help-0

[vale tundra]

I have one more q

Do you mind?

[strong compass]

Ofc ask away

[vale tundra]

No

Its E(Z) +E(w)

[strong compass]

I can’t help with this, this was removed from my specification so I haven’t learnt it

[vale tundra]

Ah

[strong compass]

Sorry

[vale tundra]

Its alrright!

[strong compass]

Someone else can

[vale tundra]

Yep!

Thank you for helping me!

I am not sure how to solve this q

.close

[lone heart]

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[uneven palm]

can someone help me with this

[uneven palm]

this absolute values really confuse me

i've been over this literaly almost 10 times

i still make the same mistakes

i know i must solve 2 equations first

[ornate condor]

hm

move the abs(2x+1) first

[uneven palm]

the first one is x-2 = -1/3x

right?

i do something like this

-1/3x<2x+1 <= 1/3x

is this correct

[ornate condor]

um

where did this come from

[uneven palm]

i saw something like this in internet

[tall topaz]

Cross multiply and split into two cases

[ornate condor]

ye

[uneven palm]

i still can't do it

so the 2 cases are

x-2 > -1/3x(2x+1)

and x-2> -1/3x(-2x-1)

[tall topaz]

Yeah pretty much

Solve them separately keeping in mind the critical values

[ornate condor]

wait what

oh yea i guess mb oops

but u shld state the conditions under which u take which equations

[uneven palm]

so for the first one the condition is x=! -1/2

right?

this results on a quadratic

[ornate condor]

um no

i think its

that the values are positive

eg

$\frac{1}{|x|}$

[uneven palm]

photomathsays that the sign of the condition changes

[uneven palm]

between 2 equations

[ornate condor]

u have

[uneven palm]

but why is that

[ornate condor]

1/x and 1/-x

right

the condition for the first is

x>0

the condition for the 2nd is x<0

both cannot have x=0

but the first ones condition is that it is positive

so the modulus is useless

[uneven palm]

but why

[ornate condor]

the 2nd is the thing is negative

thus the modulus will make it positve

[uneven palm]

how do you determine which condition is for which equation

[ornate condor]

just from defn of

well

the one thats the same equation but with mod removed

is the one that needs the modulus being positive

[uneven palm]

well let me solve this

and see if i

m correct

i guess i understod it

[ornate condor]

ok good

[uneven palm]

i just find it hard to determine the sign of the condition

[ornate condor]

well takes some time to get used to

加油

がんばて

[uneven palm]

ok than

thank you very much

[ornate condor]

npnp

[uneven palm]

closing this thanks a lot

.close

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[alpine sable]

The number of deaths of children ages 1 to 14 in 1985 compared to 1991 for 10 states was gathered.

Part A: Determine and interpret the LSRL. (3 points)

I am having troubles with part A, i had indeed learned this in class but i got sick the following week, during that time i didn't study and compleetly forgot what i learned, now im running through practice problems for my test which is tomorrow

I haven't tried anything yet because i don't know how to approach this

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

<@&286206848099549185>

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@alpine sable Has your question been resolved?

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@alpine sable Has your question been resolved?

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[turbid ember]

topic: Matrix algebra

i have no clue what the question is asking me

what does a represent?

y?

so should i just input the values of (17) into x, and then (-20) into z

and see if it works?

[waxen flame]

Actually, that would work.

[turbid ember]

ok i'll try right now

[waxen flame]

Make sure you test it with all three equations.

I suppose the correct method would be to put the equation into reduced row echelon form though.

[turbid ember]

and input the values after i row reduce them?

[waxen flame]

That depends on what type of reduced form you get.

With this, you should get one solution.

Three variables, three equations.

And I checked it, only one of them will have a valid solution.

[turbid ember]

ah so if i input 17 for x, and -20 for z

in all 3 equations

only one of them will be true

and that is my answer?

[waxen flame]

The better method would be to find the reduced row echelon form of the equation. The answer will become apparent after that.

While you can test each of the three equations, that would be tedious.

[lone heart]

@turbid ember Has your question been resolved?

[turbid ember]

im struggling to row reduce, i feel so embarrased not knowing how to do it

i'll show you what i have so far

[ionic ferry]

if you feel embarrassed check out my channel lmfao

[turbid ember]

<@&286206848099549185>

idk how to row reduce this

[turbid ember]

^

[waxen flame]

The goal is to making a leading 1 in each of the diagonal columns.

Do you know what I mean by a leading 1?

[turbid ember]

like the pivots?

[waxen flame]

Yes.

[turbid ember]

yes, though idk how to, even tho its elementary math

diagonal line of 1s

0's above and below them

[waxen flame]

You have to zero out each element of the bottom-left of the matrix one-by-one.

How would you eliminate the -1 in M_(3,1) using the allowed matrix operations?

