okay in this video I just want to do

some more examples factoring the

difference of two squares so again I

look at this and I think well hey I've

got two terms 16x squared and my others

my 25 y squared there's a I'm

subtracting them so anytime I have a

negative between two terms that always

factors and again I'm basically just

taking the square root of the first term

so the square root of 16x squared again

what multiplied by itself would be 16x

squared we would just need a 4x and a 4x

and then I do the same thing for the 25

y squared

I would simply need a 5y and 5y and

again the formula says you stick a

positive and a negative in between the

terms and again that's now how it

factors so let's do our other example

here 64 x squared minus 81 Y squared so

here I'm just trying to make a little

barrier here just to separate them so

let's see so I'm going to make my two

sets of parentheses

let's see okay it's the same thing I did

I just take the square root of 64 x

squared I would get 8 X + 8 X I do the

same thing with 81 y squared I would get

9 y + 9 Y and again to stick a positive

and a negative in between them and you

now have it factored so again notice if

you do multiply it out we'll get 64 x

squared we'll get a negative 72 X Y but

we'll also get a positive 72 X Y which

cancels and then we get our negative 9y

squared just like we want

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Factoring the Difference of Two Squares - Ex 2


all right in this video i'm going to do

another example of finding real number

restrictions and in this problem we're

going to have another fraction so again

the values that are going to be that

would have to be omitted are any values

that give 0 in the denominator

so basically i'm going to take the

denominator set it equal to zero and

solve

so notice this is a quadratic equation i

don't think it factors too nicely

at least just looking at it off the top

of my head maybe we can play with it

here and maybe

maybe actually it will work out without

too much trouble

so to get the 2x squared i'm going to

need a 2x and an x

and then i need two numbers in this case

you've got to be a little tricky you

can't just say two numbers that multiply

to negative five and add up to negative

seven like i almost said that's not

correct in this case because of the

coefficient of the two but i know on the

outsides of my parentheses i am going to

need two numbers that multiply together

to give me a negative five

so let's see what if we put um

we could try putting you know and again

one of them has to be positive one of

them has to be negative i think you

could try with different possibilities

of positive and negative 5 and i'm

simply not seeing how it factors at all

so what i do in these cases is i simply

use the quadratic formula

so x equals negative b plus or minus the

square root of b squared

minus 4ac

all over 2a

and again in this case a is going to be

the value positive 2

b is going to be the value negative 7.

c is going to be the value negative 5

and then i just plug everything in here

so we have negative of negative 7

plus or minus the square root so we

would have b squared which is negative 7

squared

minus 4 times a which is 2

times c which is negative 5. let me

extend my square root over a little bit

more

and that's all again being divided by 2

times a which is also the value 2.

and now it's just a bunch of arithmetic

simplifying this down

so the negative of negative seven is

positive seven

underneath the square root negative

seven times negative seven is forty-nine

let's see negative four times two is

negative eight negative eight

times negative five is going to be

positive forty

and that's all divided by two times two

or four

and we can keep simplifying here a

little bit more we would have seven

plus or minus underneath the square root

we would have 89

and that's divided by four

and i'm pretty positive 89 is actually a

prime number that's at least not one

that i know off the top of my head as

having any factors

so our two restrictions in this case we

would have to omit the values our two

values we would have to omit

we'd have to omit seven plus the square

root of 89

over 4 and then we would also have to

omit 7 minus the square root of 89

over 4. again kind of not nice whole

numbers but that's okay

so those would be the values that we

would have to

we would not be allowed to plug into our

original expression

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Domain of an Expression Involving Fractions - Example 3


all right so another example here of

using our sum and difference identities

um so suppose we just want to Simply

evaluate cosine of 15° well we can write

15° as 45° minus 30° and now we can use

this uh bottom identity so it says if

you have cosine of something minus

something it says you take cosine of the

first thing which is

45° you multiply it by cosine of the

second thing which is

30° um if there's a negative sign in

between a minus sign in between though

it says you turn it into a plus when you

expand it

out and then you use sign of the first

uh quantity which is

45° and then we multiply it by S of the

second quantity which is

30° okay and now it's just again kind of

knowing uh some trig values so cosine of

45° is < tk2 over2 uh cosine of 30° is

going to be < tk3 over 2 s of 45° again

< tk2 / 2 s of 30° is going to be 12 so

okay it looks like we're getting Square

< TK of 6 over 4 + 2 over the square

root excuse me the square < TK of 2 over

4 then again we can write this as a

single fraction > of 6 Plus theun of two

over 4 and I think that's actually the

uh the same value that we figured out a

second ago right we said uh I think in

our first problem we figured out that s

of 75° was also this value so does it

make sense from a geometric point of

view so you know just a little side

question does it make sense that s of 75

Dees would in fact be the same as cosine

of 15° so maybe think if that seems

reasonable to you but definitely we've

used this identity um in the correct

manner this certainly is the correct

answer

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Identities for Sum and Differences of Sine and Cosine, Example 2.  In this video, I show how we can find other values of sine and cosine (besides the commonly memorized values) using sum and difference identities.  For more free math videos, visit http://PatrickJMT.com


okay so here we're going to do a little

word problem related to the law of

cosiness see if we can squeeze it all in

here okay so the idea is we've got two

people um they're on the same side of

the river and they're 400 meters apart

uh the person on the left side is 700 m

from a house that's across the river so

there's the little house uh that

person's 700 M away and the angle formed

by the people in the house uh is 150

Dees when we use the person on the left

as the vertex we just want to know how

far the person from the right is from

the

house well okay this is really a pretty

straightforward

application of the law of

cosiness okay so this is

700 uh we know that this is

150° this is 400 again we can call this

l may be C and now we'll just use our

law of cosin so it says c^2

= 700 2ar +

2ar - 2 *

700 time

400 times cosine of the angle but we're

given that that's just 150

degrees so let's see here so 700 s

that'll be 400

90,000 uh let's see so 400

squared that would be

160,000 let's see so 2 *

700 * 400 that's going to be

560,000 let's see cosine of 15° I think

we can probably uh we can probably

figure that one out um so let's see 50°

that would put us in quadrant uh in

quadrant 2 uh the reference angle there

would be

30° so let's see if I think about cosine

of 30° cosine of 30° would be < tk3

over2 but since we're in the second

quadrant uh cosine there would be

negative so we'll get netive < tk3 over2

for cosine of

150 um so let's see here so we've got

490 uh 490,000 and 160,000 that's just

going to be

two negatives will make a positive let's

see uh 560,000 / 2 that would be

280,000 again we would have to multiply

that still by the square root of

three so let's see two let's see so uh

let's do the square of 3 S of 3 is

roughly 1.73 205 I'll multiply that by 2

80 add that to

650,000 I'm getting this to be 1 1 3 4 9

226 well the last thing we'll have to do

is just take the square root of this

number if we take the square root of all

this I'm getting this to be

1065 35 after rounding I'm going to

round that down even a little bit more

and just say it's

1,65 I believe our original units were

meters so the second person is going to

be

1,65 M away from the house roughly

Law of Cosines, Example 6

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Law of Cosines, Example 6.  In this video, we find how far a person is from a house by knowing that person's distance to someone else and the angle formed between them and the house.


all right in this video I want to take a

couple of equations and just solve for

some specified variable so in the first

example we've got that a = 12 * H * the

quantity B1 + B2 and we want to solve

that for B1 so what that means to me is

eventually we want to do a whole bunch

of algebra and get uh B1 all by

itself so there's certainly a couple

different ways we can go about doing

this um the first thing I notice is that

there's a 1/2 in here and typ typically

if I can get rid of fractions that's the

first thing I like to do so you know

multiplying by 1/2 to me is the same

thing as dividing by two so I think well

if you're dividing by two what undoes

multiplication or excuse me what undoes

division well multiplication undoes

division so since we're dividing by two

to get rid of the 1/2 I could multiply

by two but if I do it to the right side

I have to do it to the left side side as

well so on the left side we're left with

2 a the 2 * a half is just one so we can

basically just get rid of that and then

we have H * B1 +

B2 and now I want to get um there's

again a couple different ways I could do

it since I'm multiplying by H I think

well I would like to get rid of that so

again multiplying by H I'll divide both

sides by

h on the right side H divided by H is

just one on the left side there's really

nothing to simplify down so we still

have the 2 a over H you can think about

in the numerator there being a positive

one in front of the B1 + B2 if you

distribute that we'll be left with B1 +

B2 and now to Simply get B1 all by

itself since I'm adding B2 I would

subtract B2 from both

sides well on the left side they're not

really like terms so there's not a lot

we can do but we do have we could simply

just rewrite it as 2 a h minus B2 and

now we have solved for B1 it's all by

itself

so we are

finished um in the second equation we

have to solve for our variable Z so

we've got 3 Z + x - y = 2x +

4 and since there's only a single Z

floating around on the left side I'm

going to just leave that by itself so

I've really got to do two things um I've

got to get rid of the plus X and the

minus y so I can subtract X from both

sides to get rid of the X on the left

side we would have 3 Z the minus y is

still there well we would have 2x - 1X

which would just leave us with a

positive 1X and then plus

4 and again to get rid of the minus y we

can simply add add y to both

sides so on the left side we have 3

z = x + y +

4 um typically I like to stick my

variables first so I just kind of

arbitrarily put the Y after the X and

again the last thing we want to do now

is get the Z all by itself so since

we're multiplying by three we can divide

by 3 um or equivalently multiply by 1/3

so you have to divide everything on the

right side by three as well so on the

left side we just have one Z and we

could either write the right side as x +

y + 4 over 3 or equivalently we could

rewrite that as 1/3 * the quantity x + y

4 and now we've got our

solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving an Equation for a Specified Variable


