help-36

[stray stump]

I will think about it more later.

I think my brain is kinda fried.

thanks for the help.

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[hasty bear]

Im confused on this question

[hasty bear]

so pretty much i know that function f(x) = |x| will be shifted 2 units to the left and 7 units down. So part a) is easy.

But for part b) im confused because isnt the function g(x) shown right there?

g(x) = f(x+2)-7 is the equation right? So why are they asking us that

[balmy maple]

you plug in those transformations into the original equation

[hasty bear]

so it'll be g(x) = |x+2| -7?

<@&286206848099549185>

i think thats the right answer but i need someone to verify

[hasty bear]

is it correct?

[balmy maple]

yea

[hasty bear]

alr ty

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[storm orchid]

Having a a bit of trouble interpreting questions like these when there are no "top" graphs.

[storm orchid]

Would I just create two integrals and add them together?

[final saddle]

@storm orchid Has your question been resolved?

[storm orchid]

How can I create a single integral to solve for the volume of the object that would be formed by rotating the region bound by the two graphs and the y-axis, around the y-axis?

[fossil geyser]

hm, if you want to take vertical cross sections by subtracting the functions, you'll have to use the shell method

otherwise you can take horizontal cross sections (you'll need two integrals) and use the disk method

@storm orchid

I'm not sure if you've learned both of those methods yet or not, so you might not have a choice

[storm orchid]

Oh yes, that would make sense. I forgot when using the shell method you integrate perpendicular to the axis of revolution, so there would be a "top" function using that method.

[fossil geyser]

yep, but it's also totally fine to use disks and integrate with respect to y instead

[storm orchid]

I did end up doing that, but I was curious if there was a way to create one equation as the next problem references the same general problem, but rotated around x=1 and I thought it would be easier to solve if I could have them in one integral.

But the shell method here makes sense.

When rotating this about x=1 though, would that leave my equations for y as y=1-(3x) and y=1-(4-x^2)? I've only done one other problem where the axis of revolution was not the x or y axis itself.

[fossil geyser]

are you referring to the shell method?

[storm orchid]

Yes, if the axis of revolution was x=1.

I recall that I would subtract the function from the changed radius.

Or something of that nature.

[fossil geyser]

well the formula for shells is

$\int2 \pi r h dx$

[soft zealot]

tatpoj

[fossil geyser]

the height, h is the distance between your two functions, vertically

r is the radius, which in this case would be 1-x

[storm orchid]

Oh! That's right! Thank you, I was overthinking that.

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[fossil geyser]

np 👍

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[storm wyvern]

I'm assuming A. is just 5 because Sin x/x = 1?

[storm wyvern]

I also could use some help with the rest😅

[scarlet sequoia]

There's a little more to it even though your guess is correct

Since you have sin(5x) instead of sin(x), try to make sin(y)/y appear, with y in terms of x

[ancient thistle]

I remember there being a method

limit x -> 0 (sin(ax)/x = a

[scarlet sequoia]

This is the method

[storm wyvern]

make sin(y)/y appear?

[scarlet sequoia]

Yeah

You know limit of sin(y)/y = 1 when y goes to 0, whatever y may be

So maybe if you tried to find a "y" that would perfectly fit here

[storm wyvern]

Okay I think I'm really lost😬

[scarlet sequoia]

Doesn't matter

We can start a little over

So we have our initial expression, sin(5x)/x right ?

[storm wyvern]

Could I just plug in 0 to the function since 5(0) = 0

actually wait then I wouldnt get 5

[scarlet sequoia]

0/0 is undefined 😔

[storm wyvern]

OHHH right right

[scarlet sequoia]

So, sin(5x)/x looks very similar to sin(x)/x but not quite

But what if we had sin(5x)/(5x)

[storm wyvern]

OHHH

Multiply by 5/5?

[scarlet sequoia]

Yesss

[storm wyvern]

5sin5x/5x=5

[scarlet sequoia]

As a limit yes

[storm wyvern]

okay perfect thank you

[paper beacon]

5 * lim[ sin(5x)/5x ]

5 * 1

[scarlet sequoia]

This type of factorisation (for c) doesn't help for infinite limit

[storm wyvern]

Can I just plug numbers in straight away then?

[scarlet sequoia]

Not possible either, infinity/infinity is still undefined

To make sense of it, factor the numerator by the highest power of x

And do the same with the denominator

[storm wyvern]

So I should multiply the numerator and denominator by 1/x^2

[scarlet sequoia]

Yes

[storm wyvern]

Is the answer 5?

If it is then I did it right😅

[scarlet sequoia]

Yep it's 5

[storm wyvern]

Thank you so much for your help, going to try my best on my own for the last couple!

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[tranquil pine]

how can i prove that a system of equations with 3 variables has no solution

[tranquil pine]

my equations are as following

2= a + b + c
4 = a-b+c
2= a+c

[frigid hawk]

do you know how to solve systems of equations using matrices?