[turbid ember]

in the second matrix bottom left?

or starting at the first

[waxen flame]

Starting on the first.

[turbid ember]

ok

i would do R3 + R1

which would make the -1 a 0

[waxen flame]

\begin{bmatrix}
$1 & 1 & 2 & 8 \
\cancelto{0}{3}& -1 & 1 & 0 \
\cancelto{0}{-1} & \cancelto{0}{3} & 4 & 4
$\end{bmatrix}

[ocean seal]

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

[waxen flame]

Not pretty, but you want to get that bottom-left corner to be all zeroes first.

But yes, R3 + R1.

[turbid ember]

ok and then that gives me 0, 4, 6, = 12

in the bottom row

now if i were to get rid of the 4

i would need to first make the second row, column 1 position 0 as well

then use row 2 to make that 4 a 0 correct?

[waxen flame]

I'll figure Latex out one of these days. 😛

[turbid ember]

next thing im gonna do is R2 - 3R1

to get rid of the 3

[waxen flame]

Sounds good.

[turbid ember]

now i want to get rid of the 4 right

[waxen flame]

3 - 3(1) = 0
-1 - 3(1) = -4
1 - 3(2) = -5
0 - 3(8) = -24

[turbid ember]

ah

ah ok

and now i get rid of 4 on bottom row correct

ahhhhhhhhhhhh

so when row reducing i always want to work on bottom left corner in the same order we did it right?

[waxen flame]

It doesn't matter what order you do it in. Ideally, you want to choose an order that appears easy.

You can even rearrange the rows to make it easy.

[turbid ember]

ok so

i got this

does that mean that x3 = 12

or should i not worry about that and keep on row reducing

[waxen flame]

You should have gotten -12.

[turbid ember]

yes

[waxen flame]

After that, you want to reduce the second row by dividing by -4.

[turbid ember]

ohshit

i can do that?

[waxen flame]

Yes.

You should familiarize yourself with the row operations that you can perform.

[turbid ember]

dont i get a fraction then though

in the -5/4

[waxen flame]

Yes, but that will resolve itself after you zero out the top-right.

[turbid ember]

pen ran out if ink fyi xD

[waxen flame]

😄

[turbid ember]

but does that look right

[waxen flame]

Yes.

[turbid ember]

ok let me try here

for the next operaiton

sheesh idk man

[waxen flame]

Now that you have leading 1's in the diagonal, you can start eliminating the top-right elements by adding the rows the rows below them. You should first begin with M_(2,3), eliminate the -5/4 by adding 5/4 R3.

[turbid ember]

ok

[turbid ember]

multiplying it by 5/4

ah i did smtn wrong

should be positive 5/4

ok

then R1 - R2 to get rid of the 1 on top middle

[waxen flame]

Yes.

[turbid ember]

then R1 - 2R2?

and thats it

[waxen flame]

You want to eliminate the top-right elements.

[turbid ember]

before the top middle?

so i want to get rid of the 2 first?

[waxen flame]

At this point, doesn't matter what order you do it in as long as you have leading 1's in the diagonal.

[turbid ember]

so then

what would be the final answer to this question

a - none

b - ?

[waxen flame]

(11, 21, b) = (11, 21, -12)

What does b equal?

[turbid ember]

Is that what its asking me?

[waxen flame]

So you found a solution for the system of equations.

(11,21,-12)

(b) asks what does b equal if (11,21,b) was a solution to the system of equations.

11 and 21 both match up so you know that (11,21,b) works.

(17, a, -20) doesn't work because 17 doesn't match up with 11 and -20 doesn't match up with -12.

[turbid ember]

Ah I understand

Thank you so much

You’ve been a boig help

[waxen flame]

yw

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@turbid ember Has your question been resolved?

[lone heart]

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[spice wasp]

I need help finding the significant figure for [(2.05*10^6)/48.895]+134.31

[spice wasp]

I’m supposed to perform the following calculations to the correct number of significant figures

I’ve calculated it but no I don’t know how to put it into the correct sigfig

This is chemistry btw

<@&286206848099549185>

<@&286206848099549185>

????

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[clever tapir]

just reading through some lecutre notes

[clever tapir]

I just need some clarification on the red part

[naive valley]

x+3 > 0, so you can multiply both sides of the original inequality by x+3 without changing the direction of the inequality

that's where 2x+1 <= x+3 comes from

the rest is just rearrangement by adding/subtracting things to both sides

[clever tapir]

so for the case where x < -3

is it even possible?