okay in this video we're going to do an

example again related to vector addition

and scalar multiplication we're also

going to talk about sketching some

vectors the uh sum of

vectors so we've got vector v that has

components 1 3 2g5 we're going to sketch

the vector v+ W and we'll also find the

components of that

Vector so when you sketch uh the

addition of vectors um so the first

thing I'm going to do is sketch vector v

so it has components of one and three so

just from the origin I'm going to move

one to the right and up one to

three so that's going to be my vector

v now Vector W when we when we sketch

vectors um wherever the first one stops

I just stick the new one onto that so

Vector W is going to start right here

but it says basically you move two units

to the uh to the right and then we go

five units down so I think we were

already up here uh up here uh you can

think about this as being a y value of

three if we go down five units we'll be

down here at

-2 so if we go two units to the um to

the right and then five units down so 1

two 3 four five this would be my

Vector W so again there was vector v

there was Vector W now what we do is um

wherever our original Vector started um

and wherever it stops we connect

those um and that's going to be our

vector v + W okay so it starts uh at the

origin and in this case it's kind of

going a it looks like three units to the

right and then two units down is what it

looks like to me so um so to me I'm

thinking well just based on the graph

shouldn't the components be three comma

-2 uh according to our little

diagram well let's actually find the

components okay so vector v is 1 and

three uh plus W which is 2

and5 again we just add respective

components so 1 + 2 that's going to give

us 3 3 + -5 is going to give

us2 and hey so it says if you add them

the new vectors will have components 3

comma -2 two and that's in fact you know

that's what happens from our from our

graph here so let's do one more here U

got a little little runoff um we're

going to sketch vector 2v and also find

its components so um so vector v again

we said that had components 1 comma 3 so

I'll sketch it down here 1 comma

3 when you multiply a vector by a

positive uh a positive whole number

all it does is it stretches It Out by

that factor so since we're multiplying

by two you basically can stick another

vector v at the end that would have

components again you would move over one

and up three it's just going to stretch

that original Vector out by a factor of

two so again just kind of eyeballing

this new Vector again which we could

call two uh

V it looks like to me it has components

it looks like to me you're moving over

two and up six so I'm thinking well this

should have components of 2 comma 6 and

again if we just use our our rule that

says you just multiply it says you would

have two * vector v which is 1 and three

and if you distribute the two we will

get two and six for the components of

our new vector

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Vector Addition and Scalar Multiplication, Example 2.  In this video I add two vectors in component form and also sketch the vectors to illustrate how to add vectors graphically (very useful stuff!).


all right in this video we're going to

sketch the graph of a polinomial

function and here we're going to sketch

the graph of a polinomial that satisfies

all these following conditions so F of 3

= 2 frime of 1al 0 frime of -1 = -2 uh

it says the first derivative is positive

for all X greater than one the first

derivative is going to be negative for

all X less than one and the second

derivative is going to be positive for

all

X so usually when I sketch these I try

to kind of summarize everything that'll

sort of you know make more sense to me a

little bit so the first thing since F of

3 equals 2 that means that the point 3

comma 2 is on the graph and I usually go

ahead and put those dots on there just

because I know you know this points have

to be there the next thing I'm going to

think about is actually the first

derivative stuff so the first derivative

we said for all x coordinates greater

than one so for all x coordinates

greater than one the derivative is

positive

and that tells us the original function

is increasing and then it said for all x

coordinates less than one it was

negative so that means the function is

decreasing the second derivative was

greater than zero for all X that was our

last condition so that means we know

it's concave up

everywhere and already you know based on

that I'm thinking hey you know maybe a

graph kind of like a parabola would

would work here and I think it almost

will um you know so at xal 1 the slope

of the tangent line line is going to be

zero so since frime of 1 equals 0 so at

xal 1 I'm going to kind of make it you

know sort of bottom out here a little

bit so there at the x coordinate of one

there's the slope of our tangent line

equaling zero um it'll eventually we'll

have to go through that point uh 3 comma

2 and also at negative 1 we said the

slope is -2 well you can kind of imagine

you know having something that's sort of

parabolic in shape and certainly the

slope of that tangent line potentially

could equal -2 again the slope at the

bottom was a a slope of zero and I think

now we've actually got a uh a picture

that would actually agree with all this

so it goes through 3 comma 2 frime of

one that has a tangent line of

zero frime of1 equals -2 so it says at

negative 1 the slope of the tangent line

should be

-2 um we've got it increasing for all x

coordinates greater than one and we've

got it decreasing ining for all x

coordinates less than one and last but

not least it's concave up everywhere so

I think this would be a function f of x

that does satisfy all of those

conditions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Sketching a Graph Given Conditions About Derivative Requirements.  In this video, we have a few conditions about points a graph must go through, as well as conditions on the first and second derivative.  I sketch a graph that satisfies the given conditions!


all right in this video I want to do an

example of multiplying and dividing

rational expressions so initially here

we've got a little bit of division of

two fractions containing some binomial

expressions and the basic idea in this

case is to factor and cancel is really

all it comes down to so the first thing

we'll have to do here is remember if you

have a fraction divided by a fraction we

leave the first one alone

and remember we turn our division into

multiplication and then we have to flip

the other fraction so division turns

into multiplication and the one on the

right side flip flops and at this point

again it's just trying to factor things

out so I'm going to make a couple little

observations here notice there's an 8

minus K on top and there's a K minus 8

on the bottom we can't we can't cancel

those out immediately I like to stick

things in parenthesis as well just as a

personal habit doesn't change its value

again because you can imagine there's a

1 out here

notice the signs here are opposite of

what they are down here so whatever I

recognize that the idea is you can

basically just factor a negative out of

one of them so notice if I pull the

negative out I have to change the signs

to make it equivalent I can make it

negative 8 plus K because when I

distribute the negative sign the

negative and the negative will make

positive 8 and the negative times the

positive K will give me negative K so

this is kind of a useful trick factoring

out a negative okay then we have K

squared minus 64

I recognize that as being a difference

of perfect squares so that factors is K

minus 8 K plus 8 ok there's really

nothing to do with the other side k plus

k- 8 but now since we have all

multiplication in the numerator all

multiplication in the denominator we can

cancel away so again notice in this

first set of parentheses we could really

rewrite that as k+ k -8 and we have a

positive K minus 8 down here so in this

case we can cancel those out so that

cancels with that and then we have a K

plus 8 in the numerator we have a K plus

8 in the denominator so we can cancel

those out as well and I think we have

now simplified down our original our

original problem here so in the

numerator we have a negative you can

think about that as just being negative

1 and then divided by K minus 8 again

you can put this in parenthesis or not

you know algebraically it's not going to

change its value typically though I

would certainly just write it like this

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !!
Rational Expressions:  Multiplying and Dividing.  Ex 1.
In this video, I divide two rational expressions containing binomials.


okay here we're going to look at an

example involving work and rates to get

a job done so we've got latisha and

Ricky they work for a computer software

company together they can write a

program in 19 hours latisha can do the

program by yourself in 32 hours we want

to know how long it would take Ricky to

write the program all by himself the

equation that we can use here if we look

at the rate per hour of one of the

people if we multiply that by the total

time to get the job done

plus we look at the rate per hour of the

other person multiplied by the total

time that equals one so maybe the the

first person here will let that

represent latisha so by yourself latisha

can write the program in thirty two

hours so that means her rate per hour

she could do well 1 over 32 we said the

total time for them working together to

get the job done the total time was 19

hours we don't know the rate of the

second person of Ricky pull so we'll

call that 1 over X again X will

represent the number of hours it would

take for him to do it alone so just like

it took latisha 32 hours her rate was 1

over 32 his rate will be 1 over X again

we multiply that by the total time which

is 19 and that equals 1 so now we've got

our equation to solve

we have 19 over 32 plus 19 over x equals

1 so now what I'm going to do is just

subtract 19 over 32 from both sides okay

so we would have 19 over X left on the

left okay we've got fractions we can

make 1 into 32 over 32 so we've got 32

over 32 minus 19 over 32 that's going to

leave us with 13 over 32 and now what

we're going to do is cross multiply so

I'm going to take

and multiply that by 32 so 19 times 32

that gives us 608 I'm going to multiply

X times 13 which will give us 13 X and

now simply let's divide both sides by 13

we'll be left with X on the right side

and 608 divided by 13 I'm getting this

to be roughly forty six point seven

seven so it says it's going to take

Ricky about forty six point seven seven

hours if he were to do this task by

himself

Rates and Time Required To Complete a Task

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Linear Equation , Word Problem #7 :  Work, Rates, Time To Complete a Task.
  Here we are given that a person can complete a task alone in 32 hours and with another person they can finish the task in 19 hours.  We want to know how long it would take the second person working alone.