[tranquil pine]

ive heard of it

ik what a matrix is and ive done them before but i havent done it in ages

[frigid hawk]

do you have a rough idea on how it works?

[tranquil pine]

yeah i think so

[frigid hawk]

so a system is considered unsolvable when the matrix doesn’t have an inverse

do you remember how to determine if a matrix has an inverse?

[tranquil pine]

no

not at all

[barren hound]

as you try to solve them you'll end up with an equation like 5 = 0

[tranquil pine]

i tried it and got -1 = 0 after plugging in c = 3-a into the third equation

[barren hound]

yeah

[tranquil pine]

does that prove it

or do i have to do more to prove it

[barren hound]

yeah (as long as you didn't mess up the algebra)

that's enough

[tranquil pine]

yeah my algbera is good

i checked it a few times

thanks !

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[fossil arrow]

This might be a weird question

[fossil arrow]

but i need someone from texas

its for UIL

and its number sense

its like mental math

[balmy maple]

I am from texas

did number sense

can help

wsg

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[narrow thorn]

For 13 A am i multiplying 3 by 25 and 4 by 25 first

[cosmic warren]

It's a mixed fraction

[ashen pewter]

its 25.75

not 25*3/4

[narrow thorn]

just figured that out

now i just divide by 2680 x 100

should be right

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[lethal totem]

Ok

[narrow pebble]

i have no clue... how do I start
its indefinite integral is li(x)

[vital crag]

Write the integral as a sum of integrals then bound each integral

Each interval could be [1,2], [2,3], etc

[narrow pebble]

my point is if we can prove that for any arbitrarily small k, then it will work for all k greater than it

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[narrow pebble]

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[rustic tangle]

Hi all, what the 'formula' i need to follow for these type of questions. P(A) = 1/5 and P(B) = 1/4 P(A n B) = 0.1, find P(A U B)

[brave cloak]

P (A U B) = p(A) + p( B) - p(A n B)

[rustic tangle]

thank you

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[pearl flame]

Can someone please help with this math problem I've been stuck on it for an hour. The first drop down menu is linear or exponential 2nd one is asking multiplicativity or additively 3rd one is asking common difference/slop or common ratio/multiplier/base

[final saddle]

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[pearl flame]

@dusk parcel I've been stuck on this problem for more than an hour and I've asked for help twice the first time no answer waited 30 minutes second time got a weird answer saying it was neither linear or expediential when it has to be one.

[uncut dagger]

have you considered how much the money in the account changes each year

[pearl flame]

yes if it was lenear it would be 170 and 169 for 2

for exponential it would be 1.313451072, 1.460577547, 1.850680544

[uncut dagger]

geometrically, a linear function is a straight line while an exponential function would either grow increasingly faster over time or approach zero asymptotically

if you were to draw these points in a graph

would there be a lot of error if you were to draw a line between the points?

[pearl flame]

Like the ratios of the functions or the actual money in the acc (like 199.84)

[uncut dagger]

the actual money in the acc

[winged python]

In a linear line, the slope is a constant. Slope = change in y over change in x
Since change in x is constant in the given points (1, 2, 3, 4, change in x = 1), calculate the change in y between all the points and if it's the same, then they're linear, otherwise exponential

Looking at the numbers, this is definitely an odd question

[pearl flame]

The linear points are 170, 170, 169

so it would be expediential?

[uncut dagger]

It doesnt have to be exact

[winged python]

Since we're talking money, it's possible that the answer had been rounded to nearest cent

[winged python]

yes

this is why I called this an odd question

[pearl flame]

Its 169.32 which cant round to 170

[winged python]

170, 170.34, 169.32

[uncut dagger]

Its not perfect

but its close

[pearl flame]

Yes that whats I got

[pearl flame]

so it can be linear still?

[uncut dagger]

It would still be better than an exponential function

[pearl flame]

exponential function would look like 1.313451072, 1.460577547, 1.850680544

[uncut dagger]

you can see it graphically

Here you can see that the best line stays closer to the points than the best exponential function

[pearl flame]

I see. so it would be A linear function... y-values changing additively. the common difference/slope of this function____ What would the approximately of this function be though?

[uncut dagger]

it would be approximately how much the y values increase

for each x

[pearl flame]

170, 170.34, 169.32

[uncut dagger]

pick a number

that approximately describes these

[pearl flame]

170 ish?

[uncut dagger]

roughly yeah

you could also take their average

if you want slightly more precision

[winged python]

You should ask your teacher/ professor about this question when you get the chance, it doesn't feel right bc the difference is off by a decimal

[pearl flame]

Ya the teacher said these questions are weird but if the answer is only off by like .somthing then she would give credit still

So should I just put 170?

[uncut dagger]

in my opinion yes

[pearl flame]

Okay

[winged python]

good luck

hope that's right

[pearl flame]

It was right!

[pearl flame]

Can one of you guys check this one to see if I've done it right please?

