hey guys so which one of these is

greater 5 to the 100 or 3 to the 150. we

could probably just plug this into a

calculator I think it'd be more fun to

try this by hand without a calculator

upon first inspection 5 is greater than

three a one fifty is greater than 100 so

it's not immediately obvious which one's

greater I can fairly easily change this

5 into 25 and the 3 into a 27. so once

we do this it might be easier to compare

these so let's rewrite 5 to the power of

100 as 5 to the power of two times fifty

and let's rewrite 3 to the power of 150

as 3 to the power of three times fifty

and we can strategically place

parentheses here and here and so we can

see 5 squared is just 25 so this would

be to the 50 and 3 cubes is just 27 so

this would be to the 50. now we have 25

to the 50 compared to 27 to the 50. and

27 to the 50 will be the greater one so

this is the answer to the question how

exciting

Exponents Math Challenge


hey guys this looks like a fun one it

says x^2 + IX + 2 = 0 and it wants us to

solve for x this I coefficient can be

thought of two different ways you can

think of it as i^2 is equal to -1 or I

is equal to the < TK of1 it's called the

imaginary number and you can do a lot of

useful things with this I but for this

video I was wondering can we use

quadratic formula to solve this

quadratic if the coefficient is

imaginary let's try it out so a is the

coefficient for x^2 and up here that

would be an implied one so in the place

of this a we can plug in one and the

same thing for this A and B is the

coefficient for X this is where the I

comes in so in the place of this B we

can plug in I and we can do the same

thing to this B and last C is the

constant which is two up here so we can

change this C to two and now we can

clean things up let's copy down the

negative I and the plus or minus and the

square root and for this I squ we know

that i^2 is equal to -1 so I'm going to

change this to -1 and then we're going

to subtract 4 * 1 1 * 2 which is equal

to 8 and then on bottom the 2 * 1 is

equal to 2 next let's focus on this

radical we have -1 - 8 which is equal to

9 we can split up this < tk9 to beunk 1

* < tk9 and now this Ro tk1 if we look

up here is equal to I so we can change

this into I and then theun 9 is equal to

3 so now we have X is equal to i+ or - 3

i/ 2 This plus or minus is giving us two

solutions we have one for the plus and

one for the minus and now for the first

one I + 3 I is equal to 2 I and 2 / 2

cancel each other out so we have x = i

for the first one and for this one I - 3

I is -4 I and -4 / 2 is -2 so the other

solution is -2 I so the quadratic

formula did seem to work but let's check

our answers let's copy down the original

equation and in the place of both of

these X's we want to plug in -2 I for

the first term term -2 i^ 2 the squar is

going to go both to the -2 and to the I

and then for the second term we can

bring down the -2 and I * I is equal to

i^ 2 and the rest of this can be brought

down and then before we move on -22 is

the same thing as -2 * -2 which is equal

to pos4 for the next step let's copy

this down and then i^2 is equal to -1 so

in the place of both of these I squ we

can plug in -1 4 * -1 is -4 and -2 * -1

is pos2 and then -4 + 2 + 2 is equal 0

so we end up with 0 equal 0 so the first

one checks out so now let's check the

second one in the place of both of these

X's we can plug in I so we end up with

i^2 + i^2 + 2 = 0 and then i^ 2 is equal

to 1 so each of these I squar can change

into 1 and then - 1 +1 + 2 is equal to 0

so we end up with 0 equals 0 and the

second one checks out so the quadratic

formula did work for an imagin

coefficient and we now have our answers

let's put a box around them how exciting

Would Quadratic Formula Work?

I hope you guys like this one! Let me know if you have any other ideas for problems!


hey guys so we have the absolute value

of 4 - 3x is equal to X absolute value

means that whatever is inside here you

need to make it positive so there are

two ways this can be true if this stuff

inside of here is equal to X or if it's

equal tox and that's usually the way you

solve these you split it up one where

you're doing the positive of this thing

and one where you're doing the negative

of this thing and now we can just solve

each of these independently let's add 3x

to both sides so on the left hand side

we just have four and on the right hand

side we have x + 3x which is 4X and

after we divide both sides by four we

get 1 is equal to X and on the other

side we can add 3x to both sides once

again the left hand side is 4 but on the

right hand side x + 3x is 2X and after

we divide both sides by two we get X is

equal to 2 and now we just got to check

our answers so for the first one let's

plug in one and for the second one let's

plug in two inside this absolute value

we have 4 Min - 3 which is equal to one

the absolute value of one is one so that

checks out and then inside of this

absolute value we have 4 - 6 which is -2

and the absolute value of -2 is 2 so

that checks out so these are the answers

to this question x = 1 or x = 2 let's

put a box around them how exciting and

if you guys want to try more of these I

have this page on andy.com the link is

in the description

Absolute Value Equation Challenge


hey guys so we have a cube right here is

a very nice looking cube to find the

volume of the cube if we call this edge

s then it would be s cubed if you have

another cube larger its volume would

also be s cubed where s is the length of

one of these edges on this blue one if i

make this edge equal to one meter then

the volume would equal one meter cubed

it's literally a cube of dimensions of

one meter now on the red cube if i

double the length of the edge when i

calculate the volume it ends up being

eight times larger so double the edge

volume is eight times larger now let's

look at spheres the formula for the

volume of a sphere is four thirds pi r

cubed if this blue sphere has a radius

equal to approximately 0.62 meters the

volume would be equal to one meter cube

it'd be the same volume as this blue

cube up here and then for the red sphere

if i double the radius then the volume

would be equal to eight meters cubed

double the radius i get eight times the

volume so now let's look at a pizza we

can think of a pizza as a cylinder so

it's volume equals pi r squared h and

let's also look at a larger pizza which

would also be a cylinder i want to make

these very very large pizzas i'm going

to make the height of it about 5

centimeters which would end up being

0.05 i'm plugging that in for h if the

radius of this blue pizza is equal to

2.52 meters the volume of the pizza

would be one meter cubed that is a giant

pizza now if i double the radius to get

5.04 meters what is the volume gonna be

of this pizza well it's equal to four

meters cubed so when i double the radius

i got four times the volume why did i

get eight times the volume for these but

only four times volume for this answer

down below in the comments how exciting

Pizza Math Problem

For more math fun, check out andymath.com!