okay here we're going to look at solving

some more equations uh in this case

containing some integers and it'll just

be a little uh variety of uh problems uh

you know you may have to add you may

have to subtract you may have to

multiply to solve but uh we'll go

through all three of these uh in this

exam in this video so in part A we have

x - 7 = 10 well to get the X by itself

since we're subtracting seven We'll add

seven to both sides we'll be left with X

on the left side 10 + 7 is

17 and that'll be our solution to part

A okay so for Part B here we have 45 = x

- -3 well we can certainly rewrite

this we have a negative multiplied a

negative that's equivalent to a positive

so we could write x - -3 is just x + 3

and again if I want to solve for x now

since I'm adding three to both sides

what I'm going to do is just subtract

three from both sides so 45 minus 3 is

42 we're left with the X on the right

side and we have our solution to Part B

uh our solution will be the value 42

let's go ahead and look at uh part C

here as

well so we have p over 8 =

-6 I'm going to kind of do this in two

steps you could certainly uh you know do

everything here here in just a one step

but what I'm going to do is first off to

get rid of the negative on the left side

is I'm going to multiply both sides by

negative one

so a negative multiplied by a negative

is a positive so that's going to leave

us with p over 8 on the left side and -6

multiplied -1 it's going to leave us

with positive 6 and now the last thing

we're going to do

here is simply multiply both sides by

eight since we're dividing by eight on

the left side we'll multiply by eight so

the eights will cancel out we be left

with 1 p on the left side 6 * 8 is 48

and that's going to give us our solution

to uh the third equation here so our

solution will be the value P

equals 48

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Basic Equations Containing Integers.  Here we look at solving some very basic linear equations containing integers.


all right here we're going to do another

example using the direct comparison test

so here we're going to look at the

series from one to

Infinity from 1 to Infinity of sin^2 n

over 5 raised to the N power so you know

if uh if you have no feel for kind of

what's going on with a series like this

you know a lot of times I'll ask people

you know sort of what part is throwing

you off what part feels the most

confusing um I think as n increases you

know the denominator you're going to get

power of five that's pretty clear uh the

numerator well you know as n increases

you're going to have s squar of all

these different values and well what on

Earth does that really you know what

happens in that case um you know and I

tell people a lot of times just get rid

of the stuff that's throwing you off if

the S squar wasn't in the numerator we

would have 1 over 5 to the n and that's

a convergent geometric Series so I would

think maybe somehow I can somehow relate

this to a convergent geometric series

again metrick certainly uh doesn't

always work but at least it gives me um

something to think about sort of a

direction to start uh to start

going well kind of the trick for this

one is we're actually going to use you

know the direct comparison and really

kind of the squeeze serem as well you

know sign of uh sign of anything is

between positive one and negative one

well sin squar n that's what we have so

I'm going to write it as s of n quantity

squared

what two values is that in between well

don't just mechanically Square both

sides and say one and one um that's

certainly not valid if you take a number

between negative 1 and positive one and

you square it you'll actually just get

values well between zero and positive

one okay so sin squ of n is always going

to be a number between zero and one well

again we're dividing by five to the n

so 0 over 5 n would be less than or

equal to sin n over 5 n Which is less

than or equal to well 1 over 5 to the N

well on the left obviously we just have

zero um so the idea is uh the series

from nals 1 to Infinity of zero that

would be less than or equal to the

series from n = 1 to Infinity of sin^2 n

over 5 n Which is less than or equal to

the series from n = 1 to Infinity of 1 5

the N but again this is just a geometric

uh a geometric

series I guess even more important uh is

that it's a convergent geometric

Series right because here our r value uh

is going to be positive 1/5 and if our R

values between negative 1 and positive 1

uh that's I should say exclusive uh so

if if the r values less than one and

greater than negative one that's when we

have a convergent series well which we

certainly do well since sin s of n over

5 to the N that series it's always

bigger than zero greater than or equal

to zero it's always less than or equal

to this convergent geometric Series so

by the uh by the direct comparison test

our series that we're uh interested in

that must also converge

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Direct Comparison Test - Another Example 4. In this video I show that another series converges or diverges using the direct comparison theorem. I do not run over the formula/theory in this video (but do in another video, so look around if that is what you need!).


okay in this example we just want to

make a a connection between interval

notation and compound inequalities and

the question

is when we write in parentheses -2 comma

to 8 with a

bracket does that mean the same as all

the values of X that are greater than -2

but less than or equal to 8 and you know

sometimes what I do is I like to make a

little number line so here is -2 um and

since it's a parenthesis remember that

means we don't include it so I'm going

to put a parentheses or sometimes people

will put an open

circle and then it says basically to I I

think it it says use every number every

number every number up to and including

eight that's what the little brackets

mean include it and I always thought

about the brackets as being like little

hooks so they catch that number and

bring it along with us uh the

parentheses are kind of rounded so the

number gets away so yeah yes to me this

interval notation from -2 with

parentheses uh up to 8 with a bracket

that does mean use every value of x

that's strictly greater than -2 so X has

to be greater than -2 and X also has to

be less than or equal to 8 so I would

say yes that's definitely a true

statement

Interval Notation - A basic question!

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! A basic true/false question to make sure you understand interval notation!


okay in this example we have the

quadratic equation x^2 - 3x = 4 and

again it's quadratic because the highest

power that we see on X is a power of two

we want to know which of the following

is a first step that you could take to

solve for

x okay so part A says add four to both

sides of the equation B says subtract

four from both sides of the equation C

says factor to the left side is x * the

quantity x - 3 Part D says combine the

terms on the left side of the equation

so the one thing that you really need to

remember um you know if you run across a

quadratic equation you you recognize oh

that's quadratic the first thing you

want to do before you start solving it

is you want to try to make one side

equal to zero so that's that's kind of

the the thing that I'm going to keep in

mind here is I want one side of this

equation to be

zero well part A says let's add four to

both

sides well if I add four to both sides

I've got x^2 - 3x + 4 on the right side

I would have 4 + 4 which is 8 well that

doesn't really do anything to me useful

um right it doesn't help me get make one

side equal to zero all I've really done

is I've got an extra term now on the

left side this positive4 that wasn't

there before and so so that to me

doesn't really seem to be doing anything

productive so part A doesn't seem to

make sense Part B says let's subtract

four from both sides and I would think

aha you know that makes more sense again

if we're wanting to make one side of

this equation equal to zero that seems

like the reasonable thing to do in this

case so on the left side we would have

x^2 - 3x - 4 mean there's no like terms

so we can't combine anything on the

right side we would have 4 Min - 4 which

would give us

zero and now we've we've kind of done

this first step of making one side of

the the equation equal to zero so Part B

certainly looks like it's going to be

the correct answer let's look at the

other two suggestions and we'll see what

happens so part C said Factor the left

side is X multiplied xus

3 okay while it's technically correct

that you could do

that it is correct to factor the left

side is x * x -

3 again that's not doing anything in

terms of making the one side of the

equation equal to zero so this

factorization while it's correct on the

left side it's not going to help you

arrive at a solution at

all so that's also not correct let's

look at d d says combine the terms on

the left side of the

equation well you can't really combine

these terms these are not like

terms

you have one term involving X SAR and

one term involving just x to the first

power so those are not like terms which

means they

can't be

combined so a b and c all you know all

gave you things that were legal to do

it's correct part D is actually

suggesting you know is saying to do

something which isn't even algebraically

correct so all right so again it looks

like uh B was our correct Choice

subtracting four from both sides of the

equation

GED 'Most Missed' Math Practice Problems , Code A.4.a,  #1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The GED Testing Service keeps a list of the most missed math topics and can be seen here: http://www.gedtestingservice.com/uploads/files/126de0283c94ff6323b3b5b3f310da5d.pdf
This video shows how to work a sample problem using their coding system that is shown on the left using the link above.


all right in this video we're going to

look at some examples related to using

the squeeze Andor absolute value theorem

so in part A here we've got we're going

to find the limit of the sequence 1/ n

Cub Time s of n^

2 so again what we're trying to do is

we're trying to take the limit as n goes

to

Infinity of this function kind of the

problem is though you know as n goes to

Infinity n squ goes to Infinity but as

uh you know the inside uh of s gets

bigger and bigger and bigger S of that

that limit simply does not exist but

that doesn't mean that we can simply say

well this limit doesn't exist and again

what we can use in this cases is the

squeeze

theorem

so s of n s you

know sign of anything uh is always

between positive one and negative one

okay so uh we're going to try to kind of

find a function that that that's smaller

than 1/ n Cub s of n^2 and a function

that's bigger than 1 over n Cub s of n^2

and again we're going to try to show

that those limits um have the same value

and therefore the limit uh that we're

looking for will also have the same

value so this is kind of the first step

I think you know people will agree with

that okay sign of whatever is between

negative 1 and one well again 10 of the

function I want though I don't want just

s of n^ 2 I want s of n^ 2 * 1 n 3 you

know since in some positive number if we

multiply by one over n the

3 so all I'm doing is I'm

multiplying the uh Inside by 1 over n

the 3 well if I multiply the left side

by 1 over n the

3 in the right side by 1 over n the 3r

I'll now kind of get the small small

function and this big function so 1 n

Cub of n^2 is always going to be greater

than or equal to1 n cubed but it's

always smaller than a positive 1 over n

cubed well likewise if we take the limit

as in goes to

Infinity of all of

these so now here we're using the

squeeze

theorem so now we're using the squeeze

theorem and well the limit as n goes to

Infinity of1 Over N cubed that's simply

going to equal 0o likewise the limit as

n goes to Infinity of positive 1 over n

cubed is going to equal zero well we've

now shown that the limit as n goes to

Infinity of 1/ n cubed s of n^ 2 well

that's less than or equal to zero it's

greater than or equal to zero so that

implies that our limit uh our original

limit must also equal zero again using

the squeeze theorem

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Squeeze Theorem and Absolute Value Theorem, #1.  Here we look at finding the limits of some sequences by using the squeeze and / or absolute value theorems.