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[rocky cape]

how do i use this formula

[rocky cape]

like what do I plug in

[worn carbon]

think of gx as fx

[rocky cape]

right

[worn carbon]

f(x+h) just replace the x with x+h

in sqrt 2-x

so its sqrt(2-x+h)

[rocky cape]

so sqrt(2-[x+h]) - (sqrt2-x) over h?

[worn carbon]

yes

[marsh temple]

Watch your negatives

sqrt(2-[x+h])

[worn carbon]

oh yeah mb

close it if you are done

[rocky cape]

im still struggling

wait is sqrt 2 separate from -(x+h)?

[rocky cape]

is it suppose to look like this?

[worn carbon]

yes

[rocky cape]

how do i go by getting my final answer

simplifying further I just get sqrt (2-x-h) - sqrt (x-2) over h

[worn carbon]

yes

[rocky cape]

but isnt h => 0

so it just makes it all 0

[worn carbon]

no

limit is approaching, not at zero

also it would be approaching infinity

the equation is actually the derivative limit definition

[rocky cape]

from what I understand Im suppose to get rid of the h at the bottom

[worn carbon]

yes

[rocky cape]

but how

[worn carbon]

i dont think you can for this one

wait

idk

i dont have paper rn

on phone

[rocky cape]

thanks for the help ill try to figure it out

[worn carbon]

you can just take the derivative

[rocky cape]

how tho I think im just suppose to use that definition of the derivative formula given

[worn carbon]

idk man the limit equation seems impossible since you have a part without h

[rocky cape]

hm

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[forest topaz]

I need help on (b).....

Like is the question asking me any arbitrary r1/r2 vector?

Cause there are infinite vectors that start from origin to line L1

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[kindred nova]

can someone help explain limits from both sides? I don’t understand them very well or how to calculate it?

[balmy maple]

ok what happens if you plug in a number slightly less than 2 into the equation

@kindred nova

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@kindred nova Has your question been resolved?

[kindred nova]

@balmy maple undefined?

[balmy maple]

its a sqrt of a negative number right

thats undef

yea

so the limit is ...

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[kindred nova]

.reopen

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✅

[kindred nova]

I still don’t quite understand though.

When she was showing this in class she didn’t even mention that, she was getting us to write it down like

I’ll show you an example?

[balmy maple]

sur

e

[kindred nova]

Like this

i’m battery is going to die I apologize if it times out again

[balmy maple]

yeah again lets plug in a number its close to 2 but not 2 and from the left side

• means left side of 2 1.99 1.98 etc

negative not the dot

and + means 2.01 2.02 etc right

ok so if we plug like 1.99^2 into the eq

what do u egt

[magic rampart]

u can add backslash before - to remove dot point

[balmy maple]

@magic rampart ty didnt know that

[kindred nova]

isn’t that undefined

My calculator says error

unless I did it wrong

[balmy maple]

yeah

it is

so the limit is

[kindred nova]

undefined as well?

I’m not very sure.

[balmy maple]

dne

yea

It doesnt exist

[final tangle]

not quite

values there aren't in the domain of the function, so it doesn't even make sense to consider the left side limit

[balmy maple]

Yeah thats true but im just using a logical approach since domain is starting from 2 you shouldn't consider any limit below that domain

[kindred nova]

oh oops I sent the wrong photo earlier, she’s been writing it like this

I try to write down what I see because I can’t understand her accent well at all

what your saying does make sense though, to a certain point

Thank you.

[balmy maple]

btw you apply the same thing here since domain ends at 3

u dont consider anything above 3 so same thing as last problem

[kindred nova]

so then for my question, would it be x<-2 -> x-2<0?

or something like that

[cobalt lodge]

no, you cancel out the -2 on the right by adding 2 to both sides

(but yes it’s undefined for x<2)

[kindred nova]

Okay.

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[copper hull]

From a group of 8 women and 6 men, a committee con- sisting of 3 men and 3 women is to be formed. How many different committees are possible if
(a) 2 of the men refuse to serve together?
(b) 2 of the women refuse to serve together?
(c) 1 man and 1 woman refuse to serve together?

[copper hull]

How do we do C?

Ive been stuck on this question and cant seem to get it

Ive tried various methods but nothign seems to work

Ive tried a case by case method, unless i did my math wrong, I cant seem to understadn waht went wrong

[tired walrus]

let's call the particular members that refuse to serve together Alice and Bob

[copper hull]

OK

[tired walrus]

we want the committees that don't have both Alice and Bob on it

you could do it with two methods

[copper hull]

(5C3)(7C3)

[tired walrus]

either count the committees with Alice only, then with Bob only, then with neither of them, then add it up

[copper hull]

I did that third method

[tired walrus]

this is one method

[copper hull]

I eam this one

mean*

[tired walrus]

if you only counted the third kind of committees then you are wrong

[copper hull]

and for some reason, it wasn;t coming up as the coreect answe

I eam I added all 3

typo

[tired walrus]

or you can count the committees that have both Alice and Bob on them, and subtract that from the answer to part a.