[naive valley]

sure, in that case x+3 < 0, so you can multiply both sides of the original inequality by x+3 but you have to change <= to >=

however, that will lead to the conclusion that x >= 2, which contradicts the assumption that x < -3

so there are indeed no solutions with x < -3

that should be covered in "case 2" of whatever notes you're reading

[clever tapir]

its a little confusing but i understand it

is there any strategies i should use to identify all possible cases of inequalities

@naive valley

[naive valley]

well when manipulating inequalities you can add or subtract things from both sides without considering cases

you can multiply both sides by positive things without changing the direction of the inequality, but if you multiply by negative things you have to change the direction. so that's the main situation where cases come into play

"i want to multiply this inequality by x+3, but that could be either positive or negative depending on x, thus i must consider the two cases separately"

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[vale sigil]

x^(3)+15x^(2)+67x+53

[vale sigil]

i have to find zeros

[fluid basin]

normally cubics are annoying

one of the tricks, though, is to see if there's an "easy" root

[gray isle]

start by approaching with rat root theorem

[fluid basin]

first try 1 and -1

1 is not a root, but -1 is: (-1)+15-67+53 is 0

so that means you can divide out (x+1) as a factor, which leaves you with a quadratic, solve with quadratic formula

[vale sigil]

oh thanks

.close

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[vale sigil]

.reopen

[lone heart]

✅

[vale sigil]

.close

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[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

.close

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[alpine sable]

how to do this

<@&286206848099549185>

.close

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[worn vessel]

Can someone explain this?

[abstract fractal]

That's an excellent way to get help

Don't even ping helpers, just go straight to insults

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@worn vessel Has your question been resolved?

[split dagger]

🤓

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[tropic oasis]

hi guys i'm a bit stuck on this problem

i'm guessing it would be 7x^(2)-3x-9 greater than or equal to zero?

and solve from there?

if i factor it i'm not sure where to go from there

<@&268886789983436800>

some guy is selling tutoring services idk if thats allowed

[abstract fractal]

It's not

[covert shuttle]

yeah I'll take care of it

[tropic oasis]

ok ty

still need help w problem 😄

[naive valley]

any thoughts regarding what the answer might be?

[tropic oasis]

uhm

not really

i know i use greater than or equal to

and solve from there

like in my example notes

[naive valley]

do you see anything that would cause the function to be discontinuous somewhere?

[tropic oasis]

not really..

maybe i'm overthinking it

[naive valley]

you're not

it's a polynomial

polynomials are continuous everywhere

[tropic oasis]

hm

[naive valley]

so you just need to express that using an interval notation

[tropic oasis]

would that mean its -infinity to infinity?

[naive valley]

yep

[tropic oasis]

ahhh okay

so polynomials are always ganna be that

[naive valley]

yeah

as long as they're not in the denominator of a fraction

[tropic oasis]

ahhh

[naive valley]

then you will have discontinuities at the roots of the polynomial

[tropic oasis]

otherwise i use the greater than or equal to and solve for x

.close

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@crisp cloak Has your question been resolved?

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[rigid flower]

An angle is 15 degrees more four times its complement. Find the measure of the larger angle.

[young finch]

So compliment means the two angles add up to 90 degrees

[rigid flower]

so like 90-x

4 times it so

4(90-x)+15=x??

[young finch]

I think a system if equations will work well here

so let's use c as the angle we want to find

x is the smaller angle

4x+15=c

x+c=90

@rigid flower

[rigid flower]

ohh

so x = 75

[young finch]

what

[rigid flower]

wait wait

sorry

give me a sec

,calc 90-75

[ocean seal]

Result:

15

[rigid flower]

yeah larger one is 75 and smaller is 15?

[young finch]

yes

c=75

[rigid flower]

Ohh

Im so dumb

thank youu so muchh

[young finch]

No you did it right

[rigid flower]

thanks again

I'll close this one now alr?

[young finch]

Yes

[rigid flower]

.close

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[spiral jasper]

interest rate help

[spiral jasper]

@solemn juniper sorry to bother you but i have 1 more question

[naive sorrel]

just ask

[spiral jasper]

it says annual 1 year right can i also do this 3504 x 2% x 244 divided by 365 but it'll give me a whole different number then 3504 x 2% x 8 dived by 12

annum*

<@&286206848099549185>

which is right formula for 1 year

244 days divided by 365 days or 8 months divided by 12 months

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.close

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[tribal fern]

I need help how to even begin to think about this problem.
I'm supposed to find what the residual/leftover positive number is when dividing 123^456 by 100... But I have no idea where to even begin with this

[alpine sable]

the residue of division for positive x & y, is the result of the modulo operator

x mod y = (your residue value)

your question is 123⁴⁵⁶ mod 100

[tribal fern]

And division by 100 means the mod is 100

So it's 23?

[alpine sable]

the remainder from x/y is x mod y

[alpine sable]

how did you get this?

[tribal fern]

Modulo means the numbers are 100 apart right? And we start with 123, so the residual is 23? I'm just starting with this mod stuff

[alpine sable]

123 mod 100 = 23

123⁴⁵⁶ isn't just the same as 123 mod 100

to compute that, there are plenty of modular arithmetic rules you can apply

I'd recommend maybe Brilliant.org's course on modular exponentiation?

