all right in this video we're going to

do an application using the law of signs

just a little word problem so suppose

we've got a person um that H that has a

kite out on uh

1,750 ft of string and uh we've got an

observer who's watching the person with

the kite and the Observer notes that the

angle formed by the kite and the flyer

is

102° we just want to figure out how far

is the Kite from The Observer so what we

really need to figure out is uh the

distance uh along the right side of our

triangle here so maybe I'll call that D

for distance well this is one where we

can simply use the law of

signs because we can say um so D over

s of

75° the angle in the bottom left that's

going to equal um the distance the the

long distance the 1,750

ft over the S of

102° and now it's just a simple little

equation uh that we have to solve so um

to do this what I would do again is just

multiply both sides by S of

75° so s of

75° on the left side uh we'll just be

left with our D the distance that we

want and now it's just simply uh again

just kind of plugging some stuff into a

calculator so 17

75 let's do s of

102 s of 102 I'm going to round that to

978 uh we'll do s of 75 Dees s of 75 I

get that to be

966 after rounding so let's see if we do

1750 multiply that by

966 um I'm getting that to be

69.5 we'll divide that by

978 so if I divide that by

978 um I'm getting that to equal

1,728 uh.

53t um after rounding again just a

little bit so um certainly that makes

sense that the uh you know clearly The

Observer uh you know should be uh a

Little Closer closer to the kite than

the person who's flying the Kite at

least based on our picture so um

certainly The Observer is a little close

a little bit closer uh but not by too

much so again that's all there is to it

just a a nice little application using

the law of signs

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 Sines - Application/Word Problem, Ex 1.  In this video, we do a word/application problem that involves using the law of sines.


all right here we're going to look at

some examples of ordering decimals from

smallest to largest um in part A we've

got

3201 and

3.88 6 um so and I think you'll often

find that's how people actually say

decimals you know people don't typically

say we've got three

and

149,000 people will typically just say

3.49 because it's much easier to say

right so but anyway ways from least to

greatest so you know I look at the whole

number part well they've all got a three

okay so they all start with three uh no

problem there and then I just kind of

look at them you know sort of a uh I

just sort of look at one digit at a time

but I'm looking

for you know to see if there's any I'm

basically looking for the smallest

one okay so they've all got a three so

in that sense to me they're all equal

let's look at the next digit well the

next one uh um we've got oh 3.1 3.2 and

then we would have let's see if I can

make my paper here a little

smaller then we've got a 3.0 so I'm just

looking at all of these next digits so

one two um and a zero and I'm thinking

which of those digits is the smallest

well this is certainly the

smallest that's absolutely the smallest

one and then that would be be the second

smallest and that this uh 3.2 would be

the

largest so all right I've got them

listed now so um we would have

3.88 then

3149 and then

3.21 so let's do that uh with our

second second list here okay so I look

at the very first digit one one one one

those are all the same so again it

doesn't really tell me anything then I

look at the next

digit okay so I'm going to look at the

next

digit okay so my next digit this one has

a one this one has a one this one has a

four this one has a zero so zero is the

smallest of all of those so I know that

the

1.1 is going to be the very smallest

number uh the then we have ones next so

the 1. On's should come next I don't

really know which one of those comes

next yet but I know that the largest one

will be 1.4

31 so now if we move over one more digit

to kind of break the the the stal made

here again we've got a one this one is a

zero well zero is

smaller and now we've got everything

listed from smallest to greatest again

you could just kind of look at the

decimal

part and order them that way as

well so this to me kind of looks like

the number one this looks like 101 this

looks like 110 this looks like

431 well 1 101 110 431 so another way to

just kind of decide which one's the

smallest and which one's the largest

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Ordering Decimals from Smallest to Largest.  Just a couple of examples of ordering decimals from smallest to largest.


okay here we're going to look at another

example of integrating with respect to Y

so here we're going to find the area of

the shaded region so it's bounded by y

equals x plus 2y equals square root of x

and then bounded below by the x-axis so

y equals 0 and then looks like this line

y equals 2 alright so again I think if

we were going to integrate this with

respect to X notice we would have to

break this up into two separate

integrals again so to get the area of

the triangle we could really just kind

of use geometry but if we wanted to you

know think about it in terms of an

integration we would take the top

function which is well X plus 2 and then

we would subtract the bottom one which

would just be y equals 0 that would be

and then we would go it looks like from

negative 2 up to 0 but then we would

have to break it up again and go from 0

up to 4 and we would have to take the

the function on top which is y equals 2

and subtract the bottom one which would

be square root of x so we could do this

again in two pieces let's write it down

so negative 2 to 0 the function on top

is X plus 2 the bottom one would just be

y equals well zero so all of that DX and

then we would add to that the integral

from 0 to 4 again the function on top is

y equals 2 minus the function on the

bottom would be square root of x and

then we could add those together you

know it wouldn't be the end of the world

to compute it that way if we look at it

with respect to Y you know if I were to

integrate this with respect to Y to me

it's now it's always the same curve on

the top always the same curve on the

bottom and that to me suggests hey we

could actually set it up with respect to

Y and make it a single integral but you

might make it hopefully a little bit

easier to compute so again to do that we

do the right - the left curve and as we

said before we have to make these curves

we have to write them in the form x

equals x equals okay so you know we've

got y equals square root of x well we

could just square both sides so that we

give us x equals y squared so we'll use

that version and then y equals x plus 2

well equivalently we can just rewrite

that and say y minus 2 that would equal

X all right so now if we do the

rightmost curve well the rightmost curve

would be x equals y squared minus the

leftmost curve which would be x equals y

minus 2 so this is what we're

integrating with respect to Y the

smallest y-coordinate that gets used is

0 the largest y-coordinate that gets

used is 2 so now this is the function

that we're integrating so 0 2 2 y

squared M is going to distribute that'll

be minus y plus 2 dy so now if we

calculate our antiderivative we'll get Y

to the third over 3 minus y squared over

2 plus 2y and then again we're

evaluating all of this from 0 to 2 so if

we plug in 2 2 cubed is 8 over 32

squared is 4 over 2 so we'll get 2 2

times 2 will give us a positive 4 the

lower limit of integration when we plug

in our 0 will just be a bunch of zeros

so it looks like we're left with 8 over

3 so negative 2 plus 4 will give us

positive 2 over 1 well we could multiply

top and bottom by 3 so that looks like

that would give us 8 plus 6 so 8 plus 6

would be 14 over 3 and that would be the

value for our area in this case

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding Area by Integrating with Respect to Y.  In this video, I find the area of a shaded region by integrating with respect to y.  We could have also integrated with respect to x to find the area, but it would involve more than one integral to do so.


so in this question we want to know

which of the Fallen grass best shows a

strong negative association between D

and T and to have a strong negative

association it basically says is one

quantity goes up the other wing goes

down and for it to be strong we want the

dots to be bunched close together and

again you can think about negative as

being like a negative slope you know so

a line that has negative slope is

pointing down as you move left to right

so answer choice a they're all kind of

spread out there's not really you know

any sort of strong association at all

between the dots everything is kind of

all over the place answer choice B you

know that has a positive association but

again the dots are slowly trending

upwards same thing with answer choice C

answer choice D would be the correct

answer because again the dots are all

very close together and as the value on

the horizontal axis as the values of

little D get bigger the vertical axis T

those values are getting smaller smaller

smaller and again you can thank you know

if I tried to put a line you know if I

tried to fit a line somehow through

those dots that I'd probably make it

look something like that you know

roughly and again this line has negative

slope and the dots are very close to it

so i probably could have made a slightly

better looking line there but again

that's the basic idea you want them to

be bunched close together that gives it

a strong association and again you want

them to the dots to basically move down

as you move to the right so that gives

you the negative association

SAT Math - Question 5, Section 2, 2016 Edition and Beyond

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Section 2 - A Calculator IS allowed!


All right, in this video I'm going to do an example
of finding the equation for a hyperbola, and again,

kind of our two, sort of, standard forms. We can
actually write this a little more generically.

So I'll point that out here in this example.
So we're going to find the equation for this

hyperbola here, you know, in the bottom left. We
can write the equation of a hyperbola a little

more generically by saying x minus h squared over
a squared, minus y minus k squared over b squared. H

and k is going to be the point where your
asymptotes are intersecting, okay? It's going to be

the point equidistant from the vertices. So to me
it looks like the vertices here are at 1, 2, 3, it

looks like we have-- one of our vertices will be
at 4, 0, and the other one will be at negative 4, 0.

Okay, it looks like our asymptotes here are
crossing then at the point 0, 0. So that means

my h, k is going to be equal to 0, 0 so I can fill
that in at any point. Remember, a, that's going to

be the distance from your. . . from where the
asymptotes cross, over to one of the vertices. So to

me, in this case, it looks like our a value is going
to be an a value of positive 4. Again, we're just

simply moving 4 units to the right to get to
that vertice. So a is four, and the last thing

I'm going to use is remember our asymptotes are at
positive and negative for this type, the ones that

open to the left and to the right, we use positive
and negative b over a, x. But we know our a value

in this case, we've already figured it out, that's
4. We just need to figure out our b value. Well,

notice in this-- our asymptote goes through
the points 2, 1 and 4, 2. Okay, so you go up 1 over

2, up 1 over 2, so to me it looks like the slope
of this line, at least this, this asymptote with

a positive slope, to me it looks like this line
has a slope of positive 1 over 2. Ok. So again,

up 1 over 2, up 1 over 2. So our slope has-- is a
positive 1/2, so our asymptote is going to be

1 over 2x. Well, I can rewrite this asymptote y
equals 1 over 2x, I could get a denominator of

4, and if I multiply top and bottom by 2, I would
get, well, 2 over 4, x, as again this asymptote. Since

I know that my a value has-- my a value is
positive 4, well then now this numerator must be

our b value. So it says our B value is going
to equal 2, and now we've got everything we

need to fill in the equation for our hyperbola.
So it says we'll have x minus 0 squared over a

squared, minus y minus 0 squared over b squared,
which is 2 squared, equals 1. And you could always

multiply this out. It says our equation would be
x squared over 16, minus y squared over 4, equals 1.