all right in this video we're going to

look at an example related to the

derivative of an inverse function um so

here we've got the uh you know the

formula associated with finding the

derivative of an

inverse so in this one we have a

function X to the3 plus e x and we're

given that F inverse of 1al 0 we want to

find the derivative of the inverse

function at xal 1 you know one thing we

can make sure of is that this function

does have an inverse

so notice that the derivative frime of x

that would equal

3x^2 plus e to X but um notice that this

is always going to be uh greater than

zero for all X and the reason is e to

the x is greater than zero for all X if

you plug any number uh into the E

function uh the exponential function e

to the X you always get something

greater than zero you know 3x^2 is

certainly greater than or equal to

zero and when you add them together well

since this one's already positive if you

add something zero or bigger we'll

always get something greater than

zero and that tells us since our

derivative is always positive it tells

us our function uh is strictly

increasing which means it

passes the horizontal line test and that

tells us the function uh F does have an

inverse

okay so this question is you know

phrased uh properly it does in fact uh

we can ask about the derivative of the

inverse so now using our formula here it

just says well if we want to figure out

the derivative with respect to X of the

uh inverse function

evaluated um at xal 1 we'll just

use our formula

here okay okay so again we know that F

inverse of

one that's given to us as zero so really

all we're doing is we're just figuring

out 1 over F Prime of 0 well all we have

to do now is just take the derivative of

the actual

function which we've already done here

and then all we have to do is simply

plug zero into that so notice F Prime of

0 is just going to give us 3 * 0 S Plus

e the 0 which is going to give us one so

that says uh when we plug it into uh

this part we'll just be left with one

over one or we'll get one as the value

of the derivative um at that

point

Derivative of an Inverse Function , Ex 1

In this video I work a quick example of where we need to evaluate the derivative of an inverse function but we are only given the original function, not the inverse.
This is a typical topic in a calculus course.  In order to evaluate the derivative at a point we have a clever work around that uses the derivative of the original function, evaluated at the inverse.

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okay um in this video I want to talk

about finding the radius of of

convergence of a power series and I've

had a lot of people ask about this and

it's actually easy after you found the

interval of convergence so to find the

radius of convergence you first

basically have to find the interval of

convergence then you simply find the

length of that interval and divide by

two so um I've got a bunch of intervals

of convergence on the left side and the

associated radius of convergence on on

the right

side if your interval of convergence is

all real numbers negative Infinity to

positive Infinity we simply say that the

radius is equal to

Infinity if it only converges for a

single point so a trivial solution um

then we simply say that the radius is

equal to

zero

um now uh it doesn't matter whether if

you have um an interval say like in this

example negative 1 to 7 it doesn't

matter whether the end points of the

interval produce a convergent series or

not um and all of these cases um kind of

all four possible cases of the end

points either converging or not

converging again all you do is find the

length of the interval take the right

end point minus the left end point and

divide by two so if we go from- 1 to 7

hey that has a length of eight divide

div it by two and we would simply say

the radius of convergence is four so um

very easy to do um but obviously again

you have to know the interval of

convergence and that's the hard part so

all right um uh kind of short and sweet

I hope it makes sense if you have

questions let me know

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Radius of Convergence for a Power Series - In this video, I discuss how to find the radius of converge.  It is easy after you find the INTERVAL of convergence.
I get tons of questions about it though, so I thought it was time for a video. : )
Visit http://PatrickJMT.com for tons more free math videos.


all right in this video I just want to

do a quick little example about the

geometric mean and also to point out you

know what kind of how you can think

about it what it means um so the

definition the geometric mean of two

numbers A and B uh what we do is we

multiply those two numbers together and

then we take the square root of them and

that's the geometric mean so certainly

not the same as the arithmetic mean the

average I think that's the one most of

us are certainly more familiar with so

let's look at a simple example here

let's find the GE metric mean of four

and N so again what we'll do is we'll

multiply those two numbers

together and then we'll simply take the

square root of uh of those so 4 * 9 well

that's going to give us 36 so we have to

take the square root of 36 well that's

simply equal to six so we would say that

the geometric mean of the numbers four

and nine uh the geometric mean is going

to equal six so you know what what is

what's special about the geometric mean

I guess you know uh again I think an

average an arithmetic mean it's kind of

the middle point right and I think uh

you know people you know you find an

arithmetic mean if you want to find for

example your average grade on two tests

so you know what does this number six

represent in relation to four and nine

well what it has to do with it has to do

with

ratios okay so we started with four um

and n those were the numbers that we

were finding the arith excuse me the

geometric mean of and we found that

geometric mean to equal

six well notice if we look at the ratio

4 / 6 and then if we look at the ratio 6

/ 9 well 4 over 6 red reduces to 2/3 hey

6 over9 also reduces to 2/3 so this is

kind of what's special about this number

the geometric mean um if you take the

ratio of the small number to the

geometric mean and then if you look at

the ratio of the geometric mean to the

larger number those ratios or those

fractions are going to be the same okay

so that's that's that's how you can kind

of think about the geometric mean

somehow it's it's giving equal ratios of

those two values so again the arithmetic

mean is the one that I think most of us

a little are a little bit more used to

so again just to refresh you the

arithmetic mean

we would just add those two numbers

together you know 4 + 9 and then divide

by two I guess 4 + 9 that's uh 13 over

two and 13 ided two I believe that's

going to give us what uh

6.5 okay so the arithmetic mean is

6.5 the geometric mean was six and this

leads to one other uh last little remark

uh without a proof but it turns out that

the geom metric mean is always less than

or equal to the arithmetic mean okay so

one other little useful result so again

if you want to find your average grade

on two tests you're not going to use the

geometric mean you're going to use the

more familiar arithmetic mean um but the

geometric mean certainly does show up

and um again I think it's something that

people most people aren't aren't

familiar with I certainly wasn't for a

long time um so just in case you see it

that's all there is to it um and that's

it again it just really has to do with

making ratios equal to each other

The Geometric Mean

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Geometric Mean.  In this video, I show how to find the geometric mean of two numbers and show what that number means in relation to the numbers that you started with.


all right so one more example here using

sum and difference identities to

evaluate a trig function and here we're

going to find the value for cosine of PI

over 12 and again kind of the tricky

part is how to break this up I think

let's see one way we can break this up

you know and again to me to get PI over

12 that's kind of a you know a small

number I need a couple angles that are

close together let's see four PI over 12

minus three PI over 12 4 PI over 12

minus three PI over 12 that would give

us 1 PI over 12 and well let's see

cosine of 4 PI over 12 that's really the

same thing as PI over 3 3 PI over 12 we

can reduce that to PI over 4 and I know

how to evaluate you know cosine and sine

at the angle PI over 3 and PI over 4 so

this seems like a good a good way to

write it I'm going to use this bottom

identity now it says you have cosine of

something minus something so it says I

can write that as cosine of the first

thing PI over 3 times cosine of the

second one which is PI over 4 again you

just kind of forget all about the plus

of the minus and there excuse me you

forget about the positive a negative

when you're just plugging it in ok so

you just use positive PI over 3 and

positive PI over 4 let me clarify that

and again there's a negative in between

so ok here we are thinking about the

negative but it says if there's a

negative in between we use the positive

so I've got to put plus and then it says

just take sine of the first one sine of

the second one so we'll get sine of PI

over 3

times sine of pi over four and again now

it's just unit circle stuff let's see

cosine of PI over three that's going to

be I believe positive 1/2 because sine

of PI over four is going to be root 2

over 2 sine of PI over 3 that's going to

be root 3 over 2 sine of PI over 4 is

going to be simply root 2 over 2 so it

looks like we get square root of 2 plus

root 3 times root 2 is going to be root

6 again we have our denominator 2 times

2 which is 4 so again we just simply

used this difference identity for cosine

to evaluate you know a trig function to

evaluate cosine of PI over 12 again

that's one I certainly don't have

memorized off the top of my head but I

can deduce it without too much effort so

long as I remember that identity

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Sum and Difference Identities for Sine and Cosine, More Examples #2. More examples of using the sum and difference identities to find value other trig values. For more free math videos, visit http://PatrickJMT.com


so a summer camp counselor wants to find

a length x in feet across a lake as

given in the diagram and the links

AB EB BD and CD measure 1,800 ft 1400 ft

700 ft and 800 ft

respectively um segments a c and

segments D intersect at this point B and

we're given that the angle a e which is

this angle here and cdb which would be

this angle here have the same measure we

want to figure out what's the value of x

okay so I've got my my diagram here and

let me go ahead and fill in the the

values

so okay so we're told that AB measures

1,800 ft we're told that EB measures

1,400 ft BD is 700 ft so that's

700 and then we also

also are given that CD is 800 ft so

again we're told that the angle AE which

is this angle here and cdb which would

be this angle here have the same measure

now notice this angle here the angle

e and CBD those are going to be vertical

angles so those also have similar

measures so the idea is now since um two

of the Interior since two of the angles

are congruent we know that these angles

or these triangles are similar so we can

set up a proportion to solve for the

missing value so we can say that

CD ided X so again I'm kind of looking

at this angle the side opposite that's

CD and for this triangle I'm looking at

the same angle but the opposite side

that's X that's going to equal BD so BD

is the um the side length opposite this

angle that we didn't label

and we can divide that by EB which will

be again if we look at the other

triangle if we look at the angle that we

didn't label if we look at the side

opposite that'll be EB and now we can

simply fill in those values so we said

CD that was equal to

800 / X BD has a length of

700 and E we said had a length of 1400

so now all we have to do is simply solve

this for x and and we'll have our

solution

so let me take uh I guess just one or

two extra steps here for the

algebra one thing we can do is we can

simply flip both fractions so we can

write this as X over 800 =

1400 ID 700 and then to solve for x we

can simply multiply both sides by 800 so

we'll have 1400 over 700 ultip by 800

well 1400 / 700 we can cancel out the

zeros 14 / 7 is 2 so we have 2

multiplied by

800 and that's going to give us a length

of 1,600

ft so again the IDE in this one is just

recognizing that these two triangles are

in fact similar and then setting up the

corre correct proportions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 17, 2016 Edition and Beyond.
Section 1 - No Calculator allowed!