so surface area is as if we were

painting the whole thing how much area

is that so we're gonna add up the area

of all six sides

this front side is two by five so it has

an area of ten and we also have the back

side that matches this that we can't see

we'd say two times this 10

plus this top is also 2 by 5 so it's

also 10 and then we also have the bottom

so we're gonna multiply that by two then

we have this side here which is two by

two

area four and then we have it on the

other side over there so there's two of

those

so we have the front and back top and

bottom and the two sides 2 times 10 is

20 2 times 10 is 20

and 2 times 4 is 8. 20 plus 20 plus 8 is

they don't give us a label anywhere so

we can just call this unit

squared and 48 units squared is the

answer to the question if you have any

other math

questions check out andymath.com or

comment below on what you'd like me to

do next

i'll talk to you soon bye

Surface Area Fast Example #Shorts

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

#Shorts


hey guys a while ago somebody mentioned

me in this video Let's Solve IT pause it

real quick if you want to try on your

own because I'm going to solve it now

this says the five semi-circles are

identical so we know they all have the

same radius and diameter it says solve

for the diameter so I'm going to label

each of these diameters as D I know that

the distance is the same whether I

measure from the top or the bottom so

I'm going to set those two things equal

to each other so we have triple d's plus

two twelves is going to be equal to

Double D's plus double 22s plus a single

16. well Double 12 is 24 and double 22

is 44 and then we can bring down

everything else let's combine like terms

the 44 and the 16 give us 60 and bring

down everything else subtract 2D from

both sides we get a single D plus 24

equals 60 and then subtract 24 from both

sides and we get D equals 60 minus 24

which is 36. let's squish these together

and put a box around it the diameter is

equal to 36. how exciting

Canadian Math Olympiad Challenge #Shorts

For more math fun visit https://andymath.com/

#shorts


hey guys what's going on with this thing

right here how are we going to evaluate

this this is pretty confusing this thing

is quite ambiguous as far as I can tell

there's two ways to interpret this we

have this fraction on top divided by

this on bottom or we have the 4830

divided by all of this on bottom I don't

think it's clear which one of these is

intended so let's solve both of them and

see what happens I think I have a pretty

good idea and 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 do this

one first on bottom here the square root

of this fraction is equal to the

fraction of the square roots and then we

can put the two on top now we have two

fractions another way to think of this

is this fraction divided by this

fraction and we can express it like this

and now anytime you're dividing

fractions a cool way to handle this is

leave the first fraction alone change

the Divide to a multiply and flip the

second fraction on this side right here

we can break down the 805 into 23 * 35

and the square root of this product is

equal to the product of these square

roots and now we can smush it all

together the < TK of 23 on top and

bottom can cancel each other out and

then we can shift things around > 35 * >

35 is 35 and 35 * 2 is 70 4830 / 70 is

69 nice and that's the answer to our

first one so once again we have this

divided by this let's rewrite it like

this and then we can put the 4830 over

one and then we'll copy this down switch

the Divide to a multiply and flip the

second fraction once again the square

root of this fraction is equal to the

can put the two on top and now we have

another fraction divided by a fraction

let's express it like this and then once

again we can copy down this stuff change

the Divide to a multiply and flip the

divided by fraction we can smush

everything together and let's change

this 805 into a 23 * 35 and the < TK of

23 * 35 is the same thing as the < of 23

* < TK of 35 from here these two ones

ain't doing nothing so they can go away

and then rooot 35 overun 35 those can

also go away and then < tk23 * < tk23 is

equal to the whole number 23 and on top

4830 * 2 is equal to

960 and then 960 / 23 is equal to 420

and that's the answer to our second

ambiguous case so in conclusion this is

ambiguous we don't know what it is equal

to but we do know that it's equal to

either 69 nice or 420 how exciting

Some Nice Math

I hope you guys love the video!


hey guys this looks like a fun one we're

gonna find the area of the blue region

contained by these lines and the y-axis

if you want to try this on your own

pause it right now because I'm going to

solve it in three two one the equations

of these lines are given to us in

standard form I want to rewrite them as

y equals MX plus b which is known as

slope intercept form because m is the

slope of the line and B is the

y-intercept of the line let's take the

first equation X Plus y equals five and

in order to rewrite it in slope

intercept form we need to get y by

itself so we're going to subtract X from

both sides these X's are going to cancel

so the left hand side is just going to

be Y and on the right hand side we have

Negative X plus five and now we have

slope intercept form so I'm going to

rewrite X Plus y equals 5 as y equals

negative X plus five and let's do the

same thing for this lower one we'll copy

it down we're going to subtract X from

both sides left hand side is 2y right

hand side is negative X plus six and

then we're going to divide both sides by

2. left hand side is just Y and we're

going to divide each of these by two so

negative x divided by 2 is negative one

half X and 6 divided by 2 is 3. and now

we can change this lower equation up

here and do slope intercept form so now

let's use our equations to find the

y-intercepts this one up here it says

the Y intercept is five so this point is

going to be 0 5. and for this equation

the Y intercept is three so this vertex

is located at 0 3. so now that we have

these two vertices let's find this third

one and the best way to do that is set

these two equations equal to each other

and I will tell us the X and the Y point

where they're both equal to each other

so our goal is to solve for x in this

first thing I don't like fractions I'm

going to multiply both sides by two the

two will distribute so 2 times negative

X is negative 2X and 2 times 5 is equal

to 10 and this will be equal to this 2

distributed to both of these two times

negative one-half is negative one and

then times x is negative X and then 2

times 3 is equal to 6. now we want to

move both X's to the same side if I add

2x to both sides negative 2X and 2x

cancel each other out negative X plus 2X

is equal to one X and we bring down the

six now we want to get this x by itself

so we subtract 6 from both sides 10

minus 6 is equal to 4 and these cancel

leaving us with just X now we know that

the X that this intersection point is

equal to four so we can change that to a

four and then to find out what the

corresponding Y is we can plug this 4 in

4X for either one of these equations I

think it'll be easier to do this top

equation so we're going to substitute 4

into the X and negative 4 plus 5 is

equal to 1. so the Y value of this point

is one and we don't need this anymore in

fact now that we have the location of

our vertices we don't need our equations

anymore either so if this has a height

of 3 and this has a height of 5 that

means the distance is 2 in between them

and let's find this distance as well

since this x value is 4 we're going from

0 to 4 which means this distance right

here is four and now we really don't

even need these points anymore and let's

rotate this and now we want to find the

area of this triangle the area of a

triangle is one half base times height

the base of this triangle is equal to

two and the height of this triangle is

equal to four the one-half times two

just cancel each other out so we have an

area of four units squared so the area

of this blue region is equal to four

units squared now we can put a box

around that how exciting

Triangle Area from Graph


guys I think this is a cool one I want

to show you the steps of this proof that

3 is equal to Pi let's see if you can

spot the mistake when it happens so

we're going to start with x equals pi

plus 3 over 2. and first we multiply

both sides by 2. on the left hand side

we have 2X and on the right hand side

these twos cancel so we're left with pi

plus 3. and then from here let's

multiply both sides by pi minus 3. so on

the left hand side we're going to

distribute we have pi times the 2x which

is equal to 2 pi x and we have the

negative 3 times the 2x which is equal

to negative 6X and on the right hand

side this is a difference of two squares

anytime you have a plus b times a minus

B it's just a squared minus B squared or

in this case pi squared minus 3 squared

or in other words pi squared minus 9.