[copper hull]

why would I subtract from A?

[tired walrus]

oh waitn o

[copper hull]

fyi, A is 896. C is supposed to be 910

[tired walrus]

hold on sorry i thought a was unrestricted

[copper hull]

u good

[tired walrus]

then you need to find the number of committees w/o restriction and subtract from that.

[copper hull]

so subtract from unrestricted?

[tired walrus]

yes

[copper hull]

okok let me try taht

but can I type down how I did the orignal method u mentioned?

maybe I made a mistake there

[tired walrus]

sure

[copper hull]

Case 1: Alice and Bob not there

(5C3)(7C3)

Case 2: Alice is there

(5C3)(8C3)

Case 3: Bob is there

(6C3)(7C3)

Add the answers up

this is what I did

didnt work sadly

[tired walrus]

if alice is there then there are 2 female seats remaining.

not 3.

[copper hull]

hm?

oh wait...

[tired walrus]

if alice is there then you have to have 2 other women

[copper hull]

OHHHH

hold up

Wait

still doesnt work tho

It goes down, but not yet reaching the amount

I changed it for the men as wel

Case 1: Alice and Bob not there

(5C3)(7C3)

Case 2: Alice is there

(5C2)(8C3)

Case 3: Bob is there

(6C3)(7C2)

[tired walrus]

there are 8 women and 6 men right

hold on

[copper hull]

yea

[tired walrus]

then cases 2 and 3 are wrong

[copper hull]

hm?

[tired walrus]

committees with alice should be 7C2 (women) * 5C3 (men)

we want alice on but bob off, remember?

[copper hull]

oh yeee

hold up let me plug it in

Bows Head

[tired walrus]

check case 3 again

[copper hull]

https://tenor.com/view/thank-you-bow-michael-scott-the-office-gif-18616386

You are my savior

https://tenor.com/view/sending-love-gif-25755600

I was stuck on C for like 1 hour not knowing what was wrong

Thank you soooo muchh

🙂

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[real pewter]

i got the acceleration graph, but how do i go from velocity to position?

[gentle barn]

Derivation is the change of the function. Acceleration describes velocity, and velocity describes position. The area under the graph of a derivate, gives the value of the function. Area under acceleration describes velocity, area under velocity describes position.

[real pewter]

so i see i can make a triangle + a rectangle for 1-5 seconds and a rectangle for 1-15 seconds. but for 16-20, do i only do the little triangle in the poisitive?

or do i do the triangle in the negatives too

[gentle barn]

What does negative velocity mean?

[real pewter]

its position is changing in the negative direction?

[gentle barn]

That's right, so you're going to include the area under the x-axis. And remember that the area is negative, since the position is getting reduced.

[real pewter]

okay i got all of the areas! im confused on how to apply the area to the position graph. how does my area of 12 relate to the look of the graph?

[gentle barn]

What is the area in the interval [0,4] under the velocity function.

[barren hound]

so the area from 0 to 4 is 12

meaning that the difference in position from 0 to 4 should be 12

[real pewter]

so the first segment looks something like this?

[barren hound]

sure (although it's moving faster at the start so you might bend it a bit, but yeah that's fine)

[final saddle]

@real pewter Has your question been resolved?

[real pewter]

how does this look?

[gentle barn]

When a object is accelerating, the speed goes from 0 (sometimes other) to something. Does position graph in interval [0,4] look reasonable?

Otherwise, the graph looks good.

[real pewter]

Hmmm

it starts at 0

and ends at 12

does it need to be more curved?

[gentle barn]

Your curve is towards down, it maybe should point towards up.

[gentle barn]

@barren hound Is the explanation correct? I'm a bit unsure.

[real pewter]

thank you guys so much!!

[gentle barn]

If you wait someone to certify, then you're good to go.

[barren hound]

i think it's curved the other way bc it starts out fast! but that's not really a big deal rn

[gentle barn]

Oh yeah, my bad. It do starts something other than 0.

[gentle barn]

You should change the curve, my bad.

[real pewter]

Its totally okay!! I appreciate both of your help!! ❤️

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[fringe geyser]

this is an awnser my professor gave on a taylor series question, and i was wondering did he do an oopsie on the chainrule?? isnt the y"(0) suposed to be -4cos(0)=-4

[tired walrus]

indeed

[fringe geyser]

alr ty

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[formal knoll]

Hello, im having problem with Polynomial factoring (GCF)

[formal knoll]

this

<@&286206848099549185>

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[eternal wharf]

The period of the function f(x) = |sin2x|+|cos8x|

[upper kite]

What do you think?

,wolf plot |sin2x|+|cos8x| from -1 to 4

[soft zealot]

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[eternal wharf]

I don't know how to do it without desmos

[upper kite]

ah

Use Wolfram Alpha

Just kidding.

[upper kite]

So let's see

do you know the period of sin2x?

[eternal wharf]

yeah

it is pie

[charred chasm]

Obligatory @eternal wharf

[upper kite]

What does this mean?