all right in this video I'm going to

look at some examples involving

multiplication of

decimals so in part A here we've got

82.4 5 * 16 we basically just do sort of

our normal long

division and then I just count the

number of digits that are to the right

of my decimal places and that'll tell me

how far that basically helps me find

where the decimal place goes at the end

so we'll talk about that in a second so

let's do the multiplication so 6 * 5 is

30 so we'll keep the zero carry the

three 6 * 4 is 24 + 3 will be 27 so

we'll keep the 7 carry the two 6 * 2 is

12 + 2 is 14 so we'll keep the four

carry the one 6 * 8 is 48 + 1 will be

49 okay then I'm going to drop a zero

down and then I'm just going to multiply

everything by one so we'll just get uh 1

* 5 1 * 4 1 * 2 1 * 8 I do my addition 0

and 0 is 0 7 + 5 is 12 so we'll keep the

two carry the one uh 1 + 4 + 4 will be 9

9 + 2 is

11 so let's see uh 1 + 4 is 5 + 8 will

be 13 and now it's just a matter of

finding where the decimal place goes I

basically look to the right of my uh my

decimal sign in each number and I count

the number of digits so in our first

number there's two digits to the right

you could think about adding in a

decimal sign well there's really no

digits to the right of our second number

16 so I count up the total number of uh

digits that are to the right so in this

case there's only two

digits it says that's how many places we

move our decimal sign at the end so I

start all the way to the right of the

number then I just move it one two

places so what it says is it says if we

multiply 8245 * 16 we'll be left with

Adding and Subtracting Decimals, Ex 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Adding and Subtracting Decimals, Ex 3


okay so this is going to be another

volume of revolution using disks and

washers and in this one again I just

want to slightly tweak my first two

examples so again we're doing the region

bounded by X cubed and X where X is

greater than or equal to zero so that's

going to give us the region on the top

right quadrant here's the graph of y

equals x y equals x cubed you can find

the limits of integration the points of

intersection by setting them equal to

each other so that's what I did over

here off to the side and this was

explained in part one and now again we

just want to integrate we want to rotate

this region about the line y equals five

so just like in the last couple our

limits of integration will be from zero

to one and again now we have to take pi

times the outer radius squared minus pi

times the inner radius squared and then

we head agreat that with respect to X

okay so now my outer radius will be this

is usually the ones that that seemed

most tricky to people maybe maybe not so

the line y equals five is what I'm

rotating about we draw a line again

perpendicular to that so I'm going to

draw a line perpendicular to that it

just hits my region and now it's just

about to exit the region so again this

is going to be the outer radius and

that's what I plug in to the first part

this is going to be the outer radius so

again I'm trying to think what's the

length of this arrow well if it went

from y equals five to the x axis it

would have a length of five but I'm not

going all the way there I'm losing a

little bit well this little bit that I'm

leaving off again comes from the

function which is X cubed so it says

this outer radius is actually going to

be five minus that that curve which is X

cubed quantity squared and then my inner

radius I calculate that the same way

so I draw a line that's perpendicular

okay and it's just going to hit the

region but now it's going to hit the

straight line part which is the curve y

equals x so again the entire length

would be five we'd have a length of five

but now we have to subtract off that

height of X and that will be our inner

radius sorry for all the colors here

that'll be our inner radius and we've

now got our problem set up so again to

integrate I would factor out the PI

you'd have to take the five minus X

cubed and foil it out the same thing

with the five minus X collect your like

terms and then it should just be using a

bunch of the power rule stuff

Volumes of Revolution - Disk/Washers Example 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! In this video, I use the same region from part 1, but now I rotate the region about the line y = 5.
For more free math videos, visit http://PatrickJMT.com


so one thing that series are good for

are finding uh polinomial approximations

to certain functions that turns out to

be super super super useful um we're

going to kind of keep it on the easy

level here and we're just going to talk

about finding linear approximations and

this is actually just old first semester

calculus stuff they'll have you find the

equation of the tangent line or a

linearization that's all we're doing at

this case so I'm going to go through

these a little bit a little bit faster

so we're just finding the equation of a

line is all we're doing um so recall the

slope comes from the

derivative um so let's see our function

f ofx that's 2 to the X the derivative

of 2 to the x is 2 x * the natural

logarithm of 2 to get the slope of our

tangent line we just plug in whatever x

coordinate we're interested in which in

this case is three so we'll have 2 3 *

the natural logarithm of 2 two to the 3

is eight so well we're late we're left

with eight times the natural logarithm

of two notice if we plug two in or

excuse me three into our original

function to get the corresponding y

value we would get 2 the 3 and 2 the 3

is just8 so it says we would have y

minus our y coordinate well again if we

plug in three the Y value we get is8 the

slope of the tangent line is 8 * the

natural logarithm of 2 2 then we just

take x minus the x coordinate which was

given to us as three and now this is

going to be our linearization so if we

distribute out on the right side we

would have 8 * the natural logarithm of

2 * X we would get a -4 * the natural

logarithm of 2 then we could add our

postive 8 over and again this is now

this would now be our linearization

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding a Linear Approximation (Linearization, Tangent Line Approx), Another Ex 1.  Just another example of finding a linear approximation for a function at a particular point.


we're going to graph two periods of Y =

2 * cosine of 2x the first thing I'm

going to do is find the period so we'll

take 2 pi and divide that by the

absolute value of B so we'll take the

absolute value of positive2 which just

is simply equal to two so the period

will equal simply equal to Pi I like to

divide these into uh four equal pieces

and that's what I'm going to use as the

increments on my graph so I'm going to

come back to that I'll use those as the

increments the amplitude is just going

to be the absolute value of two which is

pos2 so here's my graph we've got X

we've got y the amplitude it goes from

positive2 to -2 and again the increments

I'm going to use will be pi over 4 so

again the uh the units on the x axis and

on the Y AIS are not exactly the same so

positive2 will be right

there -2 looks like it's in the right

place so let's see we'll have pi over 4

pi/ 2 3 pi over 4 we'll have

Pi 5 piun over 4 6 piun over 4 will be 3

piun / 2 we'll have 7 piun over 4 and

then 8 pi over 4 will be 2

pi so if we substitute in xals 0 we'll

get cosine of 0 which is one we'll

multiply that by two that'll give us

and after that again that's why I like

to use these increments at the next

increment it'll be at zero at the next

increment it'll be at -2 and then back

at

zero then back at positive2 then it'll

be at zero it'll be at -2 then at zero

and then back at

positive2 so let

me connect them with some uh curvy

curves and that'll be a rough little

sketch of Y = 2 * cosine of 2x

Graphing a Cosine Function  EX 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Just another example of graphing a trigonometric function


okay here we're going to look at an

sphere and the surface area we simply

take 4 * pi * the radius

squared so we're going to find the

surface area of a sphere with a radius

of 7.4 M and we'll use

3.14 for pi so again our formula 4 * pi

* the square of the

radius well we're approximating pi with

3.14 our radius is 7.4 so we simply have

to square

that and again now it's just a

matter of doing a bit of arithmetic so

7.4 *

7.4 I'm getting 54.76%

[Music]

so now I'm going to multiply so we've

got

12.56 we're going to multiply that by

54 multiply that by

54.76% round that digit up one so we'll

get

68779 again our units were in meters

since we're finding a surface area an

area is uh two-dimensional so our units

will be in meters squared

Surface Area of a Sphere

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Surface Area of a Sphere.  Here I do a quick example of finding the surface area of a sphere.  The formula is given at the beginning of the video.


okay so another example here of using

our half angle identities to evaluate a

a trig expression here so we want to

evaluate tangent of

105° well let's see again if you double

105° that would be 200 and

10° so we can write 105 as 210 over 2

and again now all I'm going to do is

just uh use our identity

here so we'll use the bottom identity

and it says we simply get 1 minus cosine

of whatever value is hanging out on top

which is going to be

210° all over s of

210° and now it's just simply a matter

of evaluating cosine of 210° and S of

210° well let's see I believe our

reference angle so if we go Halfway

Around that'll be

180° so then we would have to go another

30 to get to our angle of

210° let's see if we think about um I

guess the angle uh 30° in the first

quadrant let's see in the first quadrant

we get < tk3 over2 for the X quinate on

the unit circle and we get2 for the

y-coordinate so that means at

210° well uh they would both be negative

in quadrant three so okay so now I've

got my values that I need we'll get 1

minus cosine well cosine would be < tk3

over

2 s is going to be the value of the

y-coordinate which will simply be

one2 so let's see in the numerator we'll

just be left with 1 + < tk3 over 2 in

the denominator we have our - one2 I

think I would do the same thing as

before just uh you know get rid of the

the denominators the twos I would mult

mply the numerator by two and the

denominator also by

two so if we do that in the numerator

we'll get

two uh the twos will cancel out when we

distribute and we'll be left with the

square < TK of

three in the denominator again the twos

will cancel we'll simply be left with

Nega one you can always distribute that

negative um I'll just pull it out front

and say we

+ uh Square < TK of 3 and I think uh

there's not much else you can really do

there you can always turn it into a

decimal but this is a nice exact answer

so I think I'm going to leave it just

like that

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Half Angle Identities to Evaluate Trigonometric Expressions, Example 2. In this video, I give some half angle identities and show how they can be used to evaluate trigonometric expressions.