next let's subtract this 2 pi x from

both sides and let's add 9 to both sides

on the left hand side these cancel so

we're just left with 9 minus 6X and on

the right hand side these cancel so

we're left with pi squared minus 2 pi x

and then let's add x squared to both

sides of the equation then let's

rearrange the order of a couple of these

this is a perfect square trinomial it

ends up being x minus 3 squared if we

multiply x minus 3 times x minus 3 it

brings us back to this and same thing

here this is also a perfect square

trinomial it ends up being x minus pi

squared and if we multiply x minus pi

times x minus Pi it brings us back to

this and now from this step let's square

root both sides on the left hand side

the square and the square root cancel

each other out so we have x minus 3. and

on the right hand side the square and

the square root cancel each other out so

we have x minus Pi we end up with x

minus 3 equals x minus pi and from here

we subtract X from both sides on the

left hand side we have negative 3 and on

the right hand side we have negative pi

after we divide both sides by negative

one we get three equals pi and clearly

this is not true so where is the mistake

and if you want to try to find the

mistake on your own positive because I'm

going to show it in three two one the

mistake in this occurs right here going

from this step to this step so the

square and the square root don't just

cancel out and give you what's in the

parentheses they always give you the

positive of what's inside the

parentheses so the correct thing would

be to put both sides of the equation in

absolute value and we can actually work

with this so we do know what x is equal

to X is equal to this pi plus 3 over 2

and that is approximately equal to 3.07

so if we plug in 3.07 here this inside

will be positive which means we don't

need the absolute value symbols but if

we plug in 3.07 into this one this

inside the brackets will be negative so

before we get rid of the absolute value

symbols we need to take the negative of

this and that will turn the whole thing

positive and now let's work through this

on this side we can bring down the x

minus 3 and then on the right hand side

this negative distributes to give us

Negative X and then negative times

negative pi is equal to positive Pi

let's add X to both sides and add 3 to

both sides on the left hand side the

threes cancel which gives us 2X and on

the right hand side the X's cancel which

gives us pi plus 3 and after we divide

both sides by 2 we get X is equal to pi

plus 3 over 2. well we already knew that

so in reality we didn't accomplish

anything with all this work we just

apply to operations and then they're

inverses and it brought us back to where

we started did but we did successfully

find the mistake in this proof of 3

equals pi and that was in this step

right here how exciting

Find the Mistake in this Proof

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


hey guys this looks like a fun one we

want to solve for x and this is kind of

a staggered fraction if you want to try

it on your own pause it right now cuz

I'm going to solve it in three 2 1 the

way this is staggered you could say is

ambiguous but I'm pretty sure it's meant

to be a fraction on either side where

these are on top and these are on bottom

so let's change this four into a 4 over

one and let's change this 69 into a 69

over one so now it's a little bit less

ambiguous it's going to be this / this

equals this / this let's see if we can

make a animation for that it'll be 4

piun / 691 = 41 / x 69 dividing

fractions can be a little bit tricky but

there's a cool shortcut you can leave

the first fraction alone so we can copy

down the 4 over pi and the 4 over1 then

you change the divide into a multiply

and then you flip the second fraction so

69 over 1 becomes 1 over 69 and X over

69 becomes 69x and now multiplying

fractions is a lot easier you just

multiply across on top and on bottom and

now we can just clean things up 4 * 1 is

equal to 4 and < * 69 is 69 piun 4 * 69

is equal to 276 and 1 * X is just X now

we have a fraction equals a fraction we

can solve this by cross multiplying we

can do 4 * x = 4x and that's equal to 69

piun * 276 that's 19,44 piun and then we

can divide both sides by 4 and we get X

is equal to

4761 pi and this is the answer to the

question let's put a box around it and

that's approximately equal to

14957 one2 and let's put a box around

that too how exciting

Can we solve for x?

Is this one ambiguous or can we solve for x?


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

two R and yes radiuses is the plural of

radius we could also say radi 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 the square

minus the area of this circle now we

don't need the 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

45459 90 triangle so it ends up being <

TK of 2 * one of these sides so the

diagonal is equal to 2 R < tk2 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 < tk2 R and if we want

to know what the radius of the largest

circle is we would just divide all that

stuff by two which gives us r + < tk2 r

and now to find the area of this circle

we're going to do PI radius s 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 = 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 largest 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 s and then we

can cancel those R squs 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 ends up being

approximately

20308 how exciting

Ratio of Pink/Green

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

For more geometry challenges check out the above link!


hey guys I saw this the other day

someone wrote I've never seen this

problem before it looks interesting

enough so let's do it 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 this one is not super

difficult I think they're expecting

people to see the right triangles here

connect these and think a squared plus b

squared equals c squared and then do the

same thing here same thing here and then

maybe they expect you to connect this

and think what am I supposed to do with

this shape it can be very confusing if

you go down this route but the simple

way to do it is just to complete this

right triangle this side right here the

height of this triangle is going to be 2

plus 2 plus 2 which is six and the base

is going to be four plus two plus two

which is eight and if we call this last

side C we can do the Pythagorean theorem

we're going to plug in 6 for the a 8 for

the B and then solve for C 6 squared is

36 plus 8 squared is 64 and that's all

going to equal c squared 36 plus 64 is

equal to 100 and after we square root

both sides we get C is equal to 10 and

that's this one right here how exciting

Fun Math Problem

I hope you guys like it.