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 Equation for a Hyperbola Given the Graph - Example 1.


all right in this video we're going to

look at an example related to The

Squeeze theorem and uh we're going to

use the squeeze theorem to evaluate

these two limits so we'll do U part A

here first so we want to find the limit

as X approaches one of F ofx and we're

given the information that f ofx is

always greater than or equal to 4 and is

always less than or equal to x^2 + 6 x -

3 for all values of

X um usually kind of the tricky part of

the squeeze theorem is taking your

function and bounding it below by some

function in this case a constant

function and bounding it above by

another function here that's given to us

so it's going to make life a little bit

easier in Part B for example though

we're going to have to have to come up

with the uh kind of the smaller and the

bigger and that'll make it a little bit

trickier but um in this case if you're

given all this information the idea is

we can just take the limit of each part

so if we take the limit as X approaches

one of the number four that's going to

be well less than or equal to the Limit

as X approaches one of our function f

ofx and that's going to be less than or

equal to the Limit as X approaches 1 of

x^2 + 6 x -

3 well the limit of a constant is just

the constant so the limit as X

approaches one of the number four is

going to be

four um again so that's going to be less

than or equal to the Limit as X

approaches which is 1 of f

ofx well again to evaluate this limit um

x^2 + 6x - 3 again that's a nice little

Parabola it's continuous everywhere so

all we have to do is simply plug in our

value if we plug in one we would get 1 2

+ 6 * 1 -

3 so again we've got four is less than

or equal to the Limit as X approaches

one of F ofx and well let's see so 1^ 2

is 1 6 * 1 is going to get us uh so 1 +

6 is 7 7 - 3 is 4 okay well it says the

limit as X approaches one of f ofx has

to be greater than or equal to 4 but it

also has to be less than or equal to 4

well that would imply that the limit as

X approaches one of our function is

simply going to equal the number four

okay so again what you're kind of doing

is you're deducing information about uh

the limit of some function by knowing

information about you know a function

that's smaller than it and a function

that's larger than it so uh kind of an a

indirect way to get information about

the original function

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 for Limits, Example 2. In this video, I work a problem involving limits using the squeeze theorem.  This is an 'easy' squeeze theorem problem since the 'small' and 'large' function are both given.


okay here we're going to find the volume

of a

cylinder so to figure out the volume

again what I'm going to do is find the

area of one of the bases and then simply

multiply it by the height well to figure

out the the area of one of the bases

well since it's a cylinder that will be

a circle so the the area of one of the

faces would be pi times

the pi * the radius

squared and then we would multiply that

by the height which would be 15 but in

this case again we're approximating Pi

with

3.14 so now just a little bit of

arithmetic so

3.14 4^ SAR will be 16 and then we have

to multiply that by

15 so let's see here uh 3.14

multiplied by

16 multiplied by 15 I'm getting this to

equal

75 7 7 let me say that again

7536 and again we're using centimet so

our units will be centimet cubed or

cubic cenim

Volume of a Cylinder

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Volume of a Cylinder.  Here we look at a simple example of finding the volume of a cylinder.


okay here we're gonna look at some

examples involving integrals again with

secant and tangent and actually you know

Part A I already you know take that back

because Part A here doesn't have you

know tangent or secant in there but to

integrate this notice you know we do

have the cube root of okay so cotangent

X we'll leave that alone we really have

a 1 over sine squared and 1 over sine

that's cosecant so really we have a

cosecant squared X hanging out here as

well so kind of the moral of the story a

lot of the tricks you know that work

with the secant and tangent will work

with cosecant and also cotangent so just

keep that in mind I think this one's one

where we can do with just a nice little

substitution here we can let u equal

cotangent of X our D u will be negative

so let's see the derivative of tangent

secant squared cotangent is negative

cosecant squared X will tack on our DX

again I kind of just want a positive

cosecant squared so I'm going to

multiply both sides by negative 1 and

now I think we've got what we want so

I'm just gonna start plugging everything

in here so our cosecant squared a

cosecant squared X DX that's gonna be

our negative 1 D U and then the only

thing we would have left would be the

cube root of cotangent so we would have

you you know underneath the radical and

I'm going to go ahead and write that

it's just you to the 1/3 power alright

so if we integrate we'll add 1 to the

exponent so that'll give us 4/3

we'll divide by 4/3 or equivalently

multiplied by 3/4 plus C and again now

all we have to do is just plug in our

value so let's see let's go ahead and

just make that negative 3/4 and then our

value for u is cotangent of X raised to

the 4/3 power plus C and that will be

the antiderivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Integrating Even and Odd Powers of Cosecant and Cotangent, Ex 1.  Here I look at another random example of integrating a function involving different powers of the trigonometric functions.


all right so another example using sum

and difference identities to simplify an

expression and here we're going to

evaluate tangent of x + 4 Pi so we're

just going to uh you know simplify that

down a little bit so uh we've got our

identity here it says tangent of x + y

it says if there's a plus sign we use

tangent of X Plus tangent of Y in the

numerator but then in the denominator

you kind of take the opposite sign we'll

use 1us tangent x * tangent y so in this

case well we would simply get tangent of

X Plus

tangent of 4

pi over one but let's see so since it

was a plus originally it switches to a

negative in the denominator and then

we'll get tangent of X time tangent of 4

Pi but tangent of 4 Pi um tangent of 4

Pi that's just going to be S of 4

pi over cosine of 4 Pi well s of 4 Pi is

going to equal 0er cosine of 4 Pi is

going to equal 1 so tangent of 4 Pi

simply reduces to zero so really we're

just left with Tangent X Plus 0 in the

numerator so I'm just going to leave

that out uh well let's stick it in there

one time uh plus Z then we'll have 1us

tangent X

times tangent of 4 Pi which again is

going to give us zero well all we're

left with in the numerator is tangent X

in the denominator we're simply left

with one so that expression actually

simply reduces down to just tangent of x

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 to Simplify an Expression, Example 2.  In this video, I simply show how to simplify some expressions by using sum and difference identities.  Nothing crazy, just showing how one must know identities to make things more manageable!  For more free math videos, visit http://PatrickJMT.com


all right in this video we're going to

do an example of solving an investment

problem so pretty basic problem here

just involving simple interest so

suppose our friend Abby has 600 bucks

and she divides them between two savings

account suppose one of them ended up

earning 5% simple interest and the other

earned 3% simple

interest if she earned a $25 total um in

interest maybe over the course of that

year we want to know how much she earned

from each account okay so we don't know

how much she invested but we can easily

um you know we've got the 5% account and

we've got the 3% account we can assume

she

invested maybe X in the 5% account um

since she had 600 total if we take 600

minus X that would be the remaining

amount of money and that would be how

much she invested in the 3% account um

the simple interest uh

earned

so if you think about the interest

earned 5% is 05 so it'd be 05 * x 3% is

03 and it would be 3% of this 600 minus

X and this is going to give us our

equation because we know if we take the

interest earned

from the 5% account and add to that the

interest earned from the uh 3% account

we know that the total amount earned was

$25 and now we've got our equation to

work

with Okay so we've got

0.05x um if we take

0.003 * 600 I believe that should be

18 minus 03

x = 25 now we can combine 0 3x minus

excuse me 05x -3x will be

0.02x we can also subtract 18 from both

sides which will give us seven and now

if we

divide we get that x equal 350 and 350

was the amount in the 5% account because

again that's what we let X represent um

so since she had 350 in the 5% account

that would mean that she has 250 she had

invested $250 in the whoops the

account

Solving an Investment Problem

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 Investment Problem.  If you deposit money into two different accounts and earn so much from each one given a certain interest rate, how much did you deposit in each account?


all right here we're going to start

looking at solving separable

differential equations and the way that

I always reminded myself of this

procedure is we just separate and we

integrate so if I separate what we're

going to try to do is we're going to try

to put all the Y's on one side of the

equal sign and all the X's on the other

side of the equal sign so that's the

idea of sort of separating them I'm

going to separate the variables from

each other and we want the dy to be in

the numerator and we want the DX to be

in the numerator as well so Part A we've

got dy over DX equals x over y squared

well what we're going to do is we're

going to multiply the DX over to the

right side so we can multiply both sides

by DX based on cross multiplying that's

what I'm doing maybe that's the easier

way to think about it so x times DX

that's we're getting on the right side

we'll get dy times y squared or Y

squared dy so now we've separated and

the next thing I'm going to do is to

integrate okay so if we start with two

things that are equal we can integrate

both sides well the antiderivative of Y

squared will be Y cubed over three we

can certainly put the plus C wherever we

want to the antiderivative of the right

side will be x squared over two I'm just

going to stick the plus C on the right

side you don't have to put constants on

both sides if you wanted to as well now

we can even solve for y in this case so

we can multiply both sides by three so

three times x squared over two plus C so

three halves x squared plus three C

again that's going to be Y cubed and now

to get the Y by itself we could simply

take the cube root of both sides so if

we have Y cubed equals 3 over 2x squared

plus three C I'm just going to take the

cube root of both sides so on the left

side we'll be left with Y on the right

side we've got our

cube root so with the cube root of 3

over 2x squared plus 3c and that would

definitely be a solution now sometimes

you won't see like this 3c and I think

this is something that can be a little

misleading sometimes what you'll see

instead I equally valid okay also valid

would be to say you know suppose instead

of labeling this as a you know see

suppose I call it this little C sub 1 C

sub 1 and C sub 1 C sub 1 and C sub 1 so

okay it's just some generic constant I

could say okay let's let C equal 3 times

C sub 1 so I'm just replacing this

generic constant 3 times C sub 1 with

just the value C and this would also be

equally valid okay we could say either

one of these are going to be solutions

to the original the original

differential equation so sometimes

you'll see it written either way so

don't be thrown off by that this is

basically all that's happening probably

sometimes I'll leave it alone and

sometimes I'll change it but again it's

correct to say either one

Solving a Separable Differential Equation, Another Example #1

(https://youtu.be/nNHlSB6b1HU)
(https://youtu.be/XExEixAPK6s)
(http://www.youtube.com/watch?v=XExEixAPK6s)
Separable Differential Equations | Solving Step-by-Step
In this video, we tackle the problem of solving a separable differential equation:   dy/dx = x/y^2
We walk through the process of rearranging the equation to separate the variables and then integrate to find the general solution. Separable differential equations are a fundamental concept in calculus, and this lesson provides clear steps to help you solve these types of problems with confidence.