all right in this video we're going to

look at an example of multiplying whole

numbers

so in part a here we're going to

multiply 234

times 2 and in part b we'll do 1302

times 3. so again just like with

addition and subtraction

i tend and i think most people we'll

just write these vertically so we've got

2 3 4 times the number 2

and again now i'm just going to multiply

a digit at a time so

so i'm going to do 2 times 4 which is

going to be 8.

we stick that in the same column okay

we do 2 times 3 which is going to give

us

6 we put that again in the same column

as the 3

and then 2 times 2 is going to be 4 and

that's going to be our solution

it says we have 234 times 2 is 468

and again that makes sense you're just

kind of doubling all of the digits

all right let's do our next example so

we'll multiply that by 3

and again just like with addition and

subtraction i have all of my units lined

up

okay so three times two is going to be

six three times zero

is going to be zero three times three

is going to be nine and again we just

drop it right below

and then 3 times 1 is going to give us

3. so it says

1302 times 3 is going to give us 3

900 and six

Multiplying Whole Numbers, Example 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Multiplying Whole Numbers, Example 1.  Here we look at two examples of multiplying a two digit number by a single digit.


all right in this video I'm going to do

some examples of eliminating the

parameter so we can better describe a

curve given by the parametric equations

um I'm going to do all of these in uh

one video part A and B it's really just

going to be a matter of solving for T

and then plugging in so notice from our

first equation we can just add one to

both sides and then have x + 1 = T well

now in our second equation we'll get y =

4 * t + 1 but again I'm going to replace

the T with our x +

one well we can always distribute we'll

have 4x + 4 +

1 so we'll end up with the equation y =

4x + 5 and that'll be our answer for

part A we've now eliminated the

parameter so nothing at all crazy again

just a graph of a line

here Part B will'll do the exact same

thing um again we've got X = the < TK of

t + 1 well we can square both sides

that'll give us x^2 = t + 1 and then we

can subtract so we'll have x^2 - 1 = T

and the same thing I'm just going to

plug that in so y = 3 * t + 2 well again

we're just going to replace the T with

our x^2

and that's going to leave us with Y =

3x^2 I guess we'll have -3 + 2 which is

going to leave us with a negative one

and again we've eliminated the

parameter so one more example here part

C again I'm going to do the same thing

here I'm not going to quite solve for T

um I'm going to solve for e to the T So

notice in our first one here I can

divide by 4 x over 4 that would be e to

the T over

4 well what we can do at this point is

we can raise both sides to the fourth

power so we'll have uh X over 4 raised

to the 4th on the right side if we

multiply the exponents you know T over4

* 4 over one that's just going to leave

us with e to the T and again to me

that's good enough because our uh our

parametric equation for y involves e to

the

T so it says Y is going to equal 3 * e

to the T but again we can just replace e

to the T with our X over 4 raised to the

fourth and you can always simplify this

um you know if you want to a little bit

but again we've now eliminated the

parameter

Eliminating the Parameter , A Few Examples

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! You should ALSO think about restrictions on the value of t, based on the parametric equations for x(t) and y(t).  I do not mention that in the video though (I think).  All the same, the curves would lay on the curves I produce.


okay in this video i want to do some

slightly more complicated evaluating

exponential expressions so just some

more numbers floating around

at this point to me i really stop you

know and look at it for a second to

to make sure that i'm doing the order of

operations correctly

but typically what i'll do is i'll try

to work inside the brackets first

so and i think in this case there's

really not much else you can do

except work inside so let's see we've

got to do order of operations

so we've got multiplication but also an

exponent

so we do the exponent first so 4 squared

is 4 times

4 which would give us 16

minus okay so again you know notice that

the negative is not

included it's it's not being squared the

only thing being squared in this case

is the 5 and 5 squared gives us

25 so there is still a negative out

front

that's all being squared so let's see

we have negative 4 still out front 2

times 16 will be

32 minus 25

again that's all being squared so i

think i'm going to run into my next

problem here so let me move it

up here real quick so we still have

negative 4 out front

let's see 32 minus 25 32 minus 25

i believe would give us the number seven

still that's being squared

okay so again we have to do the exponent

first

so the negative four he's just hanging

out there

seven times seven will give us 49

and now we have to do negative four

times uh

49 and you guys can check my arithmetic

here

the way i do stuff like this is to me if

i made it 50

negative 4 times 50 would be negative

but since you only have 49 of them it's

one multiple less

so instead of negative 200 it'll give

you negative 196.

so again you can use a calculator or do

it out the long way

maybe just another way to think about it

again you know i think it's a little

tricky it takes a little bit

getting used to but that's one way that

i'm able to do the arithmetic in my head

at least all right so that'll be our

final answer for the first example

in the second one very similar problem

we have seven

and then inside of our brackets we have

two times three cubed minus

eight squared all of that being squared

okay so i'm going to do the same thing

i'm going to work on the inside

so 3 cubed that's 3 times 3

times another 3 well 3 times 3 would

give us

9 and then times another 3 would give us

minus 8 times 8 would give us

64 being squared

let's see 2 times 27

will give us 54 minus

64 again being squared

if we simplify inside the brackets

54 minus 64 will give us negative 10

squared and now so we have to take

negative 10

times negative 10 which will give us a

positive

100 the 7 is still out front

so since the negative is being included

in the exponent again it will turn

positive in this case

because it's an even exponent and then

we have 7

times 100 which is going to give us

700. so this second expression will

reduce to the value

Evaluating Exponential Expressions (Numbers Only, No Variables) - Ex 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Evaluating Exponential Expressions (Numbers Only, No Variables)


all right in this example we're going to

write the standard form equation of a

parabola with a Vertex at the origin and

a directrix of yal ne5 over 6 and again

you don't have to but I have already

sketched a little graph that corresponds

to the information that we're given so

there's our directorx yal 5 over 6 we

know that the vertex is at the origin

and well that would tell me that my

Parabola would have to be opening

upwards in this case so again kind of

the basic form that we are trying to

know fill in we have x^2 = 4 P *

Y and again the only real value we have

to come up with here is the value for p

we just have to deduce the value for p

and then um we plug it in and we're done

really not much else to it so we've seen

um that if the director okay so the

directorx had in this case we can write

it as yal

p and we know that our

focus is at the point 0 comma P okay so

basically it says to get the focus just

kind of take the opposite sign so we've

got uh -5 over 6 so our P value would

just be 5 over 6 so that tells us our

focus is at 0

56 and again you don't maybe even

necessarily have to think about that but

just to point out a couple things and

again now we have our P value mean

that's really the important part here so

our P value will be 5 over 6 again just

based on the directorx you can just take

the opposite sign and that's your P

value and now it's just a little bit of

arithmetic so x^2 = 4 * well let's plug

in 5 over six for our P value times Y

and well we can simplify this down a

little

bit let's see uh 4 over 6 would reduce

to 2 over3 2 * 5 would be 10

we would still have the three in the

denominator so it looks like we would

get x^2 = 10/3 * Y and that would be our

equation

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Conic Sections, Parabola : Find Equation of Parabola Given Directrix.  In this example, we are given that a parabola has a vertex at the origin and we are given the equation of the directrix.  From that, we find the equation of the parabola.


all right here we're going to look at

some examples of multiplication and

division of integers and the only thing

you really have to remember are just the

rules about signs so 8 * -9 we have a

positive time a negative a positive time

a negative is going to give us a

negative number and then I forget all

about the signs I just say well 8 * 9 is

72 therefore 8 * -9 is going to give us

-72 if you have two negatives you know a

negative times a negative that always

gives you a positive so 4 * 7 is going

to give us 28 so it says -4 * -7 hey

that's posi

28 exact same rules with division Okay

so we've got a 48 divided -2 so we have

a positive divided by a negative that's

going to give us a negative and then 48

divided uh 2 is going to give us 24 so

48 / --2 is going to give us

-4 so let's see Part B we've got a

negative divided by a negative it's the

same rule as with multiplication a

negative times a negative is a positive

likewise a negative divided by a

negative is a positive so it's

definitely going to be a positive number

then I just forget again about the the

signs I say well 35 / 5 that's going to

give us 7 so -35 / -5 well that also

gives us positive s

Dividing / Division of Integers

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Dividing / Division of Integers.
In this example, I do some basic examples of dividing integers.


