 and then we're going to divide by

that 3 plus 1 is 4 so both of these will

change into four and two plus one is

three so both of these will change into

three two over four can reduce into one

over two and three over three cancel

each other out leaving us with just X

cubed and then we're going to evaluate

all of this from one to three and one

and the three come from here what this

means to evaluate is we're going to plug

in 3 for the X's giving us three to the

fourth over 2 plus 3 cubed and we're

going to subtract whatever we get when

we plug in one there'll be one to the

fourth over 2 plus 1 cubed three to the

fourth is equal to 81 and 3 cubed is

equal to 27. on this side one to the

fourth is just one and one cubed is just

one and then one plus a half is equal to

1.5 81 over 2 is equal to 40.5 now we

can get rid of these parentheses 40.5

plus 27 is 6 7.5 minus 1.5 which gives

us 66. and that's the answer to our

question this integral is equal to 66.

how exciting this problem right here is

number seven on this page I have on

andymath.com if you guys want to try

more of these the link is in the

description

Quick Definite Integral


this is another Katrina a puzzle and

it's my goal to solve all of these hey

guys this looks like a fun one we got a

polygon with multiple angles given and

it wants us to find the missing angle

and we're also given that this length is

equal to this length and it's this part

that makes it a little bit trickier if

you want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 1 so first let's just go around and

Mark all the angles that we can we have

the 70° this angle right here is also

70° these are called vertical angles and

vertical angles are congruent next we

can focus right here and these are

called a linear pair these two angles

will always add up to 180° so this one's

going to be

110° and then once again because of

vertical angles this one is also 110°

next let's focus on this triangle here

the sum of the interior angles of any

triangle is 180° so 110 + 50 is equal to

160 that leaves 20° for this angle and

then we can move over to this triangle

we have 70° in 80° is equal to 150° that

leaves 30° for this angle and then next

let's go to this linear pair if this

angle is 80° that leaves 100° for this

angle and then we can look at this

quadrilateral in any quadrilateral the

sum of the interior angles is equal to

360° 110 + 100 + 60 is equal to 270 that

leaves us 90° for this last angle and

now we've pretty much found every single

angle except for the question mark and

this angle up here let's let's call this

other angle X unfortunately we can't

focus on this triangle because we have

two unknown angles in this triangle but

we still haven't taken advantage of the

fact that these two things are equal so

let's focus on the triangle that

contains these two equal lengths every

single triangle can be inscribed inside

of a circle it doesn't matter if it's

right obtuse or acute they can all fit

inside a circle and another fun fact for

any inscribed right angle triangle the

hypotenuse will be a diameter of the

circle and this is tied to the fact that

an inscribed angle is 1/ of the arc that

it subtends so for example an inscribed

angle of 30° will have a 60° Arc and an

inscribed angle of 45° would have a 90°

Arc and an inscribed angle of 90° would

have an arc of 180° and that'd be half

the circle and that's what makes this

the diameter and then since these two

portions are congruent that means this

is the center of the circle so each of

these are radi of the circle and now

let's bring all this stuff back and it

looks like this is on the circle but we

don't know that for sure let's see if we

can prove it we can focus on this

triangle right here and then this 20 and

30° is going to make this angle 50°

since both of these angles are equal to

50° that means this is an isoceles

triangle and in every Isles triangle the

sides opposite the congruent angles are

also congruent so this side is equal to

this side and since we know this side is

a radius that means this side would also

be a radius so that means this point up

here does lie on the circle so now we

have an inscribed quadrilateral we do

know one specific property about any

inscribed quadrilateral the opposite

angles add up to 180° so in these notes

a plus c equal 180 and B + D equal 180

and for our quadrilateral 50 + 90 +

question mark would equal 180 and 60 +

50 + x = 180 but we don't need this

second one because this equation will be

enough to solve for a question mark So

50 + 90 is equal to 140 and then to get

the question mark alone we can subtract

140 from both sides un left hand side

these will cancel each other out leaving

us with question mark and on the right

hand side 180 minus 140 is equal to 40

so our question mark is equal to 40

let's give it a label of degrees and

this is the answer to our question let's

put a box around it how exciting

Harder than it looks!