What You Will Learn:
How to recognize and solve a separable differential equation.
Techniques for separating variables in the equation.
Integrating both sides to find the general solution.
Great for anyone studying differential equations or looking to improve their problem-solving skills in calculus!

Support me on Patreon: https://www.patreon.com/patrickjmt?ty=c

#SeparableDifferentialEquations #Calculus #DifferentialEquations #PatrickJMT #SeparationOfVariables #MathTutorial #CalculusHelp #MathSolutions #AdvancedCalculus #YouTubeMath

Keywords: separable differential equations, solving differential equations, calculus differential equations, separation of variables, integrating differential equations, PatrickJMT, calculus tutorial, differential equations examples, math problem solving

You said:


all right so one more example here of

solving a separable differential

equation so part C we've got Dy DX =

cosine x - one I'm going to do the same

thing um as on the other ones I'm just

going to cross

multiply so let's see we would have Dy *

the quantity Yus one and I'm going to

put the Dy on the right

side or excuse me well to the right of

the y- one um on the actual right side

of the equal sign we'll have DX * cosine

X well again now what we're going to do

we've got all the Y's on one side all

the x's on the other we can just find

the anti-derivative of both sides so on

the left we'll be left with y^2 over 2 -

y on the right side the anti-derivative

of cine X will be sinx again I'm going

to stick our plus C on that side and

again you know to solve this for y um

you know our very first example we were

able to take a cube root eventually and

solve for y again here I'm not sure

exactly how to solve for y and again

sometimes you simply can't do it um so

here I'm just going to leave it alone

and say our solution to the original

differential equation would be y^2 uh

y^2 2 - y = sinx + C

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 Separable Differential Equation, Another Example #3.  In this video, I solve an example of a separable differential equation.


all right in this video I just want to

do another example of solving a linear

inequality so in this case we have X

minus 6 divided X minus 6 divided by 2

greater than 4x divided by 7 so a couple

of different ways we can start about

going about doing this problem to me

usually one thing I like to do again my

own personal preference you know the

thing I don't like about this problem or

the fractions is a - there's a 7 the way

that we can get rid of that is we can

multiply by the least common multiple of

those numbers 2 & 7 so again the least

common multiple says what's the smallest

number that is divisible evenly by both

positive 2 and positive 7 well I think

the first number that fits that bill

would be the number 14 so what I'm going

to do is I'm going to multiply the left

side by 14 and the right side also by 14

so you can really think about that as

being 14 over 1 so be careful 14 over 2

is 7 we're still going to have to

distribute that number to everything in

the parenthesis but the 14 over the 2

becomes a 7 we've got our 4x on the

right and again 14 divided by 7 is going

to be 2 and now this to me is a little

bit better I'm just going to distribute

things out like normal so I'll get 7 X 7

times negative 6 is negative 42 4x times

2 is 8 X and what I'm gonna do in this

case again I like to keep the X as

positive so I think I'm going to

subtract 7x from both sides and then

I've got negative 42 is greater than 8x

minus 7x is just positive 1x or again

equivalently I could write that as X

less than negative 42 says the exact

these inequalities are the

same say the exact same thing all my

inequalities all my problems have been

less than some negative numbers so just

a coincidence there so since it's got to

be strictly less than negative 42 I've

put a little open circle at negative 42

if I want to write this this solution

and interval notation we can write that

as parentheses negative infinity up to

but not including negative 42 and we'll

put our parentheses around it

An Introduction To Solving Linear Inequalities - Example 3

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okay here I'm going to graph two periods

of y = 3 * cent of

12x and we'll pick out the vertical ASM

tootes and zeros the only thing that's

changed on this compared to regular

cotangent is the period has changed so

to figure out the new period we'll take

pi and divide it by the absolute value

of B which is 12 that'll give us 2 pi

and since it's multiplied by three it's

stretched vertically by this factor of

three but otherwise it looks

just like regular Cent soent has a

vertical ASM toote at the Y AIS or x

equals

0 to get

to another vertical ASM toote will'll

move over the period of 2 pi and

likewise if we go to the

left that'll be the line x

= -2 pi if you look between say 0 and 2

pi the point right in the middle at Pi

that's where Co cotangent will hit zero

and since it's multiplied by three it's

just going to get stretched vertically

by that factor of three likewise at Nega

Pi it's going to have the exact same

shape and there will be two periods so

the vertical ASM tootes occur

at well at x = 2 pi and then we add 2 pi

to get to the next one so we're really

just looking at multiples of 2 pi so the

vertical ASM tootes occur when x = 2 pi

n where n is an

integer and the zeros

occur when x equals well we can pick one

of them pi to get to the next one we

would have to add 2 pi because that's

the period so the zeros will occur when

x = < + 2 piun N again where n is is an

integer

Graphing a Cotangent Function, EX 1

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I graph two periods of the function but it obviously repeats in both directions!


okay in this example we're going to look

at finding the volume of a sphere and a

separate one we'll find the surface area

so the volume is simply 4/3 * piun * the

cube of the radius the surface area is

going to be 4 Pi * the square of the

radius so nothing too

terrible so we're going to find the

volume of a sphere with a radius of 6.2

M both in terms of Pi and also so by

approximating pi with

3.14 so it says the volume is 4/3 time

Pi time the cube of the radius so we'll

have to take

cubed so let's see

um 6.2

cubed I'm getting that to be

328 and I think I'm going to start

rounding this to the nearest 100th 43 is

1.333 repeating so this is now going to

be approximately equal to so 4/3 we'll

approximate that as

1.33 we still have Pi in there and then

we've got 238 point if we round to the

nearest 100th we'll have

33 so let's see um

1.33 *

23833 I'm getting again I'm going to

round so

3169 I'm getting 978 so I'm going to

make that 98 and then we can multiply

that by pi so I'm going to make my

little equal signs a little curvy just

to indicate that we're approximating so

that would be our solution in terms of

Pi

well if we actually uh change that if we

change Pi into uh

3.14 let's see so

31698 multiplied

3.14 I'm getting this to equal

3172 and again if we round to the

nearest hundredth because of the seven

we'll have to round the uh the hundreds

digit up one so we'll have

99532 and again since we started our

original units were in terms of

meters it says our units will be in

terms of cubic meters since we're

finding a volume a threedimensional a

threedimensional volume here

Volume 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 !! Volume of a Sphere.  Here we look at a quick example of finding the surface area of a sphere.


all right so one more example here about

solving logarithmic equations and in

this case we're going to solve log base

7 of X plus 4 minus log base 7 of X

minus 4 equals log base 7 of 5 so the

first thing I'm going to do is rewrite

the left side is a single logarithm and

we can write that by saying log base 7

of the first part and since it's

subtraction this will give us division

on the right side we just have log base

7 of 5 and at this point the thing I

recognize is hey you know that we've got

log base 7 of something log base 7 of

something the idea is we can just set

the stuff inside equal to each other so

this says X plus 4 over X minus 4 equals

5 and to simply get rid of the X minus 4

I'm going to multiply both sides by X

minus 4 so on the left side we would be

left with our numerator x plus 4 on the

right side I'm going to distribute that

would give 5x minus 20 and now I'm

simply trying to solve for X we can

subtract X from both sides that would

leave me with 4 on the left and now for

X on the right minus 20 then we can

simply add 20 to both sides so if we add

20 to both sides that will give us 24

equals 4x and if well if we divide both

sides by 4 we'll get x equals 6 and now

again we have to go back and check so I

go back to the very very very original

problem I was looking at and I think

well if I plug 6 inside of here I'm

going to get the log base 7 of 10 so

that's okay

log base 7 of 10 that's a positive

number so that that number we can

evaluate we would get log base 7 of well

6 minus 4 would be 2 so that's still

good because we have a logarithm of a

positive number and well does that equal

log base 7 of 5 yea because remember

we've got subtraction so we can turn the

left side into division we can make that

log base 7 of 10 over 2

which is what we want that would reduce

to log base 7 of 5 so in fact we have

found the one solution to this

logarithmic logarithmic equation

Solving Logarithmic Equations - Example 2

Solving Logarithmic Equations Using Properties of Logarithms
In this video, I'll walk you through solving a logarithmic equation by applying properties of logarithms and using some algebra techniques. We'll specifically use the formula logA - logB = log(A/B) to combine and simplify the logarithmic terms, allowing us to solve the equation effectively. This example will reinforce how to manipulate and solve equations involving logarithms.