this is going to be a fun problem we

have a shed that has a 6 m x 6 M area

right here and then a side room that's a

3 MX 3 m then tied to this corner of a

shed is a leash for a goat this leash is

7 m long here's the goat right here and

as we all know Andy is the goat if we

wanted to analyze the amount of grass

that the goat could eat usually

something like this where it forms a

circle around where the leash is pinned

is the center of the circle and then it

goes out with a radius that's as long as

the leash normally a goat would be able

to eat all of this grass you go inside

the circle because the leash isn't stiff

but we do have these buildings in the

way so we got to calculate how much

grass can the goat eat with these

buildings that are in the way so first

off we can bring the goat over here and

we can see that he can eat all of this

very important detail here is what angle

is this this is

225° how do we know that we know the

half of the circle would be 180° and

then since this is three and we know

this whole side is six that means each

of these are equal to three and so we

have a 3X3 triangle which means this has

to be 45° so 180 + 45 is 225° so let's

calculate the area of this piece right

here well normally a circle would beunk

R 2 this is 225° out of the total 360°

of the circle and 225 over 360

simplifies to 5 over8 so this is 58 of

an entire circle with radius 7 so we can

change this R into a seven next we have

this red piece right here and it looks

like this here so what is the radius of

this well the whole length of the leash

is seven so we're using up six of it to

get to this side of the shed that means

that from here we have one more meter

that can pivot around and make a quarter

of a circle so this has a radius of 1 M

and then we can see that this makes a

90° angle once again we have the area of

a circle is PK R squ this is 1/4 of an

entire circle so we'll call it 1/4 PK R

squar and then we can plug in one for R

plug this into a calculator and get

approximately 785 next thing we have

this orange triangle and you can see

that the goat would be able to eat

everything inside of this orange

triangle so let's Analyze This each of

these two sides is three and to get this

hypotenuse here since since it's a

454590 triangle we know that that is 3 <

tk2 m and if you want to see the work

behind it here it is right here using

Pythagorean theorem you can pause it and

look at this if you want the area of a

triangle is 1/2 base time height this

triangle is upside down but this base is

going to be three and this height is

going to be three and this all

simplifies to give us 4.5 next let's do

this up here let's move the goat out of

the way and then we're looking at this

piece right here now doesn't follow all

the way we can zoom in and look at it it

goes kind of like this so it's going to

have a different radius than the circle

irle so that's why it's not following

the whole circle it's kind of like what

was going on over here with the red one

because you have a pivot point that

creates a new Mini Circle so let's look

at this pie slice right here well the

easy part is seeing that that's going to

be 45° it's the same thing as this 45

right here so we know this is going to

be 45° the trickier part is finding the

radius so we're going to call the radius

R the goat were up here we're going to

be using up a portion of the leash I'm

going to show that leash right here and

we showed earlier that that length is 3

< tk2 so we know that as far as this

goat can reach is 7 that means that 3 <

tk2 + r has to be equal to 7 that's as

far as the leash will reach we have to

subtract 3 < tk2 from both sides to find

out r r is equal to 7 - 3 < tk2 that's

approximately 2757 if we clean all this

stuff up let's find the area of this

yellow piece p r squ is the area of a

circle but we're dealing with 45° out of

360 that's going to be 1/8 of an entire

circle and then we can substitute 2757

in for the r plug this into a calculator

we get approximately

2986 so now we have the areas this red

green orange and yellow shape make up

everything that this goat can possibly

eat if we add up these four areas we'll

know how much the goat can eat and

that's the answer to the question how

exciting

Goat Math

This is a neat one. I hope you guys love it!

Subscribe for more!


hey guys this looks like a fun one it

says what's the area of the semicircle

inside this equilateral triangle so

we're given an equilateral triangle a

circle with area six and a semicircle

right here this is day 12 of our Advent

calendar for the month of December we're

going to solve 31 Kina ad puzzles in 31

days and if you want to try this one on

your own pause it right now because I'm

going to solve it in one two three first

thing we're given a six and a question

mark I think we should figure out

formulas for both of those the six is

the area of this smaller Circle and the

question mark is the area of this

semicircle let's label the radius of the

smaller Circle little r and the radius

of the bigger semicircle Big R so 6

would be equal to PI Little R 2 and the

question mark is going to be equal to Pi

big r^ 2 / 2 these look important let's

put boxes around them and our goal is to

find this question mark I think we need

to find some relationship with the Big R

and the little r and then we should be

able to do some substitution from here

let's drop this perpendicular bis sector

and let's make this black and white

first I want to focus on this orange

triangle let's zoom in right here this

line bisected an equilateral triangle so

this is normally 60° up here but this

angle will be half of that or 30° and

then this angle right here is built off

of the radius and the tangent line the

radius of any Circle will meet its

tangent line at 90° and then all three

of these angles add up to 180 so this

one would be 60°

our orange triangle is a 306090 triangle

that is a special right triangle and

it's always true that the hypotenuse is

twice as long as the shortest side so

this distance right here will be two

little r and that correlates to this

right up here this piece is a radius of

the small circle we'll call that little

r and since the whole thing is too

little r this remaining piece will also

be little r and then while we're ated

this piece right here is also a radius

let's call it little r and we can finish

this this is a radius of the semicircle

so let's call this Big R so the entire

height of our equilateral triangle is

Big R + 3 little r next let's focus on

this larger triangle and let's zoom in

on it it will also be 30° up here and

for the same reason this angle will also

be 90 de and the remaining angle will be

60 so this is another 30 6090 where the

hypotenuse is twice the length of the

smaller side so this distance will be 2

Big R so this whole distance is 2 Big R

this is one Big R R which leaves one

more Big R right here and this is the

relationship we wanted we have Big R is

the same thing as three little r this

looks important let's put an irregular

octagon around it and now we're ready

for our original equations down here in

the place of the Big R we can plug in

three little r and this exponent will

distribute to both of these to give us

3^2 little r^ 2 3^2 is equal to 9 and

now let's rearrange this a little so we

have the question mark is equal to 9 pi

little r^ 2 / 2 well Pi little r s is

equal to 6 so in the place of this we

can plug in 6 and then 9 * 6 is 54 and

54 / 2 is 27 let's give it a label of

units squared and that is the answer to

our question the area of this semicircle

is 27 unit squared how exciting this is

our answer for day 12 and this is the

puzzle for day 13 we're given the areas

of these two triangles and it wants to

know what's the area of the entire

semicircle this looks like it'll be a

fun one how exciting in

Aggvent Calendar Day 12


guys let's do some calculus this says

find dydx for y equals natural log of e

to the X over e to the x minus 1. so

this is dealing with logarithms so first

thing we want to do is apply this rule

of logarithms the natural log of a over

C is equal to the natural log of a minus

the natural log of C so in this case the

top function is e to the X and this

bottom function is e to the x minus 1.