all right so one more example here of

solving a trig equation by factoring and

here we're going to solve cosine xal

Cent X and what I'm going to do in this

case uh is simply rewrite

coent so tangent is s over cosine coent

is

cosine over s and again it might be

tempting to sort of uh divide both sides

by cosine of x to get rid of that but

again that would be a mistake as we kind

of pointed out in the last video you

don't want to do that you'll lose some

solutions so what I'm going to do is

simply subtract the cosine X over

sinx from both sides and that'll leave

me with zero on the

right well I can Factor the cosine X out

so cosine x * 1 would give me cosine

xus 1/ sin

x when I distribute these out um we'll

get our cosine over

s

so again now we have to Simply Set uh

each factor equal to zero so we'll get

cosine xal

0 and then we'll get that 1 - 1 /

sinx = 0 and well uh maybe it's pretty

clear what sin x would have to be uh we

can certainly add it over we can add the

one/ sin

x and then what I'm going to do is

simply multiply both sides by

sinx so they'll cancel on the right and

we'll be left simply with sinx = 1 so I

have to think where between 0 and 2 pi

do cosine of xal 0 well again that

happens at x = Pi / 2 and it also uh

happens at 3 pi/ 2 uh where does s of x

equal 1 well s = 1 at pi over 2 uh which

is uh I guess a bit redundant uh we've

already found that solution and that's

the only place where s equals 1 on the

interval 0 to 2 pi so we could just

simply say our Solutions are Pi / 2 and

3 pi/ 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Trigonometric Equation by Factoring, Example 3


okay here we're going to look at some

basic examples of the ratio test and for

the ratio test it says if you have some

formula you know a sub M that describes

the terms of your series it's as we look

at the limit as n goes to infinity of a

sub n plus 1 over a sub n recall if this

is less than 1 the series converges

absolutely if it's greater than 1 it

diverges and if equals 1 it's

inconclusive so here we're going to use

the ratio test and Part A to decide if

the series 2 to the N over N factorial

from N equals 1 to infinity whether or

not that converges so everywhere there's

an N we're going to plug in our we'll

plug in the n plus 1 so here we'll have

2 to the n plus 1 over n plus 1

factorial then we're going to divide

that by the original so 2 to the N over

N factorial all right well this is the

limit as n goes to infinity of 2 to the

n plus 1 over n plus 1 factorial times n

factorial over 2 to the N well let's see

we can simplify this we've got 2 to the

n plus 1 over 2 to the N that's going to

leave us with a 2 to the 1st in the

numerator and then n factorial over n

plus 1 factorial how does that simplify

well we're going to use this trick a lot

so maybe let's talk about it at least

once here so you know for example just

at random say we had 6 factorial

well that's 6 times 5 times 4 times 3

times 2 times 1 we could write it as 6

times you know we'll take 1 away from 6

hey that would give us 5 and then we'll

take two away from 6 hey that would give

us 4 and then I'm going to start putting

kind of a few arbitrary dots assuming

this number is large enough eventually

you know you're going to get every

number in between

main and then at the end we'll end up

being will always multiply by 3 times 2

times 1 you know kind of the tail end of

the sequence so in factorial I'm going

to write that as n you would just get

the very first number then we would take

one away from that then we would take 2

away from that and then eventually you

know assuming ends a big number

eventually we'll get down to 3 times 2

times 1 the same thing in the

denominator if we expand n plus 1

factorial

we'll get n plus 1 but then if we take

one away from that we'll be left with in

and then if we take one away from that

we'll have n minus 1 and minus 2 again

eventually we'll get down to 3 times 2

times 1

so notice everything is going to cancel

out except for the n plus 1 in the

denominator so in this case we've got

the limit as n goes to infinity of 2

over n plus 1 well as n goes to infinity

the denominator is going to get large 2

over a large number is 0 the limit of

that is 0 and since that number is less

than 1 that would tell us that our

series is absolutely convergent in

particular it is also just convergent so

partake we've got a nice example here of

a convergent series and almost any time

I see factorials and a series and they

want me to decide if it converges or

diverges I almost always just jump

immediately to the ratio test very

useful little test

The Ratio Test , Another Example #1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Ratio Test , Another Example #1.  Just another example of using the ratio test to determine if a series converges or diverges.


all right in this video we're going to

talk about graphing some simple polar

equations so here we're going to graph R

= 2 and Theta = pi4 there's two

different ways you can think about these

just to graph R equals 2 um what that

really says to me it says it's kind of

independent of the angle so pick any

angle that you want just go out a

distance of two from the origin well if

you do that you know if you kind of

neglect the angle if you allow it to be

whatever you want it to be but you

always go at a distance of two hey

that's just going to make a nice uh a

nice Circle so I think I put my -2 down

here a little too far let's make -2

right there it's just going to be a

circle of radius two and again you could

see that because we know that R uh we

could rewrite r as the square Ro T of

x^2 +

y^2 = 2 and if you square both sides we

get hey x^2 + y^2 = 4 and again X + y^2

= 4 that's just going to be again a

circle of radius 2 so this would be part

A Part B we got Theta = pi over 4 so

Theta = pi over 4 again you can kind of

think about this in the same way it says

go to an angle pi over 4 so pi over 4

would just be our 45 Dee line here so

there's the angle pi over 4 and then

kind of the same idea it says always let

will be pi4 but pick any r value that

you want well if R is positive we'll get

uh you know the segment of the line that

goes off into the first quadrant if R

was negative though it would reflect

down into the third quadrant um and

basically we would just get a nice

little line and again the way that you

can see that is if you take tangent of

both sides so tangent of theta equals

tangent of pi over 4 well tangent OFA

we've seen that we can write that as

YX tangent of pi over 4 that's just

equal to 1 so if you multiply both sides

you simply get the equation yal x and

again that's what yals x looks like so

two different ways to think about it you

know just uh just to think about the

equation that you're given but uh also a

very very common thing to do is to

convert them back into these uh

rectangular coordinates to where

hopefully it's something a little more

familiar

Graphing Simple Polar Equations, Ex 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing Simple Polar Equations, Ex 1.  Graphing polar curves can be tricky, but here I look at graphing some of the basic ones first!


all right in this video we're going to

look at just some questions related to

the second derivative test and these are

just sort of sort of uh

you know do you know the results type

question so suppose our function f of x

has a critical point at x equals 4

and the second derivative at 4 is

negative we want to know what can we say

about our function at x equals 4

well recall if you have a critical point

and the second derivative is negative

what we can say in that case is the idea

is in that case we're actually going to

have a local maximum

at x equals 4. so since the second

derivative is negative that kind of

is telling us that the function's

concave down

and with that being a critical point

that's going to tell us that we do have

a local maximum

so all right part b here

um so suppose g of x has a critical

point at x equals two

and the second derivative at two equals

zero

we wanna know what can we say about the

function

what's happening at x equals two well

recall in this case if the second

derivative

equals zero in that case actually our

second derivative test is inconclusive

so

what you could do in this case to figure

out if you do potentially have a maximum

or minimum

is you can go back and use like the

first derivative test do you like the

little sign

chart you know we've seen some of these

find your critical points figure out if

it's increasing

decreasing increasing for example and

then you'll know

okay i've got a local maximum or local

minimum

The Second Derivative Test to Find Maximums / Minimums, Quick Example 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Second Derivative Test to Find Maximums / Minimums.  Suppose we know that a value of x is a critical number and the second derivative is negative there.  What can we say about the point in regards to it being a local maximum or minimum?


all right in this video I'm going to do

one more example finding the general

term of an arithmetic sequence and if

you got good eyes you may notice I

changed the subscript on uh one of these

I think it's going to make the values

work out a little bit nicer but anyways

we're going to find the general term of

an arithmetic sequence of a sub 21 =

-14 and a sub 51 =

226 well okay we're just going to use

our formula here that again a subn is a

sub 1 + n -1 * D well a sub 21 we know

that that's going to be well a sub 1

plus we'll take 21 - 1 which will give

us 20 * D we know that has value

-14 likewise a sub 51 that's going to be

a sub 1 plus well we'll take 51 minus 1

which will give us

50d and we know that equals 226

so now I'm just going to uh do some

elimination by addition with the this uh

these two linear equations so what I'm

going to do is I'm going to multiply

both sides of my first equation by -1 so

that's going to give me a sub 1 -

20D =

pos4 I'm going to leave the second

equation alone so we have a sub 1 +

50d =

226 and now if we do our our elimination

by addition negative a sub 1 and

positive a sub 1 will cancel -2d + 50d

is going to give us 30 D 226 + 14 is

going to give us

240 well if we divide both sides by 30

we'll get 24 over 3 which gives us 8 so

now we know our value for

D and at this point we can use either

one of our uh original equations

uh to go back and solve for a sub one

well let's see I don't think it really

matters too much which one we use um I'm

just going to use the first one so we

had a sub 1 + 20 * D but now we know D

is equal to

8 and on the right side we have

-14 well we'll get a sub one let's see

20 * 8 is 160 = -14 if we subtract 160

from both sides

it looks like we're going to get a sub

one the first term to be -

174 well now I can just fill in my

formula so it says a subn the general

term is going to be a sub 1 which we now

know to be

-174 plus n -1 * D which is 8 and now

we've got a formula for the uh for any

uh for the nth term any general term of

our arithmetic sequence

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Arithmetic Sequences: Finding a General Formula Given Two Terms.  In this video, we have values for a_21 and a_51 and we have to find the general formula for the arithmetic sequence: a_n.


