help-23

[lean otter]

it didnt get rid of the fraction though did it

it just created a new one

10/2 x X

[worthy hemlock]

But what's 10/2?

5

[lean otter]

10 over 2

wait shit

[worthy hemlock]

So 5x

Hence simplification

I think it crashed in general, because I couldn't connect to discord for like 5 to 10 minutes

[lean otter]

holy shit bro i finally did it

alright i need to go over this in general

so basically

the law is that

when you mulitpy with the common multiple

it pushes away the x?

[worthy hemlock]

It's not a law

It's literally just clearing fractions

[lean otter]

yeah but how does it clear the fraction

[worthy hemlock]

You find the LCM, of the denominators

[lean otter]

so did i do this one correctly too

or i think that

12/3 should be 24/3 right

cause its 2y

[worthy hemlock]

What's 24/3?

Is it 6?

[lean otter]

what

dude im talking about when clearing the fractions

i did it correctly now

one sec letm e send it to you

[worthy hemlock]

You said 24/3 = 6

[lean otter]

no

[worthy hemlock]

Everything else is correct, except for that

[lean otter]

i said should i have made it 24

or 12 x 2y

and i tested

and no

i was supposed to make it 24

cause its 2y

see whati mean

i multipled 12 with 2 on the second one

[worthy hemlock]

So explain why there is a 6

[lean otter]

cause it was 2y

cause 12 divided by 2 is 6

i was wondering if the Y stays with the 2

[worthy hemlock]

You forgot about the 2, you do (12 * 2)/3

[lean otter]

yeah exactly

i didnt forget

[worthy hemlock]

It's all multiplication

[lean otter]

i didnt know i was supposed to multiply it with 12

cause i was unsure while doing that step

so i went back

and re did it

and got the correct answer

which is good

because now

i know thati need to multiply it with the 2 next time

or whatever

x

cause i wont know what the number wil be

anyways

thank you very much @worthy hemlock @blazing fiber @flint ledge

not only did you help me with the answer but i learned something new aswel

take care and thank you

how do i close the channel?

/close

.close

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[lean otter]

Write the equation using the roots

[whole cloud]

(7+3i)(7-3i)
49+21i-21i-9i^2
49-9i^2 is what i think not totally sure

[lean otter]

Mmm ngl I don't think that's right but thanks, I'll go check

[whole cloud]

im an 8th grader so dunno

[toxic needle]

i think it is x^2-14x+58, if you plug it into the quadratic formula it gives you that

ping me if u need smthn else

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[lean otter]

<@&286206848099549185>

[toxic needle]

find EDA and subtract it from 180 to find the angle beside it which is the same as the angle for the answer

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[lean otter]

<@&286206848099549185>

[toxic needle]

you want to be able to have 2 out of the 3 angles in the triangle that ACB is in, and then you can subtract those from 180 to find the angle you want

what I'm saying is, find angle A, which you can do with what you have

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[keen lance]

i literally dont understanf anything PLEAASE

[arctic osprey]

the base equation is y=x since y is the same as x on that graph

alright so consider the equation f(x)=x

uh

[keen lance]

?

[arctic osprey]

how do i explain this

so f(x)+d means that the graph goes up d units

[keen lance]

i have no idea, my teacher literally does nothing

[arctic osprey]

f(x)-d means that the graph goes down d units

so say f(x)=x, f(1)=1, f(2)=2, f(3)=3, so on

now we're considering f(x)+d
say that d is 5

f(1)+5=6, f(2)+5=7, f(3)+5=8, and so on

everything goes up by 5

[keen lance]

so

wait

hold on my brain

OH OKOK

i kind of get waht ur saying

[arctic osprey]

this isn't really hard, f(x)+d means that the graph transforms up d units

f(x)-d means that the graph transforms down

[keen lance]

i have a question tho, what is g(x)= f(x) + 6

is it around the same thing

like subsituting the x with a number

or is it differebt

[arctic osprey]

so g of x is equal to f of x but it's moved up by 6

[keen lance]

how do i solve it than?

the g thingy

[arctic osprey]

g(x) is basically f(x) but you add 6 to it

[keen lance]

so would it be

[arctic osprey]

f(x): y=x
g(x): y=x+6

[keen lance]

g(6)=f(6) + 6

how would u solve it tho

ive only done

f(x)= blah blah blah

f(x)= 6x + 4 for example

where u get a like thing you have to subsitute it wuth]

[arctic osprey]

g(x) is f(x) but you add 6 to it
if u add 6 to it the graph moves up by 6

[keen lance]

so

g(x) = f(x) + 6

would there be no solving

or

just graphing it up

[arctic osprey]

this might help u understand it

[keen lance]

oh

what about the g(x) tho it still kinda puzzles me

[arctic osprey]

there isnt really solving that i think

[keen lance]

oh

is it just graphign?