Another Catriona Agg Puzzle. This one is harder than it looks.


hey guys this looks like a fun one it's

got a rectangle with three Circles of

diameter 3 4 and six smooshed in here

and we're supposed to find the distance

from here to here if you want to try it

on your own pause it right now CU I'm

going to solve it in 3 2 1 first let us

label the center of each of the circles

let's take this diameter and extend it

up here and let's split this into a

purple distance and a green distance

let's call the purple distance X and the

green distance y so ultimately we're

trying to find x + y I want to work on y

first let's look at these two larger

circles this one has a diameter of four

which means the radius is two and this

one has a diameter of six which means

the radius is three let's bring down the

green distance and find this

perpendicular segment since this is two

right here this distance would also be

two and this whole radius of three would

be broken up into one and two next let's

connect the centers of the circles and

that'll give us the hypotenuse of our

right triangle this piece is equal to

the radius of two and this piece is

equal to the radius of three and the

entire hypotenuse would have a length of

2 + 3 or 5 now let's mess around with

our new right triangle I'm thinking

Pythagorean theorem 1^ 2 + y^2 is = 5^2

and this will clean up it gives us 1 +

y^2 is = to 25 we can subtract one from

both sides and we get y^2 is equal to 24

now to get the Y by itself we can square

root both sides of the equation since Y

is a positive distance the square otk of

y^2 is just going to be Y and then we

can simplify the square root of 24 24 is

= to 4 * 6 and the < TK of 4 * 6 is

equal to the < TK 4 * the < TK 6 and

then the < TK of 4 is equal to 2 so Y is

equal to 2 * the < TK of 6 so we can

update this green distance to be 2un 6

and we can move it back up here now

let's work on the X this circle has a

diameter of three which means the radius

is 1.5 let's connect these two centers

which will be made up of this radius of

1.5 and this radius of two so the whole

thing will have a length of 3.5 then

let's drop this perpendicular and let's

find its distance since this largest

Circle had a diameter of six that means

the height of this rectangle is also

equal to six and then we can bring this

radius of two over here so we need 2

plus the purple distance plus 1.5 to

equal 6 that means our purple distance

is going to be equal to 2.5 and now

let's play with this triangle

Pythagorean theorem tells us 2.52 + x^2

is = to 3.52 2.52 is = 6.25 + x^2 and

then 3.52 is = 12.25 we can subtract

6.25 from both sides and we get x^2 is

equal to 6 and after we square root both

sides we get X is equal 2 the < TK of 6

that means that our purple distance will

be equal to the < TK of 6 and we can

bring that back up here so now we have

the question mark is going to be equal

to the purple < TK 6 plus the green 2un

6 and < TK 6 + 2un 6 is equal to 3un 6

and that is the answer to our question

let's put a box around it how exciting I

just got a haircut what do you guys

think

Cool Math Challenge

I hope you guys like it!