What You Will Learn:

How to use properties of logarithms to combine and simplify terms.
Applying algebraic techniques to isolate variables and solve the equation.
Understanding the role of cancellation properties in solving logarithmic equations.
A practical example that demonstrates each step clearly.
📚 Check out my book: 1001 Calculus Problems for Dummies for more practice!

👍 **If you find this video helpful, please like, share, and subscribe for more math tutorials!

Support My Work:

If you'd like to support the creation of more math content, consider becoming a patron on Patreon: https://www.patreon.com/patrickjmt

:#Logarithms #SolvingEquations #MathTutorial #PatrickJMT #Algebra #Mathematics #LogProperties #LogarithmicEquations #MathHelp #Precalculus #AlgebraTechniques


all right in this video I want to talk

about finding the X intercepts and the Y

intercepts of an equation so

um a graph may or may not have any X

intercepts it may or may not have any y

intercepts if it's going to be a

function technically we should say

there's just a y intercept okay the

basic idea to find the Y intercept what

we do is we just plug in xals 0 and then

solve for y this is usually pretty easy

I think it's just arithmetic

second part's a little harder if you

want to find the X

intercepts we then plug in yal 0 and

solve for x and this usually you know

provides some sort of equation that you

have to work with so let's do two

examples here real quick um suppose we

want to know what are the X and Y

intercepts of the function y = 2 * x -1

* x + 2 * x -

3 and then in our other example down

here we'll have y = x^2 - 3x - 18 so to

find the Y intercept again we just plug

in X = 0 and if we plug in X = 0 we'll

get Y = 2 * well 0 - 1 would leave us

with - 1 0 + 2 would leave us with pos2

0 - 3 would leave us with -3 so let's

see 2 * -1 is -2 -2 * 2 is -4 -4 * -3 is

pos2 so it says the Y intercept is at at

positive 12 and we could also simply

conclude that the point

012 is on the

graph okay so not too bad so to find the

X intercepts do it down

here well now we substitute in y equal 0

and if we plug in y equal 0 in our

equation we'll have zero on the left we

have 2 on the right * x -1 * x + 2 * x -

3 the good thing is is that all of uh

the right side is factored Okay so we've

got two * some number time some number

time some number if we're going to get

zero on the right what would have to be

true is either x -1 would have to equal

zero or x + 2 would have to equal Z or x

- 3 would have to equal 0 well if I

solve the first equation I can add one

and get x = 1 the second equation I'll

get x = -2 and for the third equation

I'll get x = 3 so it actually says this

original function has three x intercepts

at 1 -2 and 3 or we could write them as

points as 1 comma 0 -2 comma 0 and 3

comma

0 so um and this other problem yal x^2 -

3x - 18 I'll do that in another video

same idea um we're going to have to do a

little extra step of factoring though

X-Intercepts and Y-Intercepts of a Functions and Finding Them! Example 1

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Alright in this video we are going to look at finding components of vectors along another vector

and projections of vectors onto other vectors

so to figure out the component of b onto a

we take the dot product a dot b and we divide that

by the magnitude of vector a

and then

to figure out the projection of vector b along

vector a, we take a dot b over a dot a

and multiply that by vector a

equivalently we can write that as a dot b over the

magnitude of a squared again times vector a

so if I leave off any of the little lines again those indicate vectors

everything I am talking about here is vectors so hopefully I wont accidentally do that.

Okay so what we want to do is we are going to find both the component of b along a

and also the vector projection of b along a for these given vectors.

alright so again to do the component

so to figure out the component of vector b along a

again we take the dot product a dot b

and then we divide that by the magnitude of vector a

so to do the dot product again we multiply respective

components so 1 times 2 will be 2

We will take 2 times -1

that will be -2

then we will add to that 3 times 4

which is going to be 12

We will divide that by the magnitude of

vector a, well the magnitude we just square the components

so 1 squared is 1

2 squared is 4

3 squared is going to be 9

and now we can just simplify this so

2 plus -2 is 0, we will have 12 on top

in the denominator looks like we are going to have

the square root of 14 so that is our component

of vector b along vector a

to do the projection of b along a

again the formula for that we just said was

a dot b and I am going to use this slightly different instead of doing a dot a

I am going to take the magnitude squared because we have already figured out the magnitude

and then we will multiply that by vector a

well again we figured out what a dot b was that is what we had in the numerator a second ago

thats just going to be 12

we can take the magnitude of vector a and square it. Well again the denominator

was the magnitude so we will just take the square root of 14 and square it.

and that will just leave us with 14

and then we multiply that by our vector a

so again vector a had components 1, 2, 3

well lets see I guess this reduces to 6 over 7

and again we are just going to multiply

so recall when you have a constant out front we just multiply each component by that number

so we will get 6 over 7 times 1 which will just be 6 over 7

we will get 6 over 7 times 2 which will give us 12 over 7

and then  we will multiply 6 over 7 times 3

that will give us 18 over 7

and now we have our vector projection.

Orthogonal Projections - Scalar and Vector Projections

Scalar and Orthogonal Projections: Quick Example

In this video, we work through a quick example of finding both the scalar and orthogonal projections of vectors. Specifically, we'll cover:

How to compute the scalar projection.
How to find the vector projection.

Using the vectors: a=⟨1,2,3⟩andb=⟨2,−1,4⟩

I will walk you through the calculations step-by-step, explaining the key concepts involved in projections and how they relate to geometry. By the end of the video, you'll have a solid understanding of how to compute both the scalar and vector projections of one vector onto another.

What You Will Learn:

How to compute scalar projections
How to compute vector projections (orthogonal projections)
Understanding of the geometric significance of projections

Make sure to leave a comment if you find the video helpful, and don't forget to like, share, and subscribe! Also, let your friends, teachers, or parents know about this channel if you think they would benefit from these lessons.

Support me on Patreon if you'd like to see more content: https://www.patreon.com/patrickjmt?ty=c

#VectorProjections #ScalarProjections #OrthogonalProjections #MathHelp #LinearAlgebra #DotProduct #Projections #PatrickJMT #MathTutorials


all right so just one more problem here

involving fractions and in this case

just a little word problem nothing too

crazy hopefully so one day we've got a

plant and it's three and three-fifths of

a centimeter tall two weeks later it's

three and seven tenths of a centimeter

tall we want to know how much did the

plant grow during those two weeks well

the growth is going to be the final

height which is three and seven tenths

and we would have to subtract away the

initial height three and three-fifths

again of a of a centimeter in both cases

well again certainly the 3s are gonna

cancel out so this is gonna be the same

thing as seven tenths minus three-fifths

and again now we have to find a common

denominator so I look at my largest

denominator which is ten I think does

the other denominator is the other

number divisible will that go into ten

as well let me say it that way well ten

is divisible by five and if we multiply

by two we're gonna get our same

denominator of ten so we have seven

tenths - it looks like six tenths well

seven minus six is gonna be one over ten

so it says the plant has grown one tenth

of a centimeter so not too much a very

very slow growing little plant there

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 Fractions / Mixed Numbers: Example 4.  Here we look at another subtraction problem involving fractions.


okay and this question we're given the

following so it says if 3x minus y

equals 12 we want to know what's the

value of 8 raised to the power of X

divided by 2 raised to the power of Y so

we have the answer choices 2 raised to

the power of 12 4 raised to the power of

4 8 raised to the power of 2 or lastly

Part D it says the value cannot be

determined from the given information so

this is one that I think you could

always start plugging in value no make

some values of x and y that give you 12

and simplify it let's do it the

algebraic way though so it's kind of

show I think what they're what they're

going for here so if we look at 8 raised

to the power of X divided by 2 raised to

the power of Y the first thing I think

the first key observation is noticing

that 8 is a power of 2 right we can

write 8 as 2 raised to the third power

because 2 times 2 times 2 times 2 is 4

times 2 will give us 8 so again 8 is

equal to 2 raised to the power of 3 so

I'm going to substitute that in so I can

have 2 to the third raised to the power

of X that's what I'm replacing my 8 with

again this is 2 raised to the power of

3x over 2 to the Y okay so hey you know

I'm seeing this 3x and this Y and that

looks you know I see this 3x minus y in

my original this original equation and

now we can use properties of exponents

and we can just subtract we can write 2

to the power of 3x divided by 2 to the

power of y as 2 raised to the power of

3x minus y and well now we know we've

got we're given this information that 3x

minus y equals 12 so we're simply left

with 2 raised to the power of 12 and

that will be our solution so answer

choice a is the correct solution you

could also do something like the

following you know again if you had to

if you weren't sure about this

you could always for example you know if

you let X equal 4 and y equals 0 right 3

times 4 minus 0 will give you 12 and

then you can substitute in you would

have 8 to the 4th over 2 to the 0 and

again you can reduce 8 to the 4th again

you have to make this observation that

you can write 8 is 2 to the power 3 but

again you could arrive at this same

exact answer by doing it that way as

well

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 14, 2016 Edition and Beyond.
Section 1 - No Calculator allowed!