[arctic osprey]

ya

basically

[keen lance]

oh

[arctic osprey]

f(x) is the base function

the +6 is what u do to it

[keen lance]

so move it up

[arctic osprey]

yes

[keen lance]

alr alr

[arctic osprey]

g(x) is f(x) but you move it up by 6

[keen lance]

ahhh

okok i think i can do the rest on my own

tysm for helping me

[arctic osprey]

btw ur teacher is probably going to do horizontal movement next

[keen lance]

oh

thx for the warning

😭

[arctic osprey]

it's going to be f(x-d) and f(x+d)

[keen lance]

i hate algebra

[arctic osprey]

if it's f(x-d) then u move the thing right by d units
if it's f(x+d) then u move the thing left by d units

[ember bough]

for horizontal movement
f(x-d) moves the graph to the right
f(x+d) moves the graph to the left

it's a bit confusing

AGH I HATE MY INTERNET THINGS TAKE FOREVER TO SEND

[keen lance]

rip

okok im gonna close this i got it now ty

.close

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[oblique nova]

not quite sure how to approach this. if I took the derivative of r(t), would i treat a and b as constants?

[pastel verge]

Treat b as constant and find the max of k

[oblique nova]

how would i go about finding the max of k

[pastel verge]

Differentiate wrt a treating b as constant and equation the differential equal to 0

[oblique nova]

sorry can't quite understand what you said, can you rephrase

[flat frigate]

it's Sam

[oblique nova]

kk

derivative of k with respect to what

t?

[pastel verge]

You will most probably get a=b

[pastel verge]

a ofc when we say f'(x) that means we differentiate with respect to whatever is in bracket

[oblique nova]

ok

[pastel verge]

f'(2x) = d(f(x))/d(2x) for example

[oblique nova]

a=b, a=-b

[pastel verge]

So from this a=-b isn't possible

As both a and b are non negative

[oblique nova]

yeah

so just a=b then

[pastel verge]

So a=b

Substitute this in k

And that will be max

[oblique nova]

tysm

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[pastel verge]

What was the answer

[safe radish]

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[wheat juniper]

does anyone know why here m4 and m5 only get grouped

why dont m4 m5 and m6 get grouped?

this is a karnaugh map

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[worthy hemlock]

Because for k maps, you circle the amount of elements based on 2^n power, so 2 elements is 2^1 and the 4 elements is 2^2

[worthy hemlock]

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[rich tiger]

Can someone explain that to me

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[quasi bison]

$5^{n+1} + 4 \times 5^{n+1} = 5 \times 5^{n+1} = 5^1 \times 5^{n+1} = 5^{n+2}$

[flat frigate]

Ann

[quasi bison]

@rich tiger is this the explanation you were looking for?

[rich tiger]

yes but how did 4 dissapear

[frozen nebula]

I think a 5^n+1 was factored making it 5^n+1(4 + 1)

[quasi bison]

it didn't disappear...

if i were to tell you x + 4x = 5x would you also ask where the 4 "disappeared" to?

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@rich tiger Has your question been resolved?

[rich tiger]

I sound so dumb rn but

How did $5^(n+1) + 4$ change to 5

[flat frigate]

Habeeb M

[quasi bison]

it did not, and there was no $5^{n+1} + 4$

[flat frigate]

Ann

[quasi bison]

$5^{n+1} + 4 \times 5^{n+1}$ means $5^{n+1} + (4 \times 5^{n+1})$, \textbf{not} $(5^{n+1} + 4) \times 5^{n+1}$

[flat frigate]

Ann

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[hasty cloud]

dont know how find the value of the a and b but they have to all add up to 360

[hasty cloud]

.close

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[hasty cloud]

need help these must equal to 360 but i dont know how to work it out??

[upbeat dragon]

The little squares there indicate right angles (90 degrees)

[hasty cloud]

oh ok

[upbeat dragon]

So that means g + 90 + 36 + 90 = 360 as you've said

[hasty cloud]

wow that was quick

thanks but i need help with this worded problem too

[upbeat dragon]

Lol I saw the channel pop up

[hasty cloud]

1. If i had a pair of complementary angles and one was 30° what would the missing angle be?

2. If I had a pair of supplementary angles and one was 80° what would the missing angle be?

[upbeat dragon]

Complementary angles are angles that add up to 90°, and supplementary angles are angles that add up to 180°

[hasty cloud]

oh thanks

just can you help me for 2 more pls

[upbeat dragon]

Sure

[upbeat dragon]

You can use this to test if they're complementary or supplementary. Add the angles they gave and see what criteria they fit into

[hasty cloud]

ok

anyways tysm @upbeat dragon

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[lean otter]

That is not fully correct

you got the -x right but missed the -1

[empty maple]

this?

like this?

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[untold yarrow]

Can someone give me a hint to evaluate this

[flat frigate]

Tim O'Brien

[static talon]

Where’s Tim

[untold yarrow]

Here

[inland ivy]

Do you agree that repeatedly multiplying an base and exponent is the same as taking the base to the power of repeatedly adding the exponents together? @untold yarrow

So like

$\prod_{s=n}^{2n-1} 10^s = 10^{\sum_{s=n}^{2n-1} s}$

[flat frigate]

PapaBread

[untold yarrow]

Not yet

Oh well kind of

Yeah I think I get it

Wait thats cool

Could you show a quick example of that thpo

Your statement there

[untold yarrow]

x

[inland ivy]

Like an equation?