[tribal fern]

Ok, well I want to solve this now and I'm sure it's been covered in a lecture I just am not seeing the connection I guess

Just tell me if I'm on the right track with finding the residue of 23^456 divided by 100

Just not sure how I handle a number that size to be honest

[alpine sable]

yup, that's correct!

• xⁿ mod y ≡ (x mod y)ⁿ mod y

but you'll need more in order to handle it's size

[sudden hinge]

A not so elegant, but easy way would be to repeatedly reduce the exponent:
123^456 = 23^456 = (23^2)^228 = ... (mod 100)
In the next step you could reduce 23^2 mod 100 and substitute it

[alpine sable]

(Very sorry if I'm getting the wrong idea, but if you don't really care about understanding how to manipulate it to get the answer & just want it done with, you could just use a calculator like Wolfram?)

[keen plinth]

this is definitely the straightforward way of doing it without having to think too much

[tribal fern]

No I just mean I don't want to go to another lecture place thing when I already have lectures on this. I'm just not seeing the connections and that's what I need help with I think.
So this nvx said, that is familiar. So it works because all congruent numbers have the same residue when divided by the mod number, this case 100

I remember that rule, just a bit shady on the application here

[alpine sable]

^ Yeah, I considered it but didn't say it

[keen plinth]

if you want to give a bit more thought to it

you can apply eulers theorem

23 is prime

so then you just need to calculate φ(100)

[tribal fern]

What does that sign mean?

I have the theorem here, it was part of a lecture the other day but I didn't quite follow. Clearly

[keen plinth]

eulers totient function

the number of integers less than n which are coprime with n

,w phi(100)

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[tribal fern]

I'm doing this in Swedish as well so the language is different

[keen plinth]

so that brings the power down to 16

[tribal fern]

things are not called the same

[keen plinth]

not sure what it's called in swedish

[tribal fern]

Here's it in my book

[keen plinth]

that's fermats little theorem

it's not as powerful as eulers theorem

[tribal fern]

Ok well that's all we covered

I googled Euler's theorem and this came up in Swedish

[tribal fern]

But going back to this, what would the next step be? I substitute (23^2)^228 by (23^4)^114? And keep going until more manageable?

[keen plinth]

you calculate 23^2

then mod it by 100

then repeat

,w 23^2

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[keen plinth]

so the residue is 29

[tribal fern]

Right right

[keen plinth]

and you do 29^228

repeat

until you exhaust the power

,w prime factorise 456

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[keen plinth]

oh gosh

there's a power of 19

,w 123^24 mod 100

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[keen plinth]

that's

real bad

wait no you just have to split it

anyway eulers theorem kills the power to 16

which is much more manageable than 456

[tribal fern]

Okay, well I will have to look up what exactly Euler's theorem is and how to use it then... Thanks for your time even if I am too dumb to grasp much of this right now..

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[reef iris]

Given two random events A and B, such that P(AB) is equal to the complement P(A'B'), and P(A) p, find P(B)

[reef iris]

I initially assumed that since P(AB) and P(A'B') do not 'overlap' at all, the only value would be zero so P(B) would equal to 1-p)

But this doesn't seem to make much sense since P(AB) could be equal to say, 0.2 or something

.close

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[alpine sable]

Hi, i want help understanding this

[alpine sable]

I just wanna to know how did they figured this ? and what they mean by infinity in the sum/why they put infinity in the sum ?

like if i have sin90 = 90 - 90^3/3! + 90^5/5! - 90^7/7! + .......

?

I mean i want to know how infinity sum = 1

I'm just can't understand it well

[trim wagon]

Lookup Taylor series expansion

[keen plinth]

we plug in radians

not degrees

[alpine sable]

ah

[trim wagon]

And this

[keen plinth]

so if you meant for that 90 to be degrees it won't work

[alpine sable]

i thought it degrees because it's x not theta

[keen plinth]

it's always radians in analysis

[trim wagon]

Well 90 could work but it won’t be =1 since sin(90rad) is not =1

[keen plinth]

[alpine sable]

yeah

[trim wagon]

,w sin 90rad

[keen plinth]

LOL

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[alpine sable]

so i should lookup to Taylor series expansion ?

[keen plinth]

yep

,w explain taylor series

[alpine sable]

I saw it once from Circle and i didn't understand it

[trim wagon]

3b1b has nice intuitive video

[ocean seal]

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[keen plinth]

wait what

[trim wagon]

https://youtu.be/3d6DsjIBzJ4

[alpine sable]

Cool

[lone heart]

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[alpine sable]

<moving to different channel>

[alpine sable]

.close

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[alpine sable]

That was so helpful thanks !

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[dark ibex]

How did this -9 get there?