[upper kite]

Alright, correct

generall for any sin(ax)

it has period 2pi/a

because if you put sin(ax) = sin(a(x + 2pi/a)) for all x due to sine repeat.

However if you have an asbolute value of a sin, the period is halfed

The period of |sin(ax)| is just pi/a.

[upper kite]

you get |sin(ax)| = |sin(a(x + pi/a))|

,wolf plot |sin(2x)| from -1 to 4

[soft zealot]

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[upper kite]

pi/2 repitition

[eternal wharf]

We are not allowed to use desmos

In the exams

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[abstract dock]

How do I solve this?

[abstract dock]

This is what i have so far

[cosmic warren]

ok good work

now try rational roots theorem , also the fact that if a is a root then (x-a) is a factor

actually rational roots might be overkill here cus personally I can kinda already find a root here just by looking at the poly

But if u don't see it immediately use that to narrow down the possible roots and try them ouf

[abstract dock]

So how do I do this really?

[cosmic warren]

well there's the cubic formula

[abstract dock]

Rational function is in the next unit

[cosmic warren]

No this is a polynomial function

rational roots just says

That if the coefficients of a poly are integers and the poly has a rational roots

the root being p/q

Then q divides the coefficient in front and p divides the constant term

So for example

6 divides 6 so q could be 6

And 36 divides 36 so p could be 36

So 36/6=6 might be a root

It doesn't say it will be

Just that it won't not be either until you've plugged it into the equation

[abstract dock]

ok

[cosmic warren]

So as another example rational roots says 5/7 is not a possible root

but yeah the goal here is to find an ez root a so you can write the polynomial as (x-a) times a quadratic

And quadratics r ez

There is a cubic formula too but it's not as nice as the quadratic formula

[sharp storm]

Don’t forget that x is equal to 0 too because of the 3x

[abstract dock]

I cant use it because it is not in the material

[cosmic warren]

yeah I didn't say to use it anyways

[abstract dock]

can you explain?

[cosmic warren]

again one of the roots here is waaaay more obvious than the other ones so don't be using weird values like 4/3 probably (even though rational roots says 4/3 is not not a possibility. Also I haven't actually checked if 4/3 is a root or not so don't quote me on this)

[sharp storm]

If you for example 10X^2 + X = 0 if you put the X out it will be X(10X + 1) = 0 and if you multiply X with everything it needs to be 0 so X must me zero

[sharp storm]

What is the answer to the question do you know?

[cosmic warren]

x=0 is a solution but it's not the only solution so "must" is kinda strong

[abstract dock]

ok

I see

are you saying I move the x over?

[sharp storm]

Possibly and you must include it as a possiblity

[cosmic warren]

no not possibility it is

[abstract dock]

it is a soultion

[cosmic warren]

possibly I mean

[abstract dock]

but not all of them

how do I find the rest

[cosmic warren]

Yeah

we'll have u been looking for a root?

well*

Like I said :p

[sharp storm]

For example X can be equal to 0, … , …

[abstract dock]

yeah how do I find the rest then

[sharp storm]

Do you know the answer?

[abstract dock]

yes I know the answers

[cosmic warren]

I told u find a root first

[sharp storm]

Like without explanation

[abstract dock]

yes

[cosmic warren]

just any root but there's a really easy root

[sharp storm]

What?

[cosmic warren]

Then you can factor out (x - a) and you'll have a quadratic left

And the quadratic is just ez factoring or quadratic formula

But find a root to the cubic polynomial I should say

[cosmic warren]

yes x =1 is the ez root

U can just add the coefficients to see why it's a root

[sharp storm]

X needs to be equal to 0 too

[abstract dock]

I just dont understand what he means by taking x=1

[cosmic warren]

Substitute x = 1

Same thing as adding the coefficients

[abstract dock]

on what basis

[cosmic warren]

Just plug in

wym what basis

[sharp storm]

It is like (6x^2 + 35x +36)(x-1) and it is easy

[abstract dock]

so hypothetically we could substitute any number in and see if it works?

[cosmic warren]

yeeeeeeees

That's what I kept telling u to do 🙃

[sharp storm]

What I saw you it equals to the equation

[abstract dock]

btw what does it mean by taking 3x to the right side?

[cosmic warren]

they factored out the 3x like you did

But they put it on the right instead of left

[abstract dock]

and it is just gone?

[cosmic warren]

no

It's a factor of the 4-th degree polynomial

just keep it in like u did

[sharp storm]

Means factored it just make it easier for you to solve and if you factored x, another answer will be x = 0

[abstract dock]

oh ok

understood

thanks

thanks guys

[sharp storm]

No problem 😉

[abstract dock]

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[unkempt viper]

I just realized there's a tangent identity tan (x-y)

What did I do wrong though

In trying to solve it

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[gilded flame]

How do I use u substitution to solve the integral of 1/(x^2+1)dx / do I even need to? It looked really easy at first but then I got stuck :/

[final saddle]

Please don't occupy multiple help channels.