all right in this example we're gonna

start looking at some some problems

involving division of two numbers so

kind of the basic idea is in all of

these as well what we're gonna do is

we're going to rewrite things and we're

gonna get rid of the decimal place and

whatever we're dividing by so to start

off by doing that well we can write 94

divided by 0.25 as you know just with a

division instead of a you know the

division sign we can use a little

division bar what we're gonna do is

we're gonna get rid of the decimal place

in the denominator so what I would like

to do is move it two places to the right

and then in a sense it would be gone

well I'm gonna do that by multiplying by

an appropriate number so to move it two

decimal places

well let me say if you multiply by 10 it

moves at one decimal place if we

multiply by 100 it'll move the decimal

place twice but of course if we do it in

the denominator we also have to do it in

the numerator so when we do our

arithmetic

well 94 times 100 is ninety four hundred

and then again we said when we multiply

0.25 by a hundred that's just going to

move the decimal place two places so

okay equivalently we have ninety four

hundred divided by 25 and now I'm just

going to do my old long division so

let's see I always think about quarters

and money so let's see so I guess that

we could do three times 25 that would

give us 75 so let's see 94 minus 75

let's see if it was 95 we would get 20

but it's one less so I guess we'll get

19 then we'll drop down our zero so

let's see I think if we had seven

quarters seven times 25 would be 175

okay and again we can just do our

subtraction so okay we've got 190 minus

175 that'll leave us with 15

and again if we drop down our 0:25 will

go into 150 exactly six times six times

twenty-five again is 150 so we don't

have any remainder so it says if we do

94 divided by 0.25 we'll be left with

Dividing Decimals, 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 !! Dividing Decimals, Ex 1


all right in this video I'm going to do

an example of solving a linear equation

um we've actually got three here but

we'll I think I'm going to break these

up into three separate parts uh so in

the first one we have negative uh the

quantity -3 + x + 4 2 * the quantity x +

3 - 6 x + 7 to me you know when you

solve uh linear inequalities the first

thing I do is get rid of all the

parentheses um so I'm going to

distribute the negative so negative ofg

-3 would be positive 3 and then we would

get our X plus 4 is just hanging out

we'll have to distribute the two to both

terms so we'll get 2 * X and then um 2 *

3 will be POS 6 we've got our NE -6x and

our

pos7 and usually I try to just simplify

everything on each side first so 3 + 4

will give us a postive

7x is still just sitting there um on the

right side we have 2x - 6X which is

-4x we have POS 6 + 7 which is going to

give me

positive3 and now my goal is I want to

put all the x's on the same side and all

the numbers on the other so I think what

I'm going to do here is I'm going to add

to move my negative -4x over I'll add 4X

to both

sides so then I've got seven on the left

x + 4x will give us positive

3x -4x + 4x is 0x and that's why we did

it we still have our positive 13 left

over and now um I'm going to move the

the seven over by simply subtracting

seven from both

sides um we'll be left with just our 3x

on the left 13 - 7 is going to give us 6

and now last but not least to get rid of

the the three since we're multiplying by

three to undo that I'm going to divide

both sides by 3 and that'll leave me

with 1X = pos2 and that'll be our

solution and again you can always check

that this is a solution by plugging two

into the original equation doing all the

arithmetic on the left all the

arithmetic on the right and verifying

that you do in fact get uh a number uh

the same number on each side of the

equal sign

Solving Linear Equations - 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 !! Solving Linear Equations - Some Basic Examples


okay in this video we're going to look

at some more examples involving implicit

differentiation so in Part A here we're

going to find the derivative 4x to the

negative 2nd minus Ln of Y to the third

equals 7

so okay I'm just going to rewrite it

down here one time I think one thing I'm

actually going to do first in order just

to make the derivative a little bit

easier I'm going to use these properties

of logarithms to bring the exponent out

front so we can write this as 3 times

the natural logarithm of y equals 7 and

now I will take the derivative of both

sides with respect to X so if we take

the derivative of x to the negative

second we'll get negative 2x to the

negative 3 and then when we take the

derivative of the natural logarithm of Y

we'll just get 1 over Y but then we have

to multiply by our dy/dx and then on the

right side the derivative of 7 is just

going to give us 0

so now all we have to do is just solve

for dy/dx I think I'm going to do one

thing first I'm going to write this as

negative 2 over X to the third minus 3

over Y dy over DX equals 0 and what we

can do at this point is add the negative

2 over X to the 3rd we'll add a 2 over X

to the 3rd to both sides so that'll give

us 2 over X to the 3rd on the right side

and then we're just left with negative 3

over Y dy over DX and now all we have to

do to get rid of the negative 3 over Y

we can just multiply both sides by Y

over negative 3 so Y over negative 3 and

then on the left side all that stuff a

just cancels out we'll be left with our

dy over DX and then on the right side

we've got 2y over negative 3 X to the

3rd and now we found our derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Implicit Differentiation - Basic Example 1 / 3 .  Just a fairly straight forward example of finding a derivative using implicit differentiation.


all right in this video I want to do an

example of actually uh determining

whether sort of a a more complicated uh

product of trig

functions uh is going to be even OD or

neither so remember we gave this

definition of even and odd um and I've

made a little handy dandy uh kind of

reference guide for me uh kind of

listing you know basically plugging in

Negative X and whether you get the

negative of the original or just the

original and that tells us if they're

even or odd so uh I'm going to you know

kind of refer to this and this is what

we just justified in the other

video so suppose we want to figure out

if see X time tangent X whether that's

even odd or

neither well okay we're just going to

use our definition and say we'll have to

look at F

ofx Well that's going to give us

secant ofx time tangent

ofx but if you remember in terms of

these being even odd or neither um we

said that secant ofx is really we can

replace that with just secant x all

right I'll do that and it says tangent

ofx it says we can replace tangent ofx

with just the negative of

tangent of

X well okay uh things everything's just

being multiplied you can pull the

negative back out front so we have

netive secant x time tangent of

X well uh the very original function was

see x * tangent X so really we have the

negative of that function so we started

with uh you know so we plugged in F of

negx we got out negative f ofx and again

in terms of our definitions it says if

you get F of negx if you start with that

and get out negative f ofx we would say

that function is odd so it turns out

that if you graph see x * tangent

X that's going to be an odd

function and again that just means when

you graph it it's going to be symmetric

about the origin

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Examples with Trigonometric Functions: Even, Odd or Neither, Example 2


okay in this video we're going to look

at an example related to finding

critical points based on a

graph and the main thing to remember

critical points happen uh there's a a

couple different conditions that we look

for so we either think about where the

derivative equals zero or where the

derivative is

undefined and the idea also we have to

keep in mind is that these points have

to be in the domain of the original

function so here our original graph

looks like a nice smooth continuous

function so uh All Points would be in

the domain and what I'm going to look at

now so we have kind of a few different x

coordinates you know -6 -1 posi 5 I've

got 10 listed here we want to know at

which of those x coordinates again do we

have a critical point well the

derivative is going to equal zero if you

think about the derivative equaling zero

that means the slope of the tangent line

has to equal zero well at -1 um it looks

like the slope of that tangent line

could certainly equal zero so I would

say at x = -1 that's where the

derivative equals zero so in that case

uh that would be one of my critical

points the other place you have critical

points are where the function there any

we call them cusps anywhere there's kind

of any sharp points any abrupt changes

in Direction so notice here there's kind

of a Sharp little point on the graph and

that would be a place where the

derivative would actually be undefined

so we would say at x = 5 the derivative

is undefined and those two points uh we

would say at x = -1 and at xal 5 those

would be the x coordinates uh those

would be the x coordinates that

correspond to points where a function

does have a critical point

Critical Points from a Graph

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Just a quick example of finding critical points from a given graph.
Recall that if f ' (x) = 0 or f ' (x) is undefined, there is a critical point.


all right in this video I'm going to

talk about rewriting some powers of I

and the main thing to remember kind of

is a cyclical nature about powers of I

so I went ahead and wrote some down here

so I to the 1 that's just I I mean

anything raised to itself is itself I 2

by definition we say that I * I is -1

well if you think about I to the 3D

power we can write that as i^ 2 * I we

know that i^ 2 is-1 so if we multiply

that by I we'll just get negative

I I 4th this is going to end up being

kind of one of the ones that I use

that's important to me so I to 4th we

can write that as i^ 2 * i^ 2 but that's

just - 1 * 1 and that gives me positive

1 and likewise notice if you do I to 5th

that's I 4th * I to the 1st which is 1

time I you get I hey we're back where we

started Okay so

again one of the things that's important

to me is if if the exponent is a

multiple of

four so for example I to the 8th or I to

the 12th or to the 16th or to the 20th

if the exponent is a multiple of four

that's always going to equal the number

one okay so that's one little trick I

use when I when I try to simplify these

down so all right so three little or

excuse me three five examples here

um again just I to kind of some big

exponents so in the first one we have I

to the -42 I'm going to write that as -1

* I to

42nd and again I'm thinking first off um

is 42 a multiple of four and well I

don't think four goes into 42 evenly but

it will go into 40 evenly it would go 10

times and then I would need an i squared

okay so what do we have here we've got

-1 notice I to the 40th that's a

multiple of four So based on what we

said a second ago that's just going to

equal the number one we still have our I

squ left over but i^ 2 is equal to the

number Nega 1 so again our I to the 40th

is turning into positive 1 and our i^ 2

is turning

into1 um and now we can just simplify it

says we have - 1 * 1 * - 1 that's simply

going to give us positive 1 as our

solution so negative I to the 42nd power

is a weird way of writing the number

one okay so I to the 31st I'm just going

to do the same trick I think does uh

four go into 31 evenly I say well no um

but I think it would go into it uh seven

time excuse me seven times which means

we could use I to the 28th power and

then if we write that as I to the 3 so

again 28 is a multiple of four it's 7 *

4 so that's just going to give us 1 * I

3r and again the way I really write this

is now this is kind of a small exponent

I just break it down so I think I to the

3 that's i^ 2 *

I but i^ 2 we said that's 1 1 * I and

that'll simply leave us with a negative

I so okay so two quick examples there um

I'm going to do the other four examples

or excuse me the other three examples if

I can count um in another video uh

really quickly

Rewriting Powers of  ' i  '  - 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 !! Rewriting Powers of  ' i  '  - In this video, I take the complex number ' i ', raise it to some different powers, and simplify!