and next is another property of

logarithms we can use that the natural

log of e to the x is equal to X so this

first term can just be simplified to X

so let's take the derivative derivative

of x is equal to one and then we're

going to subtract and then for the

derivative of natural log let's use

these notes right here it's equal to the

derivative of x over X so let's do our

fraction here where we put X on bottom

but X is this whole e to the x minus 1

and then on top is the derivative of

this and the derivative of this is just

e to the X and the derivative of

negative 1 is 0. so now we're

technically done but we can clean it up

a little bit let's give these two terms

a common denominator we'll multiply top

and bottom by e to the x minus one so

now this times one is just this and we

can smush these together these have a

common denominator so we can combine

them into a single fraction where we

just do this minus this and e to the x

minus E to the X goes away and so we're

just left with negative 1 over e to the

x minus one and that's the answer to the

question we can put a box around it you

guys want to try more calculus problems

I have a link down in the description

how exciting

Derivative Challenge


hey guys I thought this was a fun one

we're given two squares and two

triangles and it says find the area of

the red triangle I'll give you a hint

right now the method that I use to use

these notes right here we have the area

of any triangle is equal to 1/2 * one

side time another side time the sign of

the angle in between them and we also

have 306090 triangles where the

hypotenuse is twice as long as the

shortest side and the middle side is <

tk3 times the shortest side if you want

to try this on your own pause it right

now so I'm going to solve it in three 2

1 for this square up here we know all

the sides are congruent so they're all

equal to three and now looking at this

right triangle I'm noticing that the

medium side is < tk3 * the shortest side

that means it's going to be a 306090

triangle and we can align the

orientation just like this so we have

the shortest side here and the medium

side here which means this angle here

will be the 60° angle and this side is

twice as long as the shortest side so

it'll be equal to six and now looking at

this lower Square all the sides are

congruent so this last side will also be

equal to six another property of squares

all of these angles are right angles so

we can call this 90° here and in this

Square this would be 90° and let's call

this missing angle Theta now if we look

here 90 + 60 + 90 + Theta must equal

360° 90 + 60 + 90 is equal to 240 so 240

+ Theta = 360 after we subtract 240 from

both sides we get Theta equal to 120 so

we can update this up here to 120° and

now let's focus on this we're trying to

find the area of this red triangle we're

going to use these notes where the area

is equal to/ 12 * one side time another

side time the sign of the angle in

between them so our first side is going

to be three the other side is going to

be six and the angle between these two

sides is equal to 120° 6 * 3 is 18 and

half of that is 9 and the S of 120° is <

tk3 over2 so we end up with an area of 9

< tk3 over2 square units and that's

approximately equal to 7.79 square units

let's bring back all the other stuff and

we successfully found the area of the

red triangle let's put a box around it

how exciting

Area of Red Triangle


hey guys this looks like a fun one we're

trying to find the area of the red

region if you want to try this on your

own pause it right now cuz I'm going to

solve it in 3 2 one so first step let's

copy it down and let's cut it in half

with the diagonal right here and let's

make all of this one region the entire

square is 10 * 10 which is 100 so that

makes this right here 50 and we don't

know the area of these so let's give

them a variable a and let's write an

equation for this right here we know

that the 2 A's plus this x + 50 is going

to equal the entire Square which is 100

and then we can subtract 50 from both

sides and also subtract 2 a from both

sides and that gives us X is equal to 50

- 2 a and this looks pretty important

let's put it up here so all we need to

do to solve for x is figure out the area

of these two A's and I've got a good

idea how to do that let's copy the

square again but let's only include the

top semicircle the area of that

semicircle is going to be P pi r 2 divid

2 and since the sides of the square are

10 the radius is going to be equal to

five so we can update this R to be 5

then 5^ 2 is 25 < * 25 / 2 is 12.5 piun

and next let's cut this with both

diagonals each of these triangles is 1/4

of the total square so this right

triangle up here is going to have an

area of 25 and then this piece right

here is literally the exact same thing

as this piece right here so it is also

equal to a and then this is the same

size is this so we'll call it a and now

we have another equation this 25 + 2 a

makes up the entire semicircle so it's

going to be equal to 12.5 pi and after

we subtract 25 from both sides we get 2

a is equal to 12.5 piun - 25 and now we

have a value for the 2A so let's solve

for x so now in the place of this 2A we

can plug in 12.5 piun - 25 this negative

is going to distribute to give us -2.5

piun and minus - 255 is + 25 next we can

combine like terms 50 + 25 is 75 and

this is still there and now we have the

area of the red region let's give it a

label of square units and put a box

around it and here's the approximate

value of 35.7 3 square units how

exciting

The Hard Way

https://youtu.be/k4QjXIAr4W8
Here is the link to the easier version.


hey guys this is such a fun looking

problem we have two circles with a

radius of 5 m and then we have a square

stuck inside here and it wants us to

find the area of this square if you want

to try this on your own pause it right

now because I'm going to solve it in 3 2

1 first step I want to label the sides

of this Square X so the area we're going

to solve for is going to be X2 and let's

stick that right over here next let's

draw another radius from the center of

this circle to the intersection point

right here and let's draw another radius

from this point here to this tangent

point and that's going to be at right

angles each of these blue lines are both

radi of the circle so they're both going

to have a length of 5 Metter and let's

extend this line right here parallel to

this so that this will also be a right

angle so with this setup this red

portion of the line right here is the

same thing as the height of this Square

so it will also have a length of X and

since this whole thing has a length of

five this remaining blue portion will be

5 - x and then let's call this green

line Y we now have a right triangle

right here so so we can use a 2 + B2 =

c^2 we can do 5 - x^2 + y^2 = 5^ 2 5 -

x^2 is 5 - x * 5 - x which gives us 25 -

10 x + x^2 and then we still have the +

y^ 2 and it'll be equal to 5^2 which is

25 and then we can subtract 25 from both

sides and we can also subtract the rest

of the stuff from both sides on the left

hand side we only have y^2 and on the

right hand side we have 10 x - x^2 after

we square root both both sides we have y

is equal to the < TK 10x - x^2 and now

we can clean things up a little bit

let's scoot this over here and let's

duplicate the Y over here now this

distance is going to be equal to this

distance so we know that y + x + y is

equal to 5 + 5 and we can clean this up

a little bit y + y is 2 Y and 5 + 5 is

10 and now we know that Y is equal to

the < TK of 10x - x^2 so in the place of

this y we can plug in < TK of 10x - x 2

and now we just got to solve for x we

can subtract X from both sides that'll

give us 2 * the < TK of 10x - x^2 is

equal to 10 - x and then we can square

both sides of the equation this Square

will distribute to both of these so we

have 2 squar and then we have this stuff

squared 2^ 2 is equal to 4 and this

square root and this Square are going to

cancel each other out and then 10 - x^2

means this times this and it ends up

being 100 - 20x + x^2 then we can

distribute this 4 it'll give us 40x -

4x^2 and then we can subtract this stuff

from both sides so we can set it all

equal to zero let's bring down the 100

-2X - 4x is - 60x and x^2 + 4x^2 is 5

x^2 all three of these coefficients are

divisible by 5 so let's do that we end

up with 0 = 20 - 12x + x^2 and then we

can Factor this we need two numbers that

add to -12 and multiply to POS 20 well

that's going to be -10 and -2 so we have

X x - 10 * x - 2 = 0 that means that

either this equals 0 or this equals 0 so

for x - 10al 0 we can add 10 to both

sides and we get xal 10 what we can see

right here since the radius is five

there's no way that X is 10 so this is

not a valid solution and then for x - 2

= 0 we can add two to both sides and we

get x = 2 this one is a valid solution

so now that we know x = 2 the area of

our square is going to be equal to 2^2

which is 4 let's give it a label of

meters squared and put a box around it

how exciting

2 Circles 1 Square

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

I hope you guys like this one, and the several others I have on this page!