[dark ibex]

( General to standard form ellipse)

i looked at some vids and know that the 71 shouldnt change

and assume that its a mistake but i could be wrong

[tacit bobcat]

Likely from human error. Ignore it.

the left factorizations are trivial

,calc 126/9

[ocean seal]

Result:

14

[tacit bobcat]

,calc 64/16

[ocean seal]

Result:

4

[dark ibex]

Holy gaddam ivbe been sitting here looking for an explaination to that -9 for over 2 hours.

thanks

.close

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[alpine sable]

Without parts

[vale wigeon]

✨ product to sum identities ✨

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

could you show me pweassseeee

[vale wigeon]

sin(a) cos(b) = (sin(a+b)+sin(a-b))/2.

use this identity and you will get a sum of two integrals of simple sinusoids

[alpine sable]

ohhhh right

half angle formula

im so stupid tyyyyyy

.close

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[vale wigeon]

(this is not half angle formula)

[vale wigeon]

ah shit.

.close

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[alpine sable]

Hi Ann can u help me ?

[frank heath]

@alpine sable #❓how-to-get-help

[alpine sable]

#help-2

[frank heath]

my bad!

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[vivid canopy]

Can someone help me how to find the area and square of a perfect square and not a perfect square it’s 11

[ornate condor]

whats a perfect square

[vivid canopy]

A perfect square

Not a rectangle

[ornate condor]

um

there are

infinity answers

do u want a general solution?

[vivid canopy]

Do I put 11 on the outside

[ornate condor]

wdym

[gray isle]

the question doesn't make sense

what's 11

[ornate condor]

u can just take any n in R>0

then accordingly u can find a rectangle of length

1 by n^2-11

amazing

[vivid canopy]

Units?

[ornate condor]

given that n>sqrt11

wow

a general solution

im amazing

but

im sure ur q is not even related to it

[vivid canopy]

No

[gray isle]

what's 11 units

is this original question in its exact wording?

is there a diagram associated with this

[ornate condor]

paraphrased into oblivion

[vivid canopy]

Diagram and symbols

[ornate condor]

yea send it

[gray isle]

not everyone is psychic

[ornate condor]

yea

only some of us are

[gray isle]

can't help if people don't know the actual problem

[ornate condor]

but i mean

i gave a general solu to ur

Q

if it even was relavant

lol

[ornate condor]

bru what

[gray isle]

whut

[ornate condor]

wha

wot

[gray isle]

is that the original?

[vivid canopy]

I forgot to add units beside the 11s

[gray isle]

or something that you whipped up

[vivid canopy]

The question was 11

[gray isle]

11 itself is NOT a question

what is the ORIGINAL question

[vivid canopy]

It said that tho in the textbook

[gray isle]

no edits, no insertions / omissions of words

what is original problem statement word for word

[ornate condor]

q number 11 hahaha

[gray isle]

Can someone help me how to find the area and square of a perfect square and not a perfect square it’s 11
makes little to no sense, we have no idea what you actually want

[vivid canopy]

Idk even know what the question is 😭

[gray isle]

take a pic?

I forgot to add units beside the 11s
you forgot them for the 121 too

[vivid canopy]

No it’s squared tho

[gray isle]

if you just wanted clarification
a square with side length 11 units will have an area of 11 units * 11 units = 121 units^2

[wise thistle]

bruv

just

[vivid canopy]

K

[wise thistle]

take a picture

of the freakin question

in your textbook

[vivid canopy]

It’s at school

[wise thistle]

so you remembered the entire question

bruv

[vivid canopy]

Yes

[wise thistle]

obviously you didnt

[vivid canopy]

Because it was 11

[wise thistle]

your question makes no sense

peace

[vivid canopy]

😔 bye

Can someone just teach me how to do perfect squares and rectangles then?

[gray isle]

find a specific problem that you struggled with

and use the exact wording of the question

[vivid canopy]

What is the square root of 64?

[gray isle]

do you know your multiplication tables

what positive number when squared gives 64

[vivid canopy]

No I use a multiplication chart

8

[gray isle]

yes

[vivid canopy]

Ok thanks

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[torn zealot]

My current guess is A D F

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[normal hazel]

I recently solved this inequality:

[normal hazel]

Now they want me to solve this.

Where they say that "e" = a small positive number

i assume i can pick any "tiny positive number i want"? But what if i pick 0,1

Wont it be as the same question above?

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[normal hazel]

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[alpine sable]

does anyone know of any video tutorial of someone completing a logs question involving brackets in a similar manner to this?

[alpine sable]

i find video tutorials really help me learn where i went wrong

[lone heart]

@alpine sable Has your question been resolved?