all right in this video we're going to

do an example of using determinants to

find the area of a triangle and the

triangle is going to have vertices at

the point 40 72 and 23 the formula we

use is the one here in the middle this a

equal 1 12 and then we have all the

determinants the vertices have to be

listed uh in a a counterclockwise order

around the perimeter so I drew a picture

we've got 40 72 and 23 well I just uh

decided to start at this point 40 it

doesn't matter where you start so that's

going to be my X1 y1 but then if we move

to the right our X2 Y2 will be 7 and two

and then uh if we keep following this

counterclockwise motion X sub3 Y sub3

will be the point 2 comma 3 so at this

point we just have to fill in this

formula and then compute the

determinants and add them together and

multiply

by2 so let's see X1 y1 that was 40 X2 Y2

is 72

uh again X2 Y2 X3 and Y3 that's going to

be 2 and

three and then our last X sub3 and Y

sub3 that'll be 2 and 3 X1 and y1 will

be our 4

and0 all right so let's see what we get

here um we can compute these

determinants so we can do 4 * 2 which is

8 0 * 7 will be

0 our next determinant will get 7 * 3

which is 21 uh - 2 * 2 which is

4 and then our last determinant we get 2

* 0 which is 0 - 3 * 4 which is

12 all right so we get 1 12 it looks

like inside our brackets we have 8 plus

let's see 21 - 4 is going to be 17 um 0

- 12 is going to be uh -12

well 8 + 177 is 25 25 - 12 is going to

be 13 and if we multiply that by 1/2

we're simply going to get that the area

of this triangle uh with these vertices

would be uh the area would be 13 over2

uh units squared whatever your your

units are so um kind of an interesting

little formula and how you can use these

determinants I think to find um the area

of a triangle you can certainly

generalize this to a polygon of any

polygon um with with in for example in

points and we'll do an example with that

where we have something a little more

complicated than just a triangle in

another example

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Determinants to Find the Area of a Triangle - In this video, I show how one can use determinants to find the area of a triangle.
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all right so in this video I'm going to

do a couple more of the trig identities

simplifying I think I'll do uh B and C

in this example so um in B we want to

simplify see x * cosine x - cosine 2 x

well let's see see of X recall secant is

really just one over

cosine and again that's being multiplied

by cosine X and then our minus cosine 2

x we'll just leave that alone well you

can think about cosine X as being over

one so if we multiply we've got cosine

over cosine which is just one and then

we've got our 1 - cosine 2 x but weall

our uh the identity cosine 2 x + sin^2 x

1 notice if we subtract the cosine squar

we'll get 1us cosine squar and that's

going to be equal to sin

squ so to me that would be a nice uh

simplification here for Part B to say

that that's going to reduce to sin^2

X let's do

uh let's do our next example

um so let's see uh in in example C we've

got cosecant squ minus

one okay now I would kind of think maybe

about the same idea that we did before

so secant squar though um let's see what

is that that's going to be um

1/ cosine 2 x time sin^2

x all right well so cosecant squ X we'll

come back to that we'll think about that

here in a second um this would be sin

squar of X over cosine 2 X or we could

write that as Sin of X over cosine X all

squared um I'm going to do two things at

once here uh again call one of our

identities um we have that cent

squar of x + 1 that equals cose 2 x what

I'm going to do is I can simply subtract

one and say c^ 2 X = cose 2 X - 1 so I'm

going to plug that in um for the first

uh the first Factor I'll replace

cosecant 2 X - 1 with cotangent

okay well recall again that s over

cosine that's just

tangent but again it's all being squared

so we have tangent s

x well now we're almost there um recall

that uh cent of X that's really one over

tangent of X but again it's being

squared so we would have to square this

as well so cotangent squar is the same

thing as 1 over tangent squar but again

that's being multiplied by tangent s of

X so that'll just give us tangent

SAR over tangent squared which is

finally going to reduce simply all down

to just a one so um part C it says

that's really a complicated way of

writing the number one

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplifying Trigonometric Expressions Using Identities, Example 2. In this video, I simplify a trigonometry expression using some trig identities.


all right in this video I'm going to do

one more example of dividing complex

numbers and in this case we have 4 - I

over -2 - 7 I so again the basic idea is

I just look at the denominator multiply

by the complex conjugate and again the

only sign that changes is the sign in

the middle so you're still going to have

-2 that'll become a positive 7 I and if

we multiply it to the uh denominator

we've got to do it to the numerator

again I put everything in parentheses

just to remind myself I have to

distribute everything out

here all right so let's see if we can't

uh clean this up nicely so in the

numerator we have pos4 * -2 that'll give

us a 8 we have pos4 * POS 7 I that's

going to give us a positive 28 I we have

netive I * -2 so a negative and a

negative will make a positive we'll get

2 I and then we have a negative * a

positive which is a negative you can

think about there being a 1 * a 7 which

is 7 then we have I * I which is I 2 and

in the denominator okay let me get rid

of my little arrow here we have -2 * -2

which is pos4 we have -2 * POS 7 I which

will give us a -4 I -7 I * -2 will give

us a positive 14 I and then we have -7 I

* POS 7 I that'll give us a

-49 i

2 okay so in the numerator it looks like

we

have8 28 I and 2 I is going to give us a

POS 30 i - 7 i^ 2 is the same thing as

-1 and again the the imaginary all the

imaginary Parts in the denominator are

going to cancel out the4 I is gone um

the positive 14 I will cancel each other

out and then it looks like we have 4 -

49 again i^ 2 is

-1 so in the numerator we have8 we have

a -7 * a -1 those will just turn into

positives when we multiply them so8 + 7

is going to leave us with negative 1

again we still have the 30i just kind of

hanging out

divided by it looks like we have four uh

we have a - 49 * a negative 1 which is

going to again make a positive 49 and 4

is

53 again I'm just going to break this up

-1 over 53 + 30 over

53 I and again you could think you know

does this reduce it all but 53 um I do

believe is a prime number um so there's

really you know even if the top factors

and it certainly does 53 doesn't Factor

so there's nothing to cancel out so um

again at this point we're finished we

have

now divided our two complex numbers we

end up with a new complex number um and

that's it

Complex Numbers: Dividing - Ex 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Complex Numbers: Dividing - Ex 3.  In this video, I show how to divide a complex number by another complex number.


all right in this example I'm going to

do another linear equation and here

we'll solve our our second linear

equation 5 - x over3 = 4 * the quantity

x + 1 over

6 okay definitely lots of different

things that you could do here in terms

of uh getting

started what I'm going to do is I'm

going to uh do a couple things I'm going

to write five as 5 over 1 I've got uh

our X over 3 there here now notice 4

over 6 we could actually reduce that

that would reduce to

23 and then we still have our x + one

now we could start Distributing things

out but eventually I'm going to have

fractions floating around which isn't

the end of the world I mean you've got

to learn to work with fractions but

there's definitely cases where you can

avoid them so what I do is I look at the

denominators we've got a one a three and

a three and I'm thinking what's the

least common multiple of those three

numbers and the first number that they

all would divide into evenly would

simply be three so what I'm going to do

is I'm going to multiply both

sides of my

equation by the number

three and again what this does is it's

simply going to get rid of the fractions

for me and then um to me it makes the

arithmetic a little bit easier keeps me

from making mistakes so we have to

distribute 3 * 5 is 15 if we take uh the

three we'll get -3x over 3 but then the

3's just cancel so we're just left with

our Negative

X again you can think about this as

being 3 over one so I'm just going to

cancel the three and the three would

cancel we would be left with 2 * x + 1

over one which we can just write as well

2 * x + 1 and now to me we're in pretty

good shape

um I'm going to get rid of the

parentheses on the right side

Distributing we'll get

2x + 2 same thing as before I'm going to

put my x's on the same side I think here

I'm going to add X to both sides I like

to keep my x's positive if I can don't

have to um so we'll have 15 on the left

2x + 1X is going to give us

3x we still have our plus two um on the

right side and again to get rid of the

positive two I'm going to subtract two

from both sides so 15 minus 2 is going

to be 13 = 3x again since we're

multiplying by three we'll divide both

sides by three um they cancel on the

right and we're just left with X =

13/3 as our solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Basic Linear Equation - Example 2


in this video I want to do two last

examples of factoring the difference of

two squares and the two examples I'm

going to do are here at the bottom and

you could certainly do these problems in

a slightly different way uh we could

always just multiply this out x - 3 * x

- 3 collect your like terms subtract

four and then try to factor it from

there and the same thing on the other

but again really just to emphasize this

this

formula I could again write four is 2^ 2

well again to me I see two terms and I

see a negative in between them so it

says I can Factor them so it says our a

is going to be the x - 3 and then we

have our b as being the two so it says

this is simply going to factor now

as so there's our a minus our B and then

there's our a

term plus our B

and if we simplify this down if we

collect like terms we'll simply get x -

5 in the first set of parentheses and

then it looks like we'll get x uh what

is that going to be a Nega 1 in the

second set of parentheses okay and now

we have it factored so again um maybe be

a little careful you know if uh if we

had something more complicated as our B

value this negative you might you might

want to stick your B in parentheses

because the negative would have to to

get distributed out I'm not putting the

two in parenthesis just because I'm

recognizing there's nothing to

distribute with that negative so one

little place to be

careful okay so we can do the same thing

here in our other example we have x + 7^

25 okay so in the first set of

parentheses I'm going to have x + 7

there's my a term and then um minus our

B term so if you take the square < TK of

25 we just get five and then I just do

the same thing except I change this

negative into a positive so x +

five okay so there's my a term minus our

B term and then our a term plus our B

term and again we can simplify this down

a little bit so we have in the first set

of parentheses x then we have POS 7 - 5

that'll leave us with a positive2 and in

the second set of parentheses we'll have

positive 7 and positive 5 which will

give us uh x +

12 and again now we have it

factored so and you can even try you

know multiply this out collect your like

terms and try to you know try to factor

it uh and see if you don't get the same

thing in both cases so again all roads

lead to Ram different ways to do it but

in this case just trying to again really

emphasize this difference of squares

formula

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Factoring the Difference of Two Squares - Ex 3