you guys can you believe this whole

thing is equal to one let's verify it

first I want to leave this alone so

let's make a copy of it down here and

let's handle it one fraction at a time

let's do this one first in order to get

rid of the radical in the denominator

I'm going to multiply top and bottom by

the conjugate all we do is copy these

two terms but change the plus to a minus

on top is just going to be 1 times this

so we're just going to be left with root

1 minus root 2. and the square root of 1

is equal to one and on bottom we can use

the notes for the difference of two

squares so we know that root a plus root

c times root a minus root C is just a

minus C and we can see here that a is

one and C is two so on bottom we're

going to have one minus two and we can

clean up one minus 2 to give us negative

one to get rid of this negative 1 in the

denominator we can multiply top bottom

by negative one this negative is going

to distribute to both of these making it

a negative one and a positive root 2.

and on bottom negative 1 times negative

1 is just positive one and since we have

one in the denominator we don't need it

anymore all of this right here

simplified into negative 1 plus root two

in fact we don't even need this red

stuff this first fraction simplified

into negative 1 plus root 2. so now

let's focus on this next fraction we're

going to do the same thing multiply top

and bottom by the conjugate on top we're

just going to have the root 2 minus root

3. and the bottom we're going to use

these notes again but in this case a is

2 and C is 3. so on bottom we're going

to have the a minus C which is 2 minus

three and then 2 minus 3 is equal to

negative one and we can divide both

these top terms by negative one making

this first one negative and the second

one positive so we have negative root 2

plus root three if we can smush these

together and that takes care of this

second fraction now let's focus on this

third one multiply top and bottom by the

conjugate the top is just going to be

root 3 minus root four and the bottom

we're going to use these notes one more

time where a is equal to 3 and C is

equal to four so a minus C is going to

be 3 minus four three minus four is

equal to negative one these terms

divided by negative one will make the

first one negative and the second one

positive we don't need the one anymore

and we can smush this together so now we

have this expression right here this

positive root 2 and negative root 2 can

cancel each other out and this positive

root 3 and negative root 3 can also

cancel each other out and the square

root of 4 is equal to two let's bring

this down here negative one plus two is

equal to positive one that means that

this original crazy expression is equal

to one now we could put a box around it

how exciting

Beautiful Math Equation

I really like this one!


hey guys here's a fun Little Riddle That

I came up with earlier this week it's

asking you to solve for x and this looks

a lot like the quadratic formula but

it's a little different if you want to

try to solve it on your own pause right

now I'm going to show how I would

approach this in three two one so first

off this is the actual quadratic formula

I'm given that this will solve for x but

I also know for a fact this will solve

for x it looks like a has literally just

been substituted for all the other

letters and it looks like we've also

plugged in a for two so I think we can

say that a equals two I'm going to plug

in 2 for all the places of a then I can

simplify things a little bit simplify a

little bit more it's a little weird

right here because we have a negative

under the square root and the way we

handle that negative one of the square

root is y top and bottom are both

divisible by two and this is an answer

to the question and we can put a box

around that is there another approach we

can take to find another value of x how

exciting

Quadratic Challenge Problem #shorts

https://andymath.com/ #shorts


this looks like a fun one if we know

this is six this is three and this is

four how do we solve for this y there's

something I really like called the

intersecting chord theorem if you take

the distance of the intersection to the

edge of the circle and call that a call

this distance D this distance n and this

distance y it will always be true that a

times n equals D times Y no matter how

much you move the chords around now we

could put a box around this and let's

move the label to the top so this is a

typical problem for one of these the six

three four and Y and to solve this it

would just be y times three equals six

times four which gives us 3y equals 24.

divide both sides by 3 and we get y

equals eight so this is the intersecting

chord theorem how exciting

Intersecting Chord Theorem

Such a fun theorem!


hey guys this looks like a fun one it

says four consecutive positive even

integers have a product of 13440 what

are the integers so it means something

like 2 * 4 * 6 * 8 or 4 * 6 * 8 * 10

there are some four consecutive even

integers that multiply to 13440 we could

just keep guessing and checking and

trying more and more of these but I want

to see if we can do it algebraically 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 call the first even integer X

so the next even integer will have to be

two more than that so it' be x + 2 and

then we do the same thing to get the

next two integers and we need these to

multiply to

13440 next if we multiply all these out

it's going to take us a while and it

will give us this right here and then we

can subtract 13440 from both sides and

we get all of this equal to zero this

can be solved but it'll take quite a bit

of work let me show you a shortcut

instead of having the four terms be

these let's have them be these right

here they're all still two apart from

each other so these are four consecutive

even integers they're just expressed a

little bit differently and if we

rearrange them like this then we have a

difference of two squares and another

difference of two squares so x + 1 * x

-1 is just x^2 - 1 and then x - 3 * x +

3 is just x^2 - 9 and then multiplying

these out is significantly easier than

multiplying this out x^2 * x^2 is x 4 X2

* 9 is 9x2 -1 * x^2 is -1 x^2 and-1 * 9

is POS 9 we can combine like terms this

9 x^2 - 1 x^2 will become -10 x^2 and

then we can subtract 13440 from both

sides and we end up with x 4 - 10 x^2 -

13431 = 0 this is significantly easier

than this we can factor out this left

hand side we need two numbers that add

to -10 and multiply to -3 431 this will

take a little bit of trial and error but

that ends up being -21 and POS 111 and

then x^2 - 121 is another difference of

two squares that gives us x - 11 * x +

11 and then we can just copy down the

rest of this stuff and now that we have

this factored we can solve for each of

these for x^2 + 11 = 0 we can subtract

11 from both sides that'll give us x^2 =

-11 and that has two imaginary Solutions

plus or minus IUN 11 after we plug this

in up here and subtract three it will

not give us an even integer so this is

not a valid solution and then for x + 11

we can subtract 11 from both sides and

we get x = -1 11 that won't give us a

positive even integer so that is also

not a valid solution and then for the

last one x - 11 if we add 11 to both

sides we get x = 11 and x = 11 will work

for us so let's copy these down and

let's plug in 11 into each of the x's

and that gives us 11 - 3 which is 8 11 -

1 which is 10 11 + 1 which is 12 and 11

+ 3 which is 14 and these are our four

consecutive positive even integers let's

put a box around them how exciting

4 Consecutive Even Integers

https://andymath.com/consecutive-integers/

I love these consecutive integer problems. Most of them are sums, this one has an interesting twist because it is a product.


you guys it says solve for x no

calculator so step one I'm going to

cross multiply so on the left hand side

I get x minus 2 times X plus 2 and then

5 times 13 is equal to 65. so we can

multiply out the left-hand side x times

x is x squared x times 2 is 2x negative

2 times x is negative 2X and negative 2

times 2 is negative 4. and it's still

equal to 65. this negative 2X and

positive 2x are going to cancel each

other out we're just left with x squared

minus 4 equals 65. so then we can add 4

to both sides these two cancel each

other out which gives us x squared

equals and 65 plus 4 is 69. nice square

root both sides and we get x equals a

plus or minus root 69 nice and that's

the answer to the question how exciting

Rational Equation Square Root Method

Collab with Geeklyhub!