[wanton tusk]

okie let me try and find a vid for u

https://youtu.be/fnhFneOz6n8

u can watch from the 9min mark onwards if ur okay with ur basics (:

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[smoky hornet]

equilateral triangles ABE and CDE are congruent and have a common vertex E. calculate the measure of the angle ACB

[lone heart]

@smoky hornet Has your question been resolved?

[smoky hornet]

<@&286206848099549185>

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@smoky hornet Has your question been resolved?

[lone heart]

@smoky hornet Has your question been resolved?

[alpine sable]

BEC is isoceles

[smoky hornet]

Yeah i know

but I don't know its angles

[alpine sable]

@smoky hornet

from where is this question>?

there isn't enough information to answer the question
if you just imagine ABE and ECD are connected by a "hinge" at E, then you can rotate the two triangles
while rotating, the angle ACB will change

[alpine sable]

thanks its really helpful

i have another question posted in #help-16 about logs if interested

[lone heart]

@smoky hornet Has your question been resolved?

[smoky hornet]

Thanks for help i will let you know when im ask my teacher about this task

.close

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[dusty kettle]

How does the 6 show up in the numerator?

[paper holly]

$$3^(4/3)*2(7/6) = 3^(3/3)*2(6/6)*3^(1/3)2(1/6) = 63^(1/3)*2(1/6)$$

[ocean seal]

ibzi

[dusty kettle]

Thanks!

.close

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[night path]

Hello I got a 94/100 on a math test, how can I see what score I got out of 63? Kinda basic but im so stumped lmao idk why

[hard thorn]

94/100=x/63 and solve for x

x=63*94/100.

,calc 63*94/100

[ocean seal]

Result:

59.22

[rocky tide]

59.4/63

Or 59.22, yeah

[night path]

ay thanks

.close

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[rocky tide]

Np

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[strong night]

Need help on question 6

[heady pollen]

do you know how to calculate the volume of a pipe?

given that a pipe is basically a cyllinder

[strong night]

Yes

Pie x radius squared x height

[heady pollen]

$V=A\cdot h=\pi \cdot r^{2}\cdot h$, so in our case we get two volumes: $V_{1}=\pi \cdot r^{2}{1}\cdot h$, $V{2}=\pi \cdot r^{2}{2}\cdot h$
we can now add them to get $V=\pi \cdot h \cdot (r{1}^{2}+r_{2}^{2})$. we want to replace the pipes with one single pipe of radius R and same volume. $V'=\pi \cdot R^{2}$. Since the volumes are equal we can set them equal. $\pi \cdot h \cdot (r_{1}^{2}+r_{2}^{2})=\pi \cdot h \cdot R^{2}$

[ocean seal]

~Martin

[strong night]

Sorry can u put this in more simple terms

[heady pollen]

basically we calculate the volumes of the two pipes

add them together

to get the total volume

and then set that equal to the volume of the new pipe, since we want to get the same volume

[strong night]

But what’s the length of the first two pipes?

[heady pollen]

that is the nice part

we just care about the volume

and therefor just leave it as a variable

[strong night]

But don’t u need L to find volume

[heady pollen]

yes, but we dont need to find the volume

in the end, we only want to know the new radius

just keep it as a variable

in the end, when we set the new volume equal to the old one, we notice that we have h (the length) on both sides, therefore we can just divide by it

[strong night]

Okay wait can u try to solve the equation to see if u get the right answer cause I have an answer sheet I Just don’t know how to get it

[heady pollen]

sure

[strong night]

And then lemme know the steps because I find it easier to learn that way

Thx appreciate it

[heady pollen]

$V_{old}=V_{new} \ \pi \cdot h \cdot (r_{1}^{2}+r_{2}^{2})=\pi \cdot h \cdot (R^{2}) \ r_{1}^{2}+r_{2}^{2}=R^{2} \ \sqrt{r_{1}^{2}+r_{2}^{2}}=R$

[ocean seal]

~Martin

[heady pollen]

we set the two volumes equal

then solve for R

knowing r1 being 3 cm and r2 being 4 cm we get

$R=\sqrt{r_{1}^{2}+r_{2}^{2}} \ R=\sqrt{3^{2}+4^{2}} \ R=\sqrt{9+16} \ R=\sqrt{25} \ R=5$

[ocean seal]

~Martin

[last ether]

For the love of god use align

That's a lot of R's

[heady pollen]

wdym

[strong night]

Sorry what does the 2over 2 mean

Behind the radius variable

[heady pollen]

the upper 2 is an exponent

the lower 2 is an index

we have two given radius

[last ether]

$$\begin{align*}
R&=\sqrt{r_{1}^{2}+r_{2}^{2}} \ &=\sqrt{3^{2}+4^{2}} \ &=\sqrt{9+16} \ R=\sqrt{25} \ &=5
\end{align*}$$

[ocean seal]

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

[last ether]

Wait

Shit

[heady pollen]

xD

[last ether]

Ignore the random square root 25

But like

Yeah

[strong night]

I haven’t learned what an Index is 😢

[heady pollen]

what if i like the letter R

index is just "naming" a variable

we have two different radius

so calling them both "r" is not really helpful

[strong night]

Oo I c

[heady pollen]

so we call one $r_{1}$ and the other $r_{2}$

[ocean seal]

~Martin

[strong night]

Alr I’m gonna try to work this out a little and try to understand it

One sec

[heady pollen]

imagine you have 2 dogs without nice names, you dont say this is my dog and this is my dog, you could say this is dog 1 and this is dog 2

[strong night]

For the second step (second line that u wrote) y are u adding the two radius

?