hey guys let's talk about some log

derivatives so it says find dydx when Y

is equal to Ln of root X I think this is

a beautiful one first I just want to

rewrite the root X as x to the one-half

power so now we're ready to take the

derivative so we have a function inside

of a function so we're going to do chain

rule so first we're going to work with

the natural log the derivative of

natural log of anything is 1 over that

something and then because of chain rule

we're going to multiply by the

derivative of what's inside here so the

derivative of x to the one half we're

going to bring the one-half in front and

it's going to be x to the and then we do

one half minus one which gives us

negative one half this piece right here

the one-half exponent means it's going

to be the square root of x and then this

x the negative one half means we're

going to have square root in the

denominator so it's going to be one over

two root X because the 2 right here so

this 2 is left on bottom and root x

times root X is equal to X and that is

the answer to the question d y d x is 1

over 2x so this really isn't super

difficult but if you don't know what

you're doing it definitely can be tricky

Log Derivative Example


I think I already have a good idea what

this is going to be we're dealing with

some factorials usually you want to

reduce each of them in terms of the

smallest factorial so I'm going to

rewrite this 70 factorial as 70 times 69

times 68 factorial and then we'll bring

down the 68 factorial and the 69

factorial we can rewrite as 69 times 68

factorial so now we have a bunch of 68

factorials let's bring down the

numerator but on bottom I want to factor

out the 68 factorial 68 factorial times

what gives us 68 factorial that's 1. 68

factorial times what gives us 69 68

factorial that's 69. 1 plus 69 is equal

to 70. the 70s cancel and the 68

factorials cancel and that leaves us

with 69. let's use this to put it in a

box how exciting

Factorial Math Tricks #shorts

https://andymath.com/factorials/

#shorts


hey guys here's another geometry

challenge we have a square that's a two

by two square I have four circles stuck

inside of that square and there's

another Square turn 45 degrees and fit

inside those circles so that it touches

at those four points and we want to know

what is the area of this Square I think

there's two ways to solve this I kind of

want to show you both ways we'll call

one side of the square a length of s s

for side or Square then we'll draw a

little square here and we'll notice that

this square right here cuts into exactly

one-fourth of that square which means I

could probably flip it and get the same

piece right up there just like that so

this piece should equal this piece right

here if you think about how those four

squares would be cut off both of these

are exactly half of the length of s so

the two of them together should just be

S let's look at this diagonal of the

square I drew in the corner it's going

to be comprised of s plus it's a

diameter of one of these circles how

long is this diameter of this circle

well if we cut this side in half each of

these halves is equal to one and that

half is the same height as one of the

diameters so we can call this diameter

one so the diagonal of the square square

is equal to S plus one but we can pull

out of this a triangle and this green

triangle here is the same thing as this

green triangle here both of these sides

are equal to one anytime you have a

right triangle where the two legs are

equal to one the hypotenuse is always

equal to square root of two and if you

want to verify it here it is using

Pythagorean theorem so now we can plug

in square root of 2 for the diagonal and

then we can smush this together and

clean all this stuff up scoot this over

and now we can solve for S let's

subtract one from both sides and we get

S equals square root of 2 minus one s is

the side of the square to get the area

of the square that we would Square s so

we're just going to square both sides of

this equation on the left hand side

that's the same thing is multiplying

this binomial by itself and we have to

distribute so we're going to do square

root of 2 Times Square Root 2 gives us

2. root 2 times negative 1 gives us

negative root 2 and negative 1 times

root 2 gives us negative root two and

then negative 1 times negative 1 is

positive one and this is all still equal

to S squared 2 plus 1 is equal to 3 and

negative root 2 minus root 2 is minus

two root 2. so this right here is the

area of the Red Square it's 3 minus two

root 2. we can put a box around that and

let's put a box around S as well that's

seem pretty important get rid of all

this other stuff bring this down and

let's look at the other more formal way

to solve it you can find the centers of

these three circles connect the centers

we can see this piece right here is the

same thing as the side length of s

because it's the same thing as one of

the side lengths of that square then we

look up here both of these are the

radius of this top Circle so we'll call

both of those R and same thing here

these are both the radius of this circle

so we'll call those R and same thing

here both these are the Rays of that

Circle so we'll call those R now if we

extend this radius up here we'll notice

that R plus r is the same thing as one

so we just say two R equals one now

let's use Pythagorean theorem for this

right triangle if a squared plus b

squared equals c squared where a is this

bottom side here which is R plus r well

we know that R plus r is equal to one B

is this R plus r but we know that's

equal to one well C is this R plus S

Plus R and then R plus r is 2R and we

know that 2R is equal to one so we can

make that one one squared plus 1 squared

is just two then from here we square

root both sides the root and the square

cancel each other out to give us 1 plus

s when we subtract one from both sides

this side becomes root two minus one and

this side just becomes s so now we solve

for S and check this out it's the same s

that we got before so since we got the

same as both methods we know they're

both going to give us the same area so

the answer the question is the area is

equal to three minus two root 2 or in

other words

0.17157 I'm going to put units squared

so the exact value of the area of the

red square is three minus two root 2

units squared and the approximate area

of this red square is 0.17157 units

squared how exciting

2 different solutions geometry challenge


so i've seen a lot about pushing p can

you solve for p yeah we can do this so

oliver is pushing p p is equal to the

sum from one to three of one over n

minus two so we take n equals one we add

to it whatever n equals two and then we

add to whatever n equals 3 and we stop

because there's a 3 on top of this so p

is going to be equal to the sum of these

three we're going to do 1 over n minus 2

we're going to plug in 1 for the n plus

1 minus 2 plug in 2 for the n so it's 2

minus 2 and plugging 3 for the n it's 3

minus 2. bring down the p and one over

one minus two gives us one over negative

one one over two minus two gives us one

over zero and one over three minus two

gives us one over one one over negative

one is negative one so negative one plus

one is zero so these cancel each other

out so we're left with p is equal to one

over zero and that is undefined and

that's the reason why nobody can explain

to me what pushing p means that's it i'm

done thank you

Pushing P

For more check out andymath.com.