highlights from the first semester of

Algebra 2 in two and a half minutes

let's start by looking at parent

functions and we can do Transformations

on these functions we can make vertical

changes by multiplying and adding

numbers to f of x and we can make

horizontal changes by multiplying or

adding numbers to X we can also reflect

over the x-axis or y-axis and now

parabolas parabolas are very good at

modeling things like objects being

thrown in the air there are four common

equations for parabolas and these are

called quadratic equations it can come

in vertex form with an a vertex form

with a 4p intercept form or standard

form and depending on which equation you

use you can find the vertex the axis of

symmetry the focus or the directrix now

we can also find the Y intercept by

setting x equal to zero or the X

intercepts by setting y equal to zero

let's talk more about finding the

x-intercepts so they're typically four

ways to solve a quadratic equation when

it's set equal to zero the square roots

method factoring completing the square

or quadratic formula in some situations

one method is best a lot of times you

have your choice a parabola may have two

one or zero X intercepts in the

situations where there are zero

x-intercepts we still have solutions to

the equation Asians but they're complex

Solutions using I or I squared is equal

to negative one or you could think of it

as I is equal to the square root of

negative one when solving these and a

negative shows up into the square root

you know the solution will be complex

polynomial function all of these are

polynomial functions all of these are

not polynomial functions polynomial

functions have terms with coefficients

and whole number exponents being added

together we can add subtract multiply

and divide polynomials the degree of a

polynomial is the same as the highest

exponent and the number of zeros or

roots or x-intercepts of a polynomial

function is the same as that degree if

we start doing fractional exponents we

call these radical functions whenever

you have a fractional exponent the

denominator is the index of the radical

and the numerator is still the exponent

so the square root of x could be written

as x to the one half the cube root of x

could be x to the one third and the

fifth root of x squared would be x to

the two-fifths up to now we've been

working with X to some exponent what if

we go crazy and put X in the exponent

this is called an exponential function

and this is where exponential growth or

Decay comes from so we can solve for x

in exponential equations too but it

requires an inverse process this is the

reason that we created logarithms the

whole goal of a logarithm is to get the

variable out of the exponent here's a

pretty solid way to think of logarithms

and for practical purposes logarithms

have a lot of properties using these

properties we can expand logarithms we

can condense logarithms we can solve

them in our head but ultimately all this

training for logarithms is meant to

prepare us for the main event anytime we

need to solve an equation where the

variable is in an exponent and this

concludes the highlights from Algebra 2

first semester how exciting

Very High Level Algebra 2 1st Semester


hey guys we have a castle here

surrounded by a moat and the moat has a

width of 20 meters all the way around

and you happen to have two boards that

are exactly 19 meters long is there a

way to use these two boards to get to

the castle so obviously we can't do it

this way because it's not going to be

long enough to reach but what if we put

it at a 45 degree angle like this and

then put the other board right here on

top of it will this be long enough to

reach well let's zoom in on it and do

some of the math first let's look at how

far is this total diagonal distance we

need to reach and let's call that

distance D this makes an isosceles right

triangle where this side is equal to 20

and this side is equal to 20. so we can

solve 20 squared plus 20 squared equals

D Squared that's 400 plus 400 equals D

Squared which is 800 equals D Squared

and after you square root 800 you get D

is approximately equal to 28.28 let's

keep this here but get rid of everything

else so this whole board is equal to 19

which would mean that the half board

would be half of 19 which is 9.5 this

also makes an isosceles right triangle

where this this side right here is also

9.5 and this is important so let's mark

this 9.5 as red and let's get rid of

this stuff and now let's bring these

boards back well this board is one of

those original 19 foot boards with these

two boards set up like this we can

Traverse a diagonal that is 19 plus 9.5

meters and 19 plus 9.5 is equal to 28.5

well the 28.5 we can cross with our two

boards is greater than the diagonal size

of 28.28 so this method checks out how

exciting

Moat Riddle Solution & Math

I thought this was a fun one! I hope it makes sense.


hey guys this looked like a fun one it's

a lot more work than I was expecting it

to be we're trying to find the area of

this orange triangle if you want to try

it on your own pause it right now cuz

I'm going to solve it in 3 2 1 we know

the three sides of this golden yellow

triangle I want to find the area of it

first and we can do that with Heron's

formula it uses the semi-perimeter which

is half of the perimeter and then you

just plug it into this formula right

here so let's do it for our Golden

Triangle for a b and c we're going to

plug in the sides 8 4 and 6 and 8+ 4 6 /

2 is equal to 9 so in the place of all

these s's I'm going to plug in 9 and

then we'll plug in 8 for the a four for

the B and 6 for the C and we can

simplify this we can bring down the 9

and all the parentheses 9 - 8 is equal

to 1 9 - 4 is equal to 5 9 - 6 is equal

to 3 so this will equal theare < TK of 9

* 15 and we can split this up into two

separate square roots > of 9 * > 15 and

theun of 9 is equal to 3 so the area of

this triangle is 3un 15 but the area of

a triangle is also equal to/ 12 base

time height where this is the height of

the triangle right here so this will

also equal 3un 15 b stands for the base

and the base of this triangle is equal

to 6 and then 1 12 * 6 is equal to 3 and

after we divide both sides by three we

get H is equal to theare < TK of 15 and

we can update this H to be square < TK

of 15 now let's bring back our orange

triangle I want to draw a line from this

vertex up here that's parallel to this

line let's make this line more bold and

let's hide this stuff for a second these

two angles are called alternate interior

angles and since these two lines are

parallel these angles are congruent to

each other and let's make this line bold

and extend it a little bit these two

angles are called vertical angles and

they are also congruent and then this

angle and this angle are called

corresponding angles since these are

parallel these corresponding angles are

going to be congruent to each other now

we can focus on this triangle right here

we have these two base angles congruent

which means this is an isoceles triangle

so this side will also have a length of

four and since this is four and the

entire side length is eight this piece

will also be four let's label the base

of our orange triangle X and now we can

use th's theorem says if you have a

triangle and a line cutting through that

triangle parallel to the base this ratio

will be true and if we put this on its

side it'll have the exact orientation of

our triangle here and we can set up our

proportion a is going to be equal to 6 B

is going to be equal to X and C and D

are both equal to 4 well the only way 6

overx is going to be equal to 4 over 4

is if x is equal to 6 so that means the

base of our orange triangle is going to

be equal to 6 and now we're ready to

find the area of our orange triangle

this height is the same as the height of

our orange triangle the area of our

triangle is 1/2 base time height where

the base is equal to 6 and the height is

equal to < TK of 15 1 12 of 6 is equal

to 3 and that's going to be multiplied

by the < TK of 15 and that's the answer

to our question we can put a box around

it so in this problem the area of this

orange triangle is equal to 3 < TK 15

square units and that is approximately

11.62 square units how exciting this was

kind of a long one if you actually

finished it comment a TV series you'd

recommend watching I need some

suggestions

Fun Geometry Challenge


hey guys this looks like a fun one this

is another Katrina a puzzle and it's day

two of our Advent calendar we're going

to solve 31 Catrina a puzzles in the

month of December one for each day and

this is December 2nd it tells us that we

have three squares and that the smallest

has an area of seven and it wants to

know what's the total shaded area 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 let's label the sides of the