okay in this example we want to identify

two opposite rays and the definition

here we're looking for two rays with a

common endpoint that point in opposite

directions and they form a straight line

so recall the definition of array array

is a portion of a line which just starts

at a single point and then goes off in

some particular direction to Infinity

okay so that's what we're looking for in

this

case well let's see here we've got the

ray OA and OC well OA would be this Ray

that goes uh directly directly up points

in the north Direction here and

OC um it's definitely in a different

direction but again these two rays do

not form a straight line so answer

Choice one would not be

correct uh we've got

OB and O well those share a point and

they do form a straight line going an

opposite direction so certainly it looks

like answer Choice number two here is

correct let's look at the other ones as

well just to make sure we've got the ray

OE so that would be this Ray and OB

again well those do not form a straight

line so that's why answer Choice number

three wouldn't work and number four

we've got OA and OD again the same idea

those rays do share a point but again

they don't form a straight line so

answer Choice number two would make our

our little straight line here and that

would be the correct

choice

Rays : Identify Opposite Rays ( Geometry )

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Here we identify a pair of opposite rays that all start from the same point.


all right this video we're going to talk

about inconsistent systems of linear

equations and what we're doing really is

we're just trying to determine if a

system has no solutions again what we're

trying to do is we're going to use

elimination by addition to approach

these again there's certainly other

techniques to solve these systems of

equations but this is what we'll pick on

for this problem so true or false the

following system of equations is

inconsistent we've got 2x minus 2y plus

6z equals 10 3x minus 5y plus 12 Z

equals negative 6 for X minus 4y plus 12

Z equals 18 so one thing you can do is

just simply start trying to eliminate

one of your variables just you know the

normal way you'd go about doing

elimination by addition I'm going to

look at the first and third equation so

notice we've got 2x minus 2y plus 6z

equals 10

we've got 4x minus 4y plus 12 Z equals

18 well for example if I wanted to try

to get the X's to cancel out I could

multiply my first row by negative 2 and

we would have to do that to both sides

and then what will we get well our first

row would become negative 4x positive 4y

negative 12 Z and then on the right side

we would have negative 20 if we take our

original second equation we've got 4x

minus 4y plus 12 Z equals 18 and

this if we now do our elimination by

addition if we add those together we get

zero X plus zero y plus zero Z which

says you're basically just left with

zero on the left side equals well

negative 20 plus 18 is negative two so

this says hey zero equals negative two

which is clearly not the case so this

kind of a this statement that's nonsense

is clearly not true statement it's

trying to tell you something and what

it's trying to tell you is it says there

is no solution is what that means and if

there's no solution to your system of

equations we do call that an

inconsistent system it doesn't make

sense somehow okay so so I would say yes

this is true this this following system

of equations is inconsistent one way in

this you know this this isn't exclusive

but one way you can look at it and

determine that a system is inconsistent

and this is basically the trick I used

notice that the third row is a multiple

almost of the first row I mean if you

multiply the first number by two we'll

get the you know we'll get the 4x if we

multiply the negative 2y by two we get

negative 4y if we multiply 6z by two we

get 12 Z so if I multiply the first

equation by two everything matches up

until on the right side if we multiply

10 by two we don't get 18 and if

everything matches up except for the

numbers basically what's going to happen

when you do your elimination by addition

you're going to get some nonsense

statement so something like this I could

just look at make that realization to be

like oh there's not going to be any

solutions to this system of equations

therefore it would be inconsistent

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Systems of Linear Equations - Inconsistent Systems Using Elimination by Addition - Example 1


all right so a couple more examples here

dividing whole numbers so in part a

we're gonna have three 37,200 4/6 Part B

we'll have 13,000 267 divided by 18 and

I think I'll do these in separate

examples here just because mainly

because Part B can be a little bit long

so let me rewrite it real quick

so dividing by a single digit such as

six really makes life a bit easier for

sure you'll see Part B certainly is a

bit longer so six will go into three

well zero times so I'm gonna move it

over one place also in this example like

I made them up I'm pretty sure we should

have a remainder somewhere so definitely

in Part B so six will go into 37 let's

see so I guess six times so 6 times 6 is

36 really gonna know your multiplication

tables here really 37 minus 36 will be 1

I'll drop down my next digit which is a

2 well I'll say 6 will go into 12 two

times so 2 times 6 is 12 you can

subtract get our 0 I'm gonna drop down

my next digit which pay happens to be a

0 so I would say well 6 will go into 0

zero times so 0 times 6 zero again we

can drop down our last digit which is a

4 again 6 will go into 4 zero times zero

times six zero again if we subtract the

left with four we've kind of exhausted

all the digits and in this case after

we've done that this number is what's

known as the remainder so it says if we

take three seven two

Oh for and divide it by 6 it says we'll

get 6,200 and a remainder of 4 so it

says 6 will go into 37,200 for exactly

6200 times with a little bit left over

with 4 left over

Long Division of Numbers - Arithmetic Basics, 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 !! Example 4 : https://www.youtube.com/watch?v=So9YQbSh5PI
Long Division of Numbers - Arithmetic Basics, Ex 2. In this video, I show how to make a fraction into a decimal doing long division.
For more free math videos, visit http://PatrickJMT.com


all right in this video we're going to

look at integrating some composite

exponential uh functions and the

separate video we'll look at some

rational functions so um in this one in

part A we're going to integrate here sin

of x * e to the cosine X and again when

I see something like this you know kind

of the first thing I think I I notice a

couple things I see the sign and the

cosine in there and those are

derivatives you know of each other up to

you know a constant obviously when you

take the derivative of cosine you get

negative sign but you know they're

they're they're close enough to where I

could potentially make a U substitution

work and also the only formula I really

have to integrate e is if I just have e

to a single variable you know e to the x

or e to the U or e to the whatever so

that to me suggests maybe we should let

U equal the exponent here which is

cosine

X our DU would be

sinx DX well we have a positive sinx DX

so I'm just going to multiply both sides

of this by -1 which will make this now

positive sinx DX and now we can replace

everything so it says again our sinx DX

we'll replace that with -1 du and then

we would be left with e but instead of

cosine X again that's what we're calling

U so now we're just integrating netive e

to the U well uh the negative one comes

along the anti-derivative of e to the U

is just e to the

U let's stick on our plus C and now I'm

going to replace the U with our

substitution which was cosine X and now

we've got the

anti-derivative okay so let's look at uh

the second example here as well kind of

almost the same idea you know I've got e

to some stuff um I've got to make that

into a single variable so and again I

see stuff cubed and stuff squared when I

take the derivative of that I'll get

stuff squared and stuff to the first

Power which is what we have out front

oops I left off my DX again so let's

stick that in

there so I'm going to pick youu to be

the uh exponent 2x 3r +

3x^2 that would make our DU well let's

see the three would come out so that

would give us

6x2 plus 2 * 3 would give us 6X and then

all of that times DX uh notice we don't

want you know really any Co efficients

we don't want the six in front of the x

s to the X so what I can do is we can

just Factor the six

out and well to get rid of the six you

can basically just multiply both sides

by one over

six and that that will leave us with x^2

+ x DX on the right side and now again I

think we can go ahead and just you know

drop everything into there

so so our X2 + x DX it says all of that

is going to be replaced with the 1 over

du and then again we're just left with e

to the well again our exponent was 2x 3r

+ 3x^2 that's what we're we're calling U

so again we've just turned it into a

substitution into a problem where we

just have the integral of e to the U

du so if we integrate e to the U again

we just get e to the U plus C and now

the same thing before let's just plug

our u in which is 2 x 3r +

3x^2 + C and again now we have our

anti-derivative

❖ Substitution Rule / U-sub Basic 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 !! Substitution Rule / U-sub Basic Example 1.
Just a quick example of calculating an integral using a substitution.


okay in this video I want to do an

example of adding and subtracting some

rational expressions and as with you

know normal addition and subtraction of

fractions it's it's actually a little

more tedious than multiplication because

we're going to have to go through the

extra effort of getting common

denominators and uh typically that just

makes things a little tedious so the

first thing that you always have to do

is well get a common denominator and to

go about doing that typically you're

going to want to factor the denominators

of your rational expression so x^2 - 64

that factors as x + 8 * x -

8 and then uh in our our other fraction

uh that doesn't really Factor at all so

we'll just leave that

alone and now what we have to do is we

have to think of the least common

multiple of the denominators and what I

basically do when I do this is you know

I kind of look at the very first

fraction and I look at his denominator x

+ 8 x - 8 and then I look at the other

any other fractions that are present and

I look at their denominators and I think

well what are the factors in their

denominators so here there's an x + 8

and if there was not an x + 8 originally

in my first fraction I would have to

multiply top and bottom to make that X+

8 Factor

appear well the X+ 8 is there so we're

fine then I look at my next fraction I

look at his denominator x + 8 and I

think what does the other denominators

have that this one doesn't well notice

this one has an x - 8 that's not present

in this denominator so to get our common

denominator you basically multiply by

whatever is missing so I'm going to have

to multiply top and bottom by x - 8 in

our second second

fraction and now we do have our common

denominator x + 8 x - 8 we could go

ahead and stick that back together as

x^2 - 64 and then we simply do the

arithmetic in the numerator so it says

we have 7 + 3 * the quantity x - 8 and

all we'll do is just distribute this out

and just simplify it down a little

bit so it looks like in the

numerator we have seven +

3x um if we distribute the 3 to the 8

we'll get

-4 and again that's all being divided by

x^2 -

64 and again uh we can just simplify the

numerator one more time it looks like we

have 7 - 24 since the 3x is positive I'm

going to write that term

first and then if you do 7 - 24 I

believe that would give us

-17 uh the denominator he's just hanging

out and now we have simplified our

rational expression we've combined them

into a single fraction and that's what

we wanted to do

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:  Adding and Subtracting.  Ex 1.  In this video, I do a relatively simply example of finding a common denominator and adding two rational expressions.