this person requested Pythagorean

theorem so we start with the square then

we have another Square that's the exact

same size so these two squares are equal

to each other then I'll make a cut that

is the same size as a and I'll do the

same thing to the other Square do the

same thing to every side of that square

and the same thing to every side of this

square and then this other piece will be

B then if I make a cut right here on

both squares and pull this triangle out

this is the same triangle that's in our

notes up above so this side right here

is C this side is also C and let's bring

these down here next let's make another

cut here where this is a and this is B

and let's make a cut here where this is

C and then if we pull both of these

triangles out and bring them down there

these two sides should still be equal to

each other because we pulled the two

same triangles out of both of them now

let's make some more Cuts here here and

here and also here and here we've now

made more of these ABC triangles and we

can pull them out and bring them down

here and so out of both sides we've

pulled four of these ABC triangles so

they should both still be equal so this

square right here is an a by a square so

the area of it is a squ this square has

sides B so its area is b^ squ and this

square has side C so its area is c^ 2 we

can call this left hand side a^ 2 + b^ 2

cuz we're adding these two squares

together and that's going to be equal to

this right hand side which is c^2 how

exciting

Pythagorean Theorem Demo

This is an old one, I hope you guys like it!


he has welcome to this problem we're

going to factor this trinomial so

anytime there's a leading coefficient

that's greater than 1 I like to do it

this way so i'll make an X and I'll make

a box so 3 times 6 is 18 so when two

numbers that multiply to 18 and the 11

goes down here so we do is two numbers

multiply to 18 and add to 11 that's nine

and two so this 3x squared is going to

be written here the 6 is written down

here and then we're going to split this

11x up in 29 x + 2 X we're going to put

them here in here so now you just find

out what the outside would be to give

you this inside so put a 3x here and an

X here so 3x times x is 3x squared the

reason I chose put the 3x here is

because I want to line up with this 9x

so 3x times what is 9x that's positive 3

then x times what is 2x it's positive 2

so the final answer here is 3x plus 2

for this top one and for the side is X

plus 3 and that is the answer fully

factored if that makes sense talk to you

guys soon bye

How to Factor Trinomials

Factoring Quadratics Problem #2. Problem factored! For more math made easy visit andymath.com.


hey guys this looks like a fun one it

wants us to solve for x it says XX - 4 +

1 = 4X -4 if you want to try on your own

pause it right now because I'm going to

solve it in 3 2 1 so for the first step

we want to say x cannot equal 4 and the

reason for that is if x were equal to

four these would be zero in the

denominator and division by Zer is

undefined next I don't really like

fractions so let's get rid of these

denominators and we can do that by

multiplying both sides of the equation

by x - 4 on the left hand side this x -

4 will distribute to both of these terms

it'll distribute to the first term and

it'll distribute to the second term and

on the right hand side it's just going

to be 4 * the x - 4 so now we can

simplify some things for this first

fraction or rational expression we can

divide away the x - 4 from top and

bottom and then for this part anything

times 1 the times 1 isn't doing anything

and then on the right hand side this x -

4 can be divided from top and bottom so

we now have x + x - 4 = 4 we can add 4

both sides on the left hand side we'll

be left with x + x which is 2X and then

on the right hand side 4 + 4 is equal to

8 we can divide both sides by two and

that'll give us X is equal to 4 well the

problem is earlier we said X cannot

equal 4 so since X can't equal four this

is not a valid solution how exciting

Fun Rational Equation

https://andymath.com/rational-equations/

Here are some more if you want to try more!