[sweet summit]

hello there, you have a threqad open at #help-35 , we ask you don't use up mulitple threads so others can ask. Can you move your question there?

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[tranquil pine]

Hi, can someone please help me for this problem?

[tranquil pine]

I probably need to change the bounds of h(x) but am not sure where to take things

[worldly vale]

yeah try a u substitution to make the bounds 0 and x

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

How does u-sub change the bounds of an integral?

I generally only do u-sub to get an antiderivative or something

[vital crag]

Then just stick to that and substitute back in terms of x after integrating u

[tranquil pine]

Then wouldn’t I be avoiding what the exercise is asking me to do? I see a way of solving this by doing a “classic” u-sub but in that case, I wouldn’t be using the first part of the FTC

[worldly vale]

when you do a usub of a definite integral, the bounds also change in accordance with the sub

i.e. change by the same rule as the sub you choose

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

Let me try this real quick

[tranquil pine]

@worldly vale

I don’t understand what to do

I tried doing a normal u-sub but for some reason nothing works out

This integral doesn’t seem like an “elementary” integral (?)

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[turbid lotus]

stuck on 266

[turbid lotus]

idk where to start because theres a root

square the num and denom ?

[sleek abyss]

just use bobo botn eats dc

for lim

[robust mulch]

i get thats an acronym, but what does it mean

[sleek abyss]

bigger on bottom it goes to 0 bigger on top its inf if its the same power you divide the co-eff

[robust mulch]

Interesting, never heard of that

I would divide the top and bottom by x

[sleek abyss]

yeah my calc teacher said it to make it easier for us I mean its worked for like ever so :p

[cosmic warren]

U p much prove what vraj said by doing what garlic said

[sleek abyss]

yeah basically

[turbid lotus]

how do i divide the root by x

[cosmic warren]

Sqrt(1/x^2)

[turbid lotus]

what about the num

it will become 3/x then zero

[cosmic warren]

yeah

That's the point of the method

If u kept it as it was then u couldn't find a limit since the denom and num owuld go to infinfinity but by diving all the terms except the leading coefficient become 0 as x goes to infinity

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[turbid lotus]

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[tranquil pine]

Hi, are there any issues with my work here? I somehow get all different answers, this is perhaps where the “reconcile your answers” part comes in…

[barren hound]

that's exactly right

[tranquil pine]

So there’s nothing wrong with my work right?

[barren hound]

yeah your work looks good to me

[tranquil pine]

Perfect

I’m trying to reconcile my answers

But one thing

How do I eliminate the constants?

Like

How am I supposed to write out the fact that the constants can be neglected since we’re dealing with indefinite integrals

[barren hound]

like the +C?

[tranquil pine]

And hence all 3 results are equal

[barren hound]

the +C is exactly what lets you reconcile them

it may be helpful to graph the three equations

functions

[tranquil pine]

I know what you mean

I’m just wondering how to write this down

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

Still stuck 😭

Are the three different results equal only for certain values of C

[barren hound]

,w plot {0.5sin²x, -0.5cos²x, -0.25cos2x}

[soft zealot]

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[barren hound]

does this bring understanding?

[tranquil pine]

Yep so

If I understood correctly

And according to the graph

Only 1 value of C (common to the 3 results) makes it so the 3 results are equal

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[light violet]

Is there only 3 graph for a) ? and for b) how can i list out all 12 paths

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.close

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.reopen

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[prisma salmon]

I dun see any mistakes in my workings but Idk why am I not getting the answer. (1st order Bernoulli equation)

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@prisma salmon Has your question been resolved?

[prisma salmon]

<@&286206848099549185>

<@&286206848099549185>

[tranquil pine]

yes

[soft zealot]

Dyssrupt

[split acorn]

last step ^

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[manic echo]

what did i do wrong for b)

[strong mantle]

Where did you get (3,-3,2) from

[final saddle]

@manic echo Has your question been resolved?

[manic echo]

vector AB

using the coordinates

[strong mantle]

would you mind writing your work in more detail for part b?

[final saddle]

@manic echo Has your question been resolved?

[manic echo]

i can’t at the moment but it’s like $$area=|AB \cross AD|$

[soft zealot]

___
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[manic echo]

i don’t think i did the cross product wrong so idk what’s wrong with my working

[strong mantle]

What’s the answer supposed to be

[manic echo]

sqrt 307

but they use different vectors so can’t check the working

[strong mantle]

I get the idea, but if you write how you computed AB and AD, and their cross product, in more detail, then it’ll help you find the mistake

[manic echo]

oh right

[strong mantle]

it should be -3, -2, and 1, not -3, 2, and 1

this is why more steps help

I found the error

@manic echo

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@manic echo Has your question been resolved?

[manic echo]

sorry why?

ohhh

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[pulsar monolith]

I am struggling with Algebra-II. How would I solve this?
2 - 2|5x - 2| = - 18

[final saddle]

@pulsar monolith Has your question been resolved?