all right in this video I want to do

another example of solving a quadratic

equation using the quadratic formula and

in this case we're going to solve 3x^2 +

5x = 7 one thing that's important before

you use the quadratic formula is you

always need one side of the equation to

be zero so some people would just rush

in and start trying to label a b and c

but be careful um again the first thing

I always do is you want one side to be

zero so in this case we'll subtract

seven from both sides sides which will

give us 3x^2 + 5x - 7 = 0 and now I can

read off my values it says a is going to

have the value

pos3 B is going to have the value

pos5 and C is going to have the value

-7 so when we go to fill in our

quadratic formula again our netive B

plus or minus the < TK of b^2 - 4 A C

over 2 a when we do

that we'll end up getting our Solutions

as so negative of the B value which is

five plus or minus the square root it

says we take B and we Square it so we'll

have 5^

2ar - 4 * our a value which is 3 times

our C value which we said was -7

and then that's all going to be divided

by 2 * a or uh in this case 2 * 3 so now

it's just a matter again of simplifying

down this expression um and you know

just trying to reduce it if you

can so we have X = we'll have -5 plus or

minus the Square t let's see so 5 * 5 is

25 and then we have -4 * 3 *

-7 so -4 * pos3 is going to be -12 -12 *

excuse me * -7 so we have -12 * -7 will

give us a positive and then 12 * 7 is

going to give us

84 and that's all divided by 2 * a or

6 so we have -5 plus or minus the square

Ro T so let's see 20 5 and 84 so I

believe that's going to give us a

postive 109 when we add those together

and all of that's going to be divided

Again by

six and if you wanted to again we can

kind of break this down into our two

separate

statements so we've got -5 + the sare <

TK of

109 / 6 as one solution our other

solution will be -5 minus the square <

TK of

109 divided 6 and here we could start

trying to break down um 109 trying to

reduce it a little bit but I think if

you play with it a little bit um you

know maybe you'll find something that

divides into it but to me I'm pretty

sure that 109 is actually a prime number

so this isn't going to factor any

further um we're not going to be able to

reduce it in any kind of meaningful way

so to me these would simply be our two

solutions -5 + < TK 109 6 and -5 minus

the < TK of 109 over 6

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Quadratic Equations using the Quadratic Formula - Example 3


okay in this video I'm going to talk

about using What's called the method of

undetermined coefficients to solve

non-homogeneous second order linear

differential equations and this is the

basic form of such an equation a y prime

plus b y Prime + C yals some function of

x to do this we're first going to solve

What's called the complimentary equation

and basically we just replace our

function G ofx with a zero and solve

that first so this is is going to be

part one um so cuz I think it can be a

little bit of a long process um so the

first part is just the theorem here it

says basically okay we're trying to

solve um one of these non-homogeneous

second order linear

equations it says basically to get the

solution to that you have to look at two

different things it says um you can look

at what's called so it says the solution

is going to be What's called the

particular solution y of P um and also

kind of What's called the uh the general

solution so it says y of p is going to

be a particular solution of this um

original equation we're trying to solve

and Y if C is going to be a solution to

the complementary equation that we're

trying to solve okay so let's try to

make some sense out of all this um

solving this last part isn't too bad I

mean you just have to remember some of

the other formulas for um differential

equations but

hopefully not not too bad to start off

with Okay so let's solve this

corresponding um let's try to solve

this so we're going to have y

prime plus 3 y

Prime + 2 y =

X2 okay so the first thing what we're

going to do again is just turn this into

our complementary equation and we're

going to solve first off y Prime + 3 y

Prime + 2 y =

0 okay now we turn this into What's

called the auxiliary equation R 2 + 3

R um+ 2al 0 and then based on the

solutions to this auxiliary equation

it's going to tell us Solutions um to

our homogeneous second order

differential equation that we have here

okay so I think this is going to factor

relatively

easily looks like we'll just get r + 1

um * r + 2 = 0 so that means our

solution Solutions are going to be

ral1 and also Ral -2 so this is a case

where we have two distinct real roots

and remember if you have two distinct

real Roots it says basically your

solution is going to look something like

just going to be some uh

coefficient e to the first root um it

doesn't matter which one so one of the

roots time X Plus again this coefficient

C2 e to the 2 root * X so okay well we

know our Roots here all we have to do is

plug those in so our general solution is

going to be um C1 e to

thex plus C2 e to

-2X okay so this is going to incorporate

be incorporated into part of our

solution at the end so um and I think

I'm going to stop this part here in the

second part I'm going to solve the

um the particular solution part using

the method of um of undetermined

coefficients and we'll try to put it all

together into some coherent solution so

feel free to take a look at that it

should be um posted nearby

Method of Undetermined Coefficients/ 2nd Order Linear DE

Solving Second Order Linear Non-Homogeneous Differential Equations
In this video, you'll learn how to apply the method of undetermined coefficients to find the general solution of second-order linear non-homogeneous differential equations. We'll break down the key concepts of these types of equations, walk through an example, and introduce the theorem for finding general solutions. Perfect for students learning about differential equations and their applications!

(https://www.youtube.com/watch?v=XPwSR3cqGi4&feature=youtu.be)

What You Will Learn

The basics of second-order linear non-homogeneous differential equations.
How to solve the complementary (homogeneous) equation.
How to use the method of undetermined coefficients to find a particular solution.
Steps to combine the complementary and particular solutions for the general solution.
The method of undetermined coefficients is a useful technique for finding particular solutions to certain types of non-homogeneous differential equations.

This video covers the procedure step by step, beginning with the associated homogeneous equation, then applying the method to find a particular solution for the given non-homogeneous term, and finally combining them to form the general solution. By the end, you'll have a solid understanding of how to tackle these problems and find their solutions.

If you find this content helpful, please like, comment, and subscribe! Share it with friends, classmates, or teachers studying differential equations.

Support my work on Patreon: https://www.patreon.com/patrickjmt?ty=c

#DifferentialEquations #UndeterminedCoefficients #SecondOrderDE #NonHomogeneousEquations #MathTutorial #PatrickJMT #Calculus #SolvingDEs #Mathematics #ComplementarySolution #GeneralSolution #MathForStudents #MathHelp #MathematicsLearning #DEExamples #CalculusConcepts #MathEducation #CalculusProblems #LearnDifferentialEquations #Education


all right in this video we're going to

start looking at some examples of

solving trigonometric equations by

factoring and we're just going to find

the Solutions in the interval 0 to 2 pi

for these so uh let's look at 2 sin^2 =

1 + cosine X here first so um you know

initially if I see something like this

uh you know I see signs and cosiness I

don't really see you know much to factor

but we can certainly use our trig

identity so recall that s^ squ plus Co

cosine s = 1 so if we solve for sin s we

would get 1 - cosine 2

x so I'm just going to basically try to

turn this problem into I just want

everything to involve cosiness is all

I'm doing so I'm going to distribute the

two so 2 * 1 will be two we'll get -2

cosine 2

x = 1

+ cosine X

and I think what I'm going to do is um

I'm going to add the 2 cosine squ to the

right

side we've already got our positive

cosine

X I'm also going to subtract two from

both sides so 1 minus 2 would simply

give us -1 and then we'll be left with

zero um on the left side and now

hopefully um the right side factors

without too much trouble so

so I think if we use 2 cosine X and

cosine X that's certainly going to give

us our 2 cosine squ X um to get a

negative one I know I've got to use uh a

positive one and a negative one um I

think the positive would have to go uh

with our second Factor since uh when we

multiply it out we'll get 2 cosine ^ 2

plus 2 cosine x minus cosine X that

would give us a positive cosine X and

then we've got negative 1 left over so

now once we have it factored we just set

each uh each factor equal to

zero well here I can add one and divide

by two and then we'll be left simply

with cosine X um if we set our second

Factor cosine x + 1 = 0 we'll simply get

cine x = -1 and now I have to remember

um where cosine x = 12 let's see cosine

x = 1 I believe at pi over 3 that would

be one of our

Solutions and also in Quadrant 4 cosine

X would be 12 and that would be at the

angle 5 pi over 3 where does cosine x

equal ne1 well I believe cosine x equals

1 at the angle of

Pi so those would be our solutions for

uh the first equation we've now found

all Solutions in the interval from 0

to 2 pi

Solve Trigonometric Equation by Factoring, 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 !! Solving a Trigonometric Equation by Factoring, Example 1


okay here we're going to look at finding

some expressions for the remainder of a

tailor polinomial and these uh are

actually not too bad at all all we're

really going to have to do is just uh

fill in this expression with the N plus

first derivative um at Z and then just

plug in where we're centering it and

calculate a a few things so let's look

at an example here so suppose we want to

approximate uh the function e to the X

near xal 1 with the polom the tailor

polom e + e * x -1 + e over 2 * x -1

2 plus e over 6 * x -1 Cub + e over 24 *

x -1

4 okay so um in our case um so we're

figuring out the remainder associated

with a fourth degree polinomial Okay so

we've gone out to uh to degree 4 so it

says the remainder and we need to put

this in absolute value here it says

it'll be the well the n + one so the

fifth derivative evaluated at Z over

well again we'll have five factorial and

then we'll take x minus where we're

centering it which for us is at

positive1 and then all of that will be

raised to the fifth

power well in this case our function f

ofx is just e to the X well all the

derivatives for f ofx or E to X So

eventually the fifth derivative for X

well that's also going to be e to the

X well that means the fifth derivative

evaluated at Z well that's simply going

to equal e to the Z and now that's all

we have to do is is just simplify that

uh and uh plug that in and simplify the

expression so um so again the uh F the

fifth derivative of our function

evaluate it at Z that's going to be e to

the Z let's see 5 factorial I believe

that's 120 so 5 * 4 * 3 * 2 * 1 that's

20 and another six that's 120 then we'll

have x -1 raised to the 5th power and

that will now be an expression for our

remainder

📚 Finding the Remainder of a Taylor Polynomial – Example Using Taylor’s Remainder Theorem 📚

In this video, I walk through Taylor's Remainder Theorem and show how to find an expression for the remainder when approximating f(x) = e^x using a fourth-degree Taylor polynomial centered at x = 1. We show how the remainder formula applies to this specific example.