hey guys welcome to this video we're

going to solve and graph these the way

you handle this type of problem is

solved for x for each of these three add

5 for X is less than 40 and divided by

4x is less than 10 we'll do the same

thing here minus 4 negative X is less

than or equal to negative 14 divided by

negative 1 and we get x + 14 and anytime

you multiply it or divide by a negative

number you switch the sign so you can't

forget to switch the sign and then we

saw the word and here so and means we

want to know when they are both true so

let's graph this tried to show zero to

10 and the 14 not to scale and I'm sure

many other numbers this pretty much

tells the story 10 is an open circle

because it does not include it and it

says X is less than 10 so I'm going to

go this way and then X greater than or

equal to 14 I'm going to shade the

circle in because it's greater than or

equal to so the equal to means is shaded

in and I'm going to go this way because

X is greater than it so the word and

means we want both of them to be true in

this case both are never true so the

answer is no solution or it could also

be empty set depending on your teacher

so let's answer the question hope that

makes sense please like and subscribe

check out aunty Matt Kong for more

videos and notes and I'll talk to you

guys soon bye

you

you

you

you

Solving Compound Inequalities


area of a circle is pi r squared why is

that how did they figure this out

you can split the circle up into pieces

i split it up into ten pieces here

as we separate them we start to rotate

them like this and line them up and if

we bring them together

i just took all these pieces of the

circle and turned it into a

parallelogram now it's true these are

little curves

the more slices you cut it into the

straighter those curves will be and if

you cut it into an infinite number of

slices

all those curves will go into a straight

line so essentially the circle is the

same thing as this parallelogram the

area of a parallelogram is base times

height so what is the base

and what is the height if we pull our

parallelogram apart you'll notice that

this base

is half of the circumference and

circumference is 2 pi r

so half of the circumference would be pi

r next let's talk about the height if we

pull out one of the slices you'll notice

that the height

is the same thing as the radius of the

circle so we have a base of pi

r and a height of r therefore the area

of this parallelogram

is pi r squared area of a circle is pi r

squared

Why Area of Circle is pi*r^2

Check out my website!
https://andymath.com/

#Shorts


hey guys this looks like a fun one it

says solve without logarithms this is

part two in part one I solved it with

logarithms this is what it looked like

right here a lot of people mentioned

there was an easier way to solve it the

reason I did it this way because I

wanted to show what it would look like

if the answer wasn't clean like x equals

6 if x had been some crazy irrational

number you'd have to go through all

these steps but since X was 6 there is a

way to do it without logarithms if you

want to try it on your own positive

right now cuz I'm going to solve it in 3

2 1 first want to express everything in

their prime factors so the N9 can be

broken down into 3^ SAR and the four can

be broken down into 2^ s and then to

break down the 144 let's make a factor

tree 144 breaks down into 12 * 12 and

then each 12 will break down into 6 * 2

so we have 6 * 2 * 6 * 2 and then each

of these sixes will break down into 3 *

2 so we have 3 * 2 * 2 * 3 * 2 * 2 in

this we have 3 * 3 which is 3^ 2 and we

have 2 * 2 * 2 * 2 which is 2 4

so 144 and its prime factors is 3^ 2 * 2

4 so we can change this 144 to 3^ 2 * 2

to 4th now we have everything broken

down into their prime factors we can

clean things up a little bit let's bring

this exponent out here and we can bring

this exponent out here and then this two

will distribute to both of these so this

would make it 2x - 10 and then down here

this two can distribute to both of those

so it'll be 2x - 16 and then next I

don't really like fractions so let's get

rid of this denominator we'll move this

over here and then we'll put in a one

placeholder and then we can focus on

this piece with the denominator using

these notes anytime you have 1/ n that's

the same thing as n^ the1 power so we

can rewrite this as this to the 1 power

and then next we can distribute this

negative 1 to both of these- 1 * POS 2x

is -2X and - 1 * -16 is pos6 and then we

can copy down the rest of it now we have

3 to some power * 2 to some power = 3 to

a power * 2 to a power let's make these

exponents for the 3es blue and we'll

make the exponents for the twos red if

this exponent were to equal this

exponent and this exponent were to equal

this exponent then this equation would

have to be true so let's solve for x for

both of these for this first equation

let's add 10 to both sides on the left

hand side we'll have 2X and on the right

hand side 2 + 10 is equal to 12 and then

to solve for x we can divide both sides

by two and we get X is equal to 6 and

then for this equation we can subtract

16 from both sides on the left hand side

we'll have -2X and on the right hand

side 4 - 16 is equal to -12 and then to

solve for x we can divide both sides by

-2 and that'll leave us with X is equal

to 6 so if x equal 6 this will equal

this and this will equal this which will

make the equation true that is the

answer to our question let's put a box

around it how exciting here's the next