[pulsar monolith]

Not yet, I'm trying to solve it again though

[gritty elbow]

Well that’s the uh….answer on the first picture, but how you got -10 during your substitution there….I’m unsure about

Yeah, 12/5 and -8/5 are correct after my check, but let’s go back to the start of your substitution process.

[pulsar monolith]

Solved it

[gritty elbow]

Oh

Nice

[pulsar monolith]

I messed up the negatives 💢💢

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[pulsar monolith]

Totally not like I have eight more of those to solve or anything!

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[thorny chasm]

Help

[thorny chasm]

Bearing

[split acorn]

[thorny chasm]

@split acorn ^

[split acorn]

Sorry, I dont help with geometry questions, I am not good at those.

[thorny chasm]

It’s good but do u know anyone who’s good

[split acorn]

Someone else will be able to help with those, wait here.

[thorny chasm]

Bcus I’m abut to go to sleep and I have a test tmwr

And I already feel like I will FUXK it up

[split acorn]

yeah dw, i'll try

[thorny chasm]

Tysm

[tranquil pine]

Only if you could mark those bearing things

[thorny chasm]

I don’t fucking understand the whole

Thing

[tranquil pine]

Hmm nice me too

Why couldnt they just have given the angles like

bearings

Take clockwise angle from north direction, and they wont even write 75° as 75°

They would make it what, 705 ?

[thorny chasm]

Huh

[tranquil pine]

I just read google for bearings

And it fucked my mind too

[thorny chasm]

💀💀

[severe hawk]

Bearing angles are just taking a different convention for the starting point.
In mathematics, typically, angle 0º is the positive X axis, you count counterclockwise, and the positive Y axis is 90º.
In bearings typically, angle 0º is the positive Y axis (north), you count clockwise, and the positive X axis is 90º

There are other ways to measure both, these are just the most common

[tranquil pine]

Then what is B to A and A to B??

[split acorn]

here

[split acorn]

after finally understanding what bearings are, here is the labelled diagram

[tranquil pine]

Is CD||North??

[split acorn]

no

[little python]

not truly अक्लमंद बिल्ली

[split acorn]

#chill please

[tranquil pine]

Oh right doesnt need to be

[little python]

sorry, old guy

[tranquil pine]

OH I GOT IT

Oh wait no

Dumb

Okay i got it now

[tranquil pine]

From this drawing

• drawing line parrallel to North at C
• Extending AD till it cuts the previous line

This will give you a triangle, then you can do it

[split acorn]

so complex why

[tranquil pine]

Less brain used

And i did it fast , cuz, less brain used

Idk

If theres an easier method

But

• 70° from Interior Alternate angle
• Find x, y by Linear Pair
• a and z , from Angle sum property of triangle

Answer you need that bearing shit is just (180+a+z)°

@thorny chasm

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@thorny chasm Has your question been resolved?

[thorny chasm]

Yes Tysm love 🫶🏻🫶🏻

[tranquil pine]

You're welcome

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[teal cypress]

What do i do to get my limit if I have an unknown in my equation

[strange shore]

its just a constant, work with it

[teal cypress]

I got this. Idk where to go from here

[rustic wedge]

It should be 2*(x+3)-C for the numerator no?

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[gray rover]

Yo

[gray rover]

How does dividing fractions work in terms of mathermatical reasoning

[vital crag]

!15m

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[round anchor]

you multiply it by the reciprocal

[gray rover]

Yes but how why does that work

Isnt there some type of reason

<@&286206848099549185> i dont understand how multiplying fractions is done by logic

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[gray rover]

Nah but its ok

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[indigo plover]

Hopefully this work now

Ok

I believe i do some mistake idk which

[vital crag]

what happened after this

show how you got this step

[indigo plover]

uh

I just factored out the x^2-4

cancel x+2 with x+2

idk why i cross x-2 but that stays

Uhhh then x-2 is the only one left

Sot aht is the denominator

and then numerator x^2-x+1

Xd idk

so i made x^2-x+1 = x+1

so my final answer was

x+1/x-2

[vital crag]

yea you don't have to keep uploading the same image

[vital crag]

what cancelling are you talking about, there's no equation?

[indigo plover]

oh god

im an idiot

i even wrote this down and still forgot

rational expression no canceling

OK i redo

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[fierce orchid]

vietaas

or quad formula

[warped summit]

isnt that only sum of roots

vietas

[fierce orchid]

and product

but there is a better way, you can factor the quadratic

[warped summit]

how do you get difference from that

ill try quadratic

[fierce orchid]

no factor

sqrt((a+b)^2-4ab)=a-b

[warped summit]

alr ty

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[molten mesa]

I have 0 idea

Status is 1

<@&286206848099549185>

?

Dude seriously?

<@&286206848099549185>

[hoary kernel]

Yes

Alright

Let me take a look

[molten mesa]

Alr thx

[molten mesa]

Do u have an idea?