larger Square X and let's label the

sides of this smaller Square y we can

now write a formula for the Shaded area

the Shaded area will be equal to the

large Square which is X2 minus this

Square which is y^ 2 minus this Square

which is 7 this is what we're trying to

solve for let's put a box around it so

to move forward I'm going to make one

assumption that isn't explicitly stated

and that is that this semi circle is

intersecting this square at these

Corners I don't know how I would solve

it without that assumption so I'm going

to make that assumption comment if you

have another way to solve it that

doesn't use that assumption using that

assumption I'm going to draw the line

from here to here this is the radius of

this semicircle the semicircle has a

diameter of X which means the radius

would be half of that or X over 2 and

then next let's draw this perpendicular

line this is the same length as the side

of this square and this square has an

area of 7 so the side length is square <

TK of 7 which means this has a length of

square < TK of 7 and last let's connect

this right here this distance is half

the side of this Square so it's going to

be y/ 2 now let's play around with this

right triangle as usual we're going to

use the Pythagorean theorem which is A2

+ B2 = c^2 in the place of the a we'll

plug in this side right here < TK 7 in

the place of the B we'll plug in this

side which is y over 2 and in the place

of the C we'll plug in this side which

is X

over2 for the first term the square root

and the square will cancel each other

out so we're left with seven for the

next term the exponent distributes to

both the top and the bottom giving us

y^2 over 2^2 and same thing over here

this changes into x^2 over 2^2 and then

let's simplify this 2^2 is equal to 4

2^2 is equal to 4 and let's give this 7

a common denominator 7 is going to be

the same thing as 284 and now that these

have the same denominator we can write

them as a single fraction 28 + y^2 over

4 next I don't really like fractions so

let's get rid of this denominator we can

multiply both sides of the equation by

four this four and this four will cancel

each other out and this four and this

four will cancel each other out so now

we have 28 + y^2 is equal to x^2 let's

subtract y^2 from both sides on the left

hand side these y^2 will cancel each

other out so we'll have 28 and that'll

be equal to x^2 - y^2 which is x^2 - y^2

and check this out we needed x^2 - y^2

in the place of this x^2 - y^2 we can

plug in 28 28 28 - 7 is equal to 21 and

there's no units in the original problem

so we'll give it a label of units

squared this is the answer to our

question it's already in a box how

exciting this was day two of our agvent

calendar this will be day three this is

the video that we'll be doing tomorrow

we want to know what is the area of this

shaded region divided by the area of the

entire circle how exciting

Aggvent Calendar Day 2

https://andymath.com/agg-vent-calendar

Here is the link to the aggvent calendar if you'd like to try some of them!


had some crazy polynomial function like

this and you need to know what is f of

2. traditionally the way you'd solve

this is you take all this and plug in 2

for the x's and then bring down all the

other stuff figure out what each of

these terms is equal to and then add or

subtract them all together and you get

three it would have taken a little bit

of time to figure it all out there's

another method you can do called

synthetic substitution you make a little

bar like this but the number you're

going to plug in outside here and then

put the coefficients including the

negative values right here so this is

how you set up the synthetic

substitution now to perform this

synthetic substitution all we're going

to do is add multiply add multiply add

multiply add multiply three plus zero is

three two times three is six negative

six plus six is zero two times zero is

zero negative one plus zero is negative

one two times negative 1 is negative two

three plus negative two is one two times

one is two negative four plus two is

negative two two times negative 2 is

negative four seven plus negative 4 is 3

and that is our answer and that's the

same answer we got up here this is

synthetic substitution for polynomials

it will always work and for large

polynomials it's usually easier than

this method how exciting

Synthetic Substitution Example #shorts

https://andymath.com/synthetic-substitution/

#shorts


hey guys I made a video about this the

other day and I kind of like the way

that I did it but several people

requested me to solve it a different way

where I repeatedly Square both sides of

the equation I thought it could be cool

to see what that looks like so let's do

it for the first step let's Square both

sides on this left hand side the square

and the square root will cancel each

other out and on the right hand side I

don't really want to do 32 squar but we

can change this 32 into 2 to 5th and now

to do 2 the 5th squar we just multiply

the 5 * the 2 to get 2 10 and next we

Square both sides of the equation on

this side the square will go both to the

numerator and to the denominator it's

going to look like this and in the

denominator the square and the square

root will cancel each other out and on

the right hand side we can multiply

these exponents to give us 2 to the 20

and then once again we're going to

square both sides of the equation this

squared can distribute to this term this

term and this term it'll look like this

and then x^2 s is X to 4th this guy's

good and on the very bottom this square

root and this Square will cancel each

other out and then on the right hand

side we're going to multiply these

exponents again to give us 2 to 40th and

next step we're going to square both

sides of the equation on the left hand

side this Square will distribute to all

four of these terms it's going to look

something like this this x 4th squar

will become X to 8th the x^2 SAR is

going to become X to 4th this will be X2

and on the very bottom the square root

and a square will cancel each other out

and on the right hand side this 40 will

multiply by the two to give us 2 to the

80th power and now let's clean up this

left hand side let's focus down here x^2

/ X will simplify to just X and then X

4th / X will simplify to X cubed and

then x 8 / X Cub we can just subtract 8

- 3 to give us X 5th and now we have X

5th = 2 8th we can take the fifth root

of both sides on the left hand side we

just have X and on the right hand side

we can do this 80 / 5 to give us 2 to

the 16 and 2 the 16th power is equal to

65,536 and that's the answer to our

question we can put a box around it if

you want to see the first video I have a

link in the description how exciting

Radical Math Problem Method 2

https://youtu.be/LAy7emYBsjg
Here is part 1.


start with Y it's clear and true and

divide this log by r s that's what we do

proceed with care for this thing is

layered next step we multiply both sides

by r s the best way to weather this

equational storm change it from

logarithmic to exponential form r s is r

and multiply M to left and right the

result is Merry Christmas and to all a

good night

Ho Ho Ho Exciting!

Christmas Math Poem!