🚀 What’s covered:

Explanation of Taylor's Remainder Theorem and its importance in approximating functions.
Step-by-step calculation of the Taylor polynomial for f(x) = e^x at x = 1.
How to find the remainder using the general remainder formula and a detailed breakdown of the process.
This is an essential technique for students studying calculus and higher-level math, helping you understand the accuracy of Taylor series approximations and how to handle errors in your estimates.

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all right in this video we're going to

talk about finding a unit vector and the

unit the unit Vector for a vector

v all we do is we take our vector v and

we divide it by its magnitude or its

length and what a unit Vector is it's

simply a vector that points in the same

direction as the original Vector but uh

we're going to kind of uh alter its

magnitude or length so that it has a

magnitude or length of

one so nothing too table here let's uh

we got three examples here let's find uh

the components of a unit Vector in the

direction of each of these given vectors

so just a unit Vector here for for each

one well so we've got the vector uh 3

comma 4 with components 3 comma 4 let's

find its

magnitude um you know maybe we'll call

this again vector

v so the magnitude of vector v again we

just take each

component and we Square it we add them

together and square root it so 3^ squ is

n 4 squ is 16 that's going to give us

the square root of 25 which is simply

going to give us five when we take the

square root of

it so really um it says the unit

Vector would be this Vector uh three

with components three four but now we're

dividing by five

and remember all that happens here you

can really think we're multiplying by a

scalar which we are we're just

multiplying by

1 so we'll have

1 uh being multiplied by the vector with

components 3 comma 4 and recall uh if

you have a a real number out front all

we do is just multiply it in so we'll

get three fifths and four fifths and

that'll give us our unit Vector U again

it's going to be pointing off in the

exact same direction as the original

again but now it simply is going to have

a length of one

Finding a Unit Vector, 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 !! Finding a Unit Vector, Ex 1.  In this video I discuss the idea of a unit vector and show how to find it (divide the vector by its magnitude!).


all right in this video I want to do

another example of multiplying a couple

rational expressions in this case we

have quadratic trinomials in both the

numerators and the

denominators so again the main idea on

these types of problems is just to

factor and cancel that's usually that's

what people that's usually the gist of

the problems so I'm going to give myself

a little room and try to factor and

cancel again you know keep in mind too

that usually these problems are set up

so that things will cancel so

so for example okay uh let me Factor

Okay so we've got 3x^2 + 7 x + 2 to me

that one's the worst one I'll come back

to that one in the denominator we have

x^2 - 2x - 8 so since the coefficient on

the x^2 is a 1 remember the rule is we

basically look for two numbers that

multiply

to8 but add up to -2 well I think -4 and

pos2 would do that and then to get the

X2 we simply use an X and an

X okay so I've got an x - 4 and an x + 2

in the denominator I'm already thinking

you know there's a good chance maybe

there's an x - 4 or an x + 2 somewhere

in the numerator if things are going to

cancel out so at least it helps me you

know sometimes guess what the factors

might be so I'm going to do my x^2 - 3x

- 4 I believe that would Factor as well

so to get the X squ we need an X and an

x two numbers that multiply to4 4 but

add up to -3 I think we could use -4 and

pos1 hey there's our x - 4S showing up

x^2 + 3x + 2 I think that one's not too

bad either x + 1 x + 2 again if we

multiply we'll get pos2 and those will

add up to

positive3 um and then 3x^2 + 7 x + 2 you

could do this by factoring by grouping

this is one that I would certainly just

kind of do by trial and error assuming

it factors nicely so to get 3x^2 we'll

need 3x and X really the only good way

to get positive2 is to have a positive

one and a positive two but it definitely

matters in which set of parentheses you

put the positive one or the positive2 I

think we would need the positive2 to go

on the second Factor the positive one to

go on the first because if you multiply

that out we'll get

3x^2 + 6x + 1 x which gives us our 7 x

and then our plus

two all right and now I'm just going to

cancel everything out that I can so it

looks like we have an X+ 2 and x + 2 it

looks like we have an x - 4 and x - 4

will cancel we've got an x + 1 and an x

+ 1 and what we're left with we've got

3x + 1 in the

numerator and we've got an X +2 left in

the denominator and we have now

simplified down our original

multiplication of our rational

expressions

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 2.  In this video, I divide two rational expressions containing quadratic trinomials.


so on a Saturday afternoon Armand sent M

text messages each hour for five hours

and Tyrone sent P text messages each

hour for 4 hours we want to know which

of the following represents the total

number of messages sent by arand and

Tyrone on Saturday afternoon so we have

the choices 9 multiplied by m multiplied

by P 20 multiplied by m multiplied P 5 m

+ 4 p and then 4 m + 5 p

so you

know you could probably you know maybe

you can reason this one out without um

you know even even using some real

numbers but let's suppose that Arman

just for to give some concrete numbers

suppose Armand sent um suppose he sent

three text messages each

hour well he did that for 5 hours total

so he sent three plus another three plus

another three plus another three plus

another three for a total of 15 messages

well the way that we got that is we

would say again if he sent three

messages he sent three messages

multiplied by 5 hours so 3 multiplied 5

is going to give us

15 okay so again instead of three though

that was our the value we used for m so

we would have M multiplied 5 or 5 m

message messages total sent by Armand so

Armand sent 5 multipli by m messages and

the same reasoning Tyrone sent P

messages for 4

hours so Tyrone sent P messages for 4

hours so that means he we would take

four and multiply that by P to get the

number of messages that Tyrone sent so

the total in this case we would just add

those two values together we would have

5 m + 4 p and those are not like terms

because we have the variable M and we

have the variable P um so we can't

combine that any further but that would

be our solution in this case so it looks

like answer Choice C here would be the

correct choice

SAT Math - Question 3, 2016 Edition

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 3, 2016 Edition
Section 1 - No calculator allowed


all right so another couple examples

here involving reading and writing whole

numbers and what we're going to do is

we're just going to take some uh you

know kind of some English and translate

it into actual you know actual numbers

so we're going to write uh

2,500 and four in standard form so just

you know write that out

numerically so again just as a little

reminder

here remember if we have digits I'm just

going to write things at random here

uh uh

three and of course you could make

numbers bigger than this but um again

just to remind you about the the the

names so the eight here is in the one's

place the seven is in the

tens the six is in the

hundreds let's see the five is in the

thousands the four would be in the

10,000s uh our three next would be in

the 100

THS uh and if we kept going uh this uh

our two will be in the millions

place and then our next digit would be

in the 10 Millions okay so I'm just

going to kind of reference this this on

my next two examples here real

quickly so it says we have well

254 well 2,000 you know I find the

thousand's place which would basically

be uh right there and the thousand's

place it says we have well 2,000 so

instead of my five I'm simply going to

have a

two and then the number uh you know the

next part is kind of smaller than the

Thousand so we have

500 so I think well the hundreds uh

instead of a six since it's 500 we're

going to put a five right there and then

the last part of the statement it says

uh we also have the number four well the

number four there's no tens right you

know it's not 1 four or two four or 3

four because that would be 14 24

34 so since it's just Plano four we can

write four as well 04 and now we've got

the number 2,500

4 in standard

form so one more example

here um and this is why I went out to

Millions because I knew my next example

did go out to Millions so it says uh

what is uh what's the following number

in standard form

386 so

again kind of the same thing you know

I'm going to think well it says six

million so I'm going to find the the

million's place

well there's the millions place so we

have six millions so I'm just going to

put a six

there it says we have 2

100,000 well there's the 100, thousand's

place uh the 100, thousand's place is

you know the very next one it says we

have

200,000 so I'm going to put a two in the

100,000 place um it says we have

200,000 and then

so going to put it a couple zeros so

there's our

200,000 and now it says we have 386 so

again in the 100 spot is our

three and then our 86 we can just write

that as 8 and six and now we've got our

number again written in standard form

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing Whole Numbers in Standard Form (English to Number!).  Here we have a number written out in English and we write it in standard form.  Two examples are shown.


all right so another example here using

the alternating series test so we've got

the series from Nal 1 to infinity1 to

the n * n/ 5n + 6 so in this case again

if we kind of get rid of the alternating

alternating part we're left with n over

5n + 6 so two things we have to look at

for this case we have to think you know

is it

decreasing and the other one is does the

limit as n goes to Infinity of this

function does that equal

zero

well I'm going to answer the second part

first um you know if we take the limit

as n goes to Infinity uh of this

function since the degree of the

numerator equals the degree of the

denominator again we've seen these cases

you can always just do lal's rule on

this problem as well because we have

Infinity over infinity but again if the

degree of the numerator equals the

degree of the denominator we can just

use the ratio of those coefficients so

our limit's simply going to equal 1/5

well that's not zero so in this case the

conditions aren't satisfied and we would

simply say that this series diverges and

you can do the same thing at the very

beginning you could simply just take the

you know use the test for Divergence and

uh when you take the limit of this stuff

it doesn't equal zero so again that

automatically says our series diverges

Alternating Series - Another Example 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Alternating Series - Another Example 2.  In this video, I show that an alternating series converges or diverges.