[ornate condor]

this is billy and this is augustine the 4th

[strong night]

@heady pollen

[heady pollen]

i used the distributive rule

$\pi \cdot h \cdot r_{1}^{2}+\pi \cdot h \cdot r_{2}^{2}=\pi \cdot h \cdot (r_{1}^{2}+r_{2}^{2})$

[ocean seal]

~Martin

[heady pollen]

ab+ac=a*(b+c)

[strong night]

Wow u are a life saver

Thanks so much

U mind if I add u @heady pollen?

[heady pollen]

sure

[strong night]

Thx so much

@heady pollen I may have one more question within the next 15 mins

[heady pollen]

sure

[strong night]

Trying to figure one out on my own first

[lone heart]

@strong night Has your question been resolved?

[strong night]

@heady pollen

O sorry not 9

Just 10

[heady pollen]

do you know the formula for the volume of a sphere and that of a cube?

[strong night]

Ya

[heady pollen]

$A_{cube}=a^{3} \ A_{sphere}=\frac{4}{3} \cdot \pi \cdot r^{3}$

[strong night]

Correct

[heady pollen]

we know that the sphere fits exactly into the cube

what does this tell us?

[strong night]

The diameter is the same amount as one of the sides of the cube?

Not quite sure

[lone heart]

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[strong night]

.reopen

[lone heart]

✅

[heady pollen]

exactly

sadly i cant do shading here

[strong night]

Lol

[heady pollen]

so we get

[edgy hearth]

j

[ocean seal]

~Martin

[heady pollen]

$V_{cube}=a^3 \ V_{sphere}=\frac{4\pi}{3}\cdot (\frac{a}{2})^3$

[ocean seal]

~Martin

[heady pollen]

this is because r=a/2

where a is the side length and the diameter

$V{sphere}=\frac{\pi a^{3}}{6}$

[strong night]

Isn’t the pi symbol supposed to be beside the fraction and not a part of it?

[ocean seal]

~Martin

[heady pollen]

that does not really matter

$\frac{a}{b}\cdot c=\frac{ac}{b}$

[ocean seal]

~Martin

[strong night]

O right

[heady pollen]

now we have our 2 volumes

we want the ratio

$\frac{V_{sphere}}{V_{cube}}$ if I'm not mistaken

[ocean seal]

~Martin

[heady pollen]

plugging in our formulas we get

$\frac{a^3}{\frac{\pi a^{3}}{6}}$

[ocean seal]

~Martin

[heady pollen]

simplifying we get $\frac{6}{\pi}$

[ocean seal]

~Martin

[strong night]

For the bottom equation I don’t understand how it simplified to pi times “a” cubed/ 6

Doesn’t the a/2 cubed fraction stay alone since it’s. A fraction with an exponent?

[heady pollen]

$\frac{4\pi}{3}\cdot (\frac{a}{2})^{3}=\frac{4\pi a^{3}}{3\cdot 2^{3}}=\frac{4 \pi a^{3}}{24}=\frac{\pi a^{3}}{6}$

[ocean seal]

~Martin

[strong night]

Okay but when u did this, the cube was “a” cubed and the rest was a part of the spheres volumes. Therefore did u just compare the sphere volume to itself since the “a” cubed was cancelled out

[heady pollen]

i compared the volume of the cube to the volume of the sphere

V1/V2

[lone heart]

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[lone heart]

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[keen seal]

I can’t remember what order I need to plug in these values

[abstract fractal]

Wdym what order

[keen seal]

Like
Is the point plugged in to f(a) and a?
Is it done before we plug in x to a?

[abstract fractal]

I don't get what you're asking

Can you provide a concrete example of the different orders?

[keen seal]

Like
((x^2-8)-8)/x-4

[abstract fractal]

What do you get if you "order" it differently

[candid torrent]

first you should find the derivative of the function

using the definition of the derivative

[abstract fractal]

You just reworded what the problem was asking

[candid torrent]

yeah and?