problem I plan on working on if you guys

want to try it out I'll post the

solution as soon as I solve it how

exciting

Part 2 The Easy Way

Part 1 is here: https://youtu.be/liJ9jTSYouc


hey guys let's visualize sine squared

plus cosine squared equals 1. this right

here is the graph for sine of x and this

red one is sine squared of x you'll

notice how everywhere where the sine is

negative the sine squared becomes

positive and you can tell here in the

positive area that the shape of the

graph changes a little bit as well so

the sine squared of pi over two is equal

to one we can tell that because the

height of the graph is one at x equals

pi over two we can shade from the x-axis

to the height of the sine squared of x

graph the reason that's nice is because

if i shift it upwards it's no longer the

height on the y-axis that tells me the

function value is the height of the

shading at each point that tells me the

function value i could even turn upside

down like this and it really wouldn't

matter here at x equals pi over 2 we can

see that this height is still 1. next we

can look at the graph of cosine of x and

then in blue here we have cosine squared

of x let's get rid of this cosine of x a

shape between the x-axis and the y-axis

so that the height of the shaded area

gives us the function value and now

let's bring down our sine squared of x

and you see they fit together perfectly

and that gives us sine squared plus

cosine squared is equal to one how

exciting

Pythagorean Identities Visualized #shorts

https://andymath.com/trigonometric-identities/ #shorts


he guys let's say we wanted to find the

rectangle where the perimeter was equal

to 420 and the area was equal to 69 we

need to find the base and the height

that would make this true if you want to

try it on your own pause it right now

cuz I'm going to solve it in 3 2 1 so

let's called the base X and the height y

we know that the perimeter is going to

be 2 X's + 2 y and that needs to equal

420 and the area is going to be the base

time height or in other words x y and

that needs to equal 69 so right now we

have two variables X and Y and we have

two equations let's solve for x and y so

I'm going to clean things up a little

bit on top we can simplify things by

dividing everything by two and on bottom

we can get the Y by itself by dividing

both sides by X up here on the left hand

side these twos will cancel each other

out and on the right hand side 420 / 2

is 210 and then down here on the left

hand side these X's will cancel each

other out so we're left with Y is equal

to 69 /x let's take this X+ y = 210 and

copy it down here and we know that Y is

equal to 69 /x so in the place of this y

let's plug in 69x and I don't really

like fractions so let's get rid of this

X in the denominator we can do that by

multiplying both sides of the equation

by X on the left hand side this x will

distribute to give us x * x and x * 69

/x and on the right hand side we get 210

* x x * X gives us x^2 and this X and

this x will cancel each other out this

is a quadratic so I want to set it equal

to zero so let's subtract 210x from both

sides on the right hand side these will

cancel each other out so it's equal to

zero and on the left hand hand side we

have x^2 - 210x + 69 now let's solve

this using the quadratic formula the

leading coefficient is a one so both of

these A's are equal to 1 and then the

second coefficient is -210 so in the

place of both of these B's we can plug

in -210 and last the C is equal to 69

now we can clean this up -210 is equal

to postive 210 -210 2 is equal to

44,100 and 4 * 1 * 69 is = to 2

276 and on bottom 2 * 1 is equal to 2

44,100 - 276 is equal to

43824 and

43824 can we split up into 16 *

2739 and then we can give each of these

their own square root 16 is a perfect

square so we can change this into four

and this two in the denominator can be

split up for both of these terms 210 / 2

is 105 and 4 < TK 2739 / 2 is equal to 2

< tkk 2739 and now we have our values of

X they're 105 plus or- 2 < TK 2739 and

this plus or minus can be split up into

two different numbers we have the plus

version and the minus version and now

for each of these X's we have to find

the corresponding y let's use this x + y

is equal to 210 we can subtract X from

both sides these two x's will cancel

each other out and now we know that Y is

equal to 210 - x so in the place of both

these y's let's plug in 210 - X for the

first box X is equal to this so let's

plug this in for x and for the second

box X is equal to this so let's plug in

this for X both of these are positive so

the negative will make them both

negative and aside these parentheses the

first one's positive and the second

one's negative after we subtract they're

going to switch the first one will be

negative and the second one will be

positive now for the first box 210 - 105

is equal to 105 so we have the value of

y that corresponds to this X and then

for the second box 210 - 5 is 105 so we

have the corresponding value of y for

this value of x so all these boxes are

doing are just swapping the X and the Y

so we can just get rid of the second box

so now we have the value of our base is

105 + 2 < TK 2739 and the value of our

height is 105 - 2 < TK 2739 so now we

can put these values in for the base and

the height and create a problem that

says find the area and perimeter and if

you want to see the video solving this

problem the link is in the description

how exciting

Nice Rectangle Part 2

https://youtu.be/-f1ozWwRDSc
Link to Part 1