[hoary kernel]

Possibly

[molten mesa]

I got (x/x) + 2

[hoary kernel]

That seems correct to my work that I did for the problem

[molten mesa]

Aight th

Ty

[molten mesa]

I have a few other ones as well

Can I send them here

[hoary kernel]

Sure

[hoary kernel]

That looks like a bingo to me

[molten mesa]

Also this one jeje

Hehe

This one is spot on I’m so happy :)

[hoary kernel]

Your good to go

[molten mesa]

Ty

[hoary kernel]

Cross your fingers for a good score

[molten mesa]

I have more tho lol

[hoary kernel]

Oh

Send them

[hoary kernel]

Have you solved this?

I’m so sorry but I have to go

At least I helped you with the other ones

Have a good day

[molten mesa]

Tysm cya

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[tranquil pine]

ik its simple but my brain isnt working correctly rn 😭

[supple dove]

@stable kraken

Need help badly

[fossil geyser]

@supple dove This channel is already taken, please open your own #❓how-to-get-help

[supple dove]

My bad

[fossil geyser]

Consider this quadrilateral

Can you name any one of the four vertices?

[tranquil pine]

wait wdym 😭

like the YZKJ?

[fossil geyser]

well yeah I'm just using this rectangle as an example

but like, the top right corner

[tranquil pine]

ohh okok

[fossil geyser]

the top right corner is (1,2), right?

[tranquil pine]

mhmm

[fossil geyser]

now if you rotate the rectangle 180 degrees

where is that point going to land?

[tranquil pine]

thats the part i need help on, i forgot how to rotate shapes 😭

like how the point rotates and stuff yk

[fossil geyser]

pretend it was like a real thing you could pick up

turn it around 180 degrees, so it's upside down

and then put it back down

[tranquil pine]

ohh okkk

let me graph it rq

[fossil geyser]

I mean like

If the red point is here, at (1,2)

then if you rotated this rectangle 180 degrees, wouldn't it look like this?

like if you were actually able to pick it up and turn it around

[tranquil pine]

OHHH

ok that makes more sense 😭

[fossil geyser]

so it went from (1,2) to (-1,-2)

[tranquil pine]

ohh okk

so would it be (x, y) —> (-x, -y) because both points changed to negative?

[fossil geyser]

yeah exactly 👍

That's what would happen no matter what point you chose to start with

[tranquil pine]

alright tysmm

[fossil geyser]

no problem 👍

[tranquil pine]

.close

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[soft zealot]

aguaman

[weary canopy]

we have y = e^x/(1+2e^x)

swap the x and y positions

x = e^y/(1+2e^y)

what should I do next?

hmmmm

possibly take ln of both sides?

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.close

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[topaz goblet]

Hwo do I do this

I tried making cases

X<=-1 implies (x³+1)(y³+1) < 0

But x =0 => y = 1 and (x³+1)(y³+1) =2

So no max case then

X,y>0 implies xy <= 1/4

[vital crag]

Can you just use derivatives

[tranquil pine]

just

implicitaley derive

however you spell that

find dy/dx

wait nvm

i read wrong

im unsure

maybe its an optimization or modeling problem

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[azure iron]

Can anybody please explain how to solve this problem

[azure iron]

Its from Barrons AP calculus review

Im completely lost

[tranquil pine]

apply leibniz rule for 1 and 3

idk 2

[azure iron]

should i know how to do that in calc 1

[tranquil pine]

no idea tf is calc 1

leibniz is diff of int

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@azure iron Has your question been resolved?

[tranquil pine]

i think

the second one is related to leibniz too

gotta check the

nature of the graph

[gritty zodiac]

Cant u just integrate it and plug in 2 and 6 for second one?

Youll get values

[azure iron]

yeah thats what i did

its I and III that trip me up

idk how to get those

becuase what if i do a test and i have to find those out

[tranquil pine]

test?

[azure iron]

Yes im reviewing for a test

like an examination

[tranquil pine]

idk that looked kinda hard to intg and easier to look at the graph

theres x going in the lims 😵😵

[tranquil pine]

just diff it

[azure iron]

wdym by differentiate it

dont i have to solve the integral firsy

then differentiate

[soft zealot]

Dyssrupt

[tranquil pine]

@azure iron

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[thin urchin]

hi guys sorry to bother again but in this why do we need to multiply by n at the end

[lime summit]

The n was always there the solution was just highlighting the other part of the limit

[tranquil pine]

I kind of dislike this solution. It's using the product of limits rule where one of the limits does not exist (∞). Correct way is to do polynomial division and get it in the form f(n) + r(n)/g(n) where deg(r)<deg(g) and it follows the limit diverges to ∞

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@thin urchin Has your question been resolved?

[thin urchin]

So whether i said 3 or + infinity it didn't change stuff?

Both are correct?

[lime summit]

The limits answer is +infinity not 3

[thin urchin]

Why tho

Why multiply by n after all the factorisation

If u did that every time there would be no constant likit

Limit