[abstract fractal]

OP already stated what their issue was, and it's something to do with how something is ordered. Rewording the problem doesn't get to the bottom of what they mean by that

[candid torrent]

understandable

i'll leave you guys alone then

[abstract fractal]

Don't forget what a is

It's not lim x->a

a has a value

[keen seal]

Oh
4

So you get 8

And the normal deriv of x^2-9=8 is 2x, plug in 4, you get 8

So it works

[abstract fractal]

Yep

[keen seal]

Thanks

@lone heart done

Figured I'd try

[abstract fractal]

Say ".close"

[lone heart]

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[livid cove]

sec^-1(3s^3 + 7)

[livid cove]

find derivative

so derivative of arcsecant is 1 / |x|(x^2 - 1)^1/2

I got 6s / |3s^2 + 7| ((3s^2+7)^2 -1)^(1/2)

but it says its wrong

[pastel bane]

Can you take a picture of your work?

[livid cove]

this is the answer the derivative calculator got

how did it get the fraction under the square root in the denominator

and I thought it was X^2 - 1 under the square root in the denominator

but they have it as 1 - 1/X^2

[pastel bane]

You’re thinking of arcsin or arccos

[abstract fractal]

√(1 - 1/x²) = √((1/x²)(x² - 1))

Just factor out the 1/x² from inside the square root

[livid cove]

where is it?

[abstract fractal]

Wdym where is it

[livid cove]

the factored out

[abstract fractal]

1 + 1/a = (1/a)(a + 1)

[livid cove]

is this the correct arcsec derivative?

[lone heart]

@livid cove Has your question been resolved?

[solar tapir]

Yes it is

[livid cove]

so why is there a fraction in the answer

in the denominator square root

[solar tapir]

Use product rule for derivatives

[livid cove]

why is there a fraction in the square root in the denominator

this is the question I am working on and I don't know how they got that

[solar tapir]

Because of the pythagorean identity $\sec^2 x = 1 + \tan^2 x$

[ocean seal]

Max Hetfield

[livid cove]

where does sec come from

I don't see sec in there

[solar tapir]

From the second term, left to right

[livid cove]

3s^2 + 7?

why would that be in there

why arent they using the last term

[solar tapir]

You need to focus in the first image you shared

before applying that to this one

[livid cove]

yea I looked at it

why wouldn't they use the finished derivative, the last term?

[lone heart]

@livid cove Has your question been resolved?

[livid cove]

?

can someone help me with this arcsec derivative

[lone heart]

@livid cove Has your question been resolved?

[livid cove]

anyone help me with this derivative?

[livid cove]

??

[livid cove]

need help with arcsec derivative

[alpine sable]

$\arcsecx=y$

[ocean seal]

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

[alpine sable]

uh

how does this work

anyway

y=arcsecx

secy=x

differentiate both sides

secytany * dy/dx = 1

dy/dx= 1/secytany

@livid cove

@heavy otter close this channel this guy

isnt closing it

[lone heart]

@livid cove Has your question been resolved?

[trim wagon]

close this channel this guy

[lone heart]

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[minor needle]

make them linear firstly, then solve the system

[vital atlas]

this is what i did

i got x=42

which is not what the answer key says

[minor needle]

ok nvm, I see

you need to have log(...) = log(...) to remove logarithms

use property on RHS then continue

[vital atlas]

oh

okay this makes more sense now

tysm <33

.close

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[lean lily]

Anyone please ?

m is a positive integer

[echo socket]

Maybe integration by parts? Not so sure but this is would be my 1st attempt

[lean lily]

Alr tried

[ornate condor]

oh hey its u

[keen plinth]

uh oh the baseline is misaligned

[ornate condor]

fix preamble

[keen plinth]

annoying

[lean lily]

Contour ? 😭

[keen plinth]

oh no

the

smh

[keen plinth]

the only pole is at 0

the residue isnt too hard to compute

[lean lily]

we haven't studied contour integrals yet

[keen plinth]

rip

[lean lily]

i'm supposed to find it using the basic integration stuff LMAO

[keen plinth]

theres probably a better way with induction

i just havent bothered

residue theorem just does it

[lean lily]

I was certain there was a way to write U2m or Um+2 or Um+1 using Um

But i couldn't find it

like Wallis's integral way

[keen plinth]

yes ive done those before

[lone heart]

@lean lily Has your question been resolved?

[lone heart]

@lean lily Has your question been resolved?

[lone heart]

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[alpine sable]

I need to check if my answer is correct in the following:

[alpine sable]

,rotate

[alpine sable]

I am requested to find the tangent equation for a function.

Is my answer correct?

Ignore the function at the bottom (f(x) = x²)

[wanton tusk]

yesh seems right to me (:

also u have beautiful handwriting

[alpine sable]

Haha, tyyyy

[keen plinth]

y'all both have beautiful handwriting lmao

[wanton tusk]

but look at this omg

woah 🤯

but tyty HAHAH

[alpine sable]

May I close this help channel?

[frank heath]

yea:)

[alpine sable]

.close

[lone heart]

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