[untold yarrow]

Yeah

[inland ivy]

I'm really bad at simplifying sums but

If you had a super basic product like

$\prod_{n=1}^{5} 5^n$

[flat frigate]

PapaBread

[inland ivy]

You could rewrite it as 5 to the power of this whole sum

$\sum_{n=1}^{5} n$

[flat frigate]

PapaBread

[inland ivy]

Which is super easy to numerically prove

5^(1+2+3+4+5) = 5^1 * 5^2 * 5^3 * 5^4 * 5^5

Ofc

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[formal mirage]

Hey. If you have a unit circle with an angle between 2pi and (5pi)/2 does it lie in the same place as an angle thats between pi and (5pi)/4?

[formal mirage]

The measurements are in radians btw

wait nvm

i forgot about refrence angles

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[untold yarrow]

Oh yeah because $a^b\cdot a^c = a^{b+c}$

[flat frigate]

Tim O'Brien

[untold yarrow]

Thanks bro

[lean otter]

Yes

[untold yarrow]

.clost

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[untold yarrow]

I forgot you can't chat in these anymore

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[red peak]

Hey guys

[red peak]

Can someone help me with this question

I have the solution on me and im talking about b.

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@red peak Has your question been resolved?

[red peak]

<@&286206848099549185>

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@red peak Has your question been resolved?

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@red peak Has your question been resolved?

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@red peak Has your question been resolved?

[red peak]

<@&286206848099549185> help please?

[idle dove]

For (b) you're solving the differential equation with the initial condition y(0) = f(0) = 2

If it's easier to see in y instead of f(x), you should be able to do the integral of y'/y^3 * dy

[dire pelican]

does anyone know the formula to find the intensity(Amps) for a resistive/inductive circuit in parallel??

[worthy hemlock]

Google probably does

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[red peak]

Yessir

.close

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[gentle nest]

help

[gentle nest]

i need to find DE

thats 15 deg. btw

not 150

basically this is what i tried

no

CDE is 15 degrees

im focusing on the bottom triangle

since the entire triangle was a 90-75-15 triangle

i split it into a 30-60-90 and a 15-15-150 iscoseles

(dunno if this is the right step)

i labeled the two legs of the isosceles triangle as a

no this is competitive math

i got this

using triangle ratios

since the side length equals 10

i can find a

(lemme find it rq)

but idk what to do from now

basically trying to find x

oh

yes

ok

i can use trig on this

but i cant

its amc 10

im not allowed to use trig

and no calcs btw

k can anyone help i found a

need to find x

wait

on

oh

AHHHH

im overthinking it

i can just use pythag

lmao im dumb

thx

cant believe im doing math on a napkin rn

literally dont have paper...

lmao yeah

lmao im being dead ass

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[worn laurel]

can someone help me with this question please?

[lean otter]

is the answer 15* (2)^n-1 @worn laurel ?

[worn laurel]

I'm not sure

[pulsar condor]

What's the 1st term?

Didn't ask you

• don't give answers

[lean otter]

Didn't know, sorry

[worn laurel]

30/2=15

[pulsar condor]

Yes

So you can just plug your a and R into the general term

[worn laurel]

15*2^n-1

is my formual correct?
@pulsar condor

[pulsar condor]

Yes

[worn laurel]

how to work out the first term?
@pulsar condor

[pulsar condor]

You... did it already

You also don't need to ping in every message

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[worn laurel]

。reopen

.reopen

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✅

[worn laurel]

how to calcualte the third term?

15*2^3-1?

[pulsar condor]

Yes.

[worn laurel]

okay thank you

.close

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[solemn willow]

heelllooo

[solemn willow]

i dont understand the bracket part

like what does it mean by all of that

can someone show a proper equation T_T

[brave edge]

[
\frac{p}{q} = 2^{1/3} \times \frac{p^3}{q^3} = 2p^3 = 2q^3
]

[solemn willow]

u.u

[flat frigate]

Epsilia aka Epe

[solemn willow]

i dont understand how can 2 cube times p^3/q^3

equals to that

[brave edge]

p and q are numbers such that gcd(p,q)=1, from the definition of irrational numbers, right ?

[solemn willow]

yeappp

[brave edge]

So you can not find any integer factor in p or in q such that something cancels out at the top and at the bottom

[solemn willow]

oh....

[brave edge]

Since p^3 and q^3 are both multiplication of prime numbers

They don't share any common factor

[solemn willow]

ah i see.... well i kinda dont get it by visual, but maybe i get it from words hehe 😅

[solemn willow]

how about the 2nd bracket

[brave edge]

So $gcd(p^3,q^3)=1$ too

[flat frigate]

Epsilia aka Epe

[solemn willow]

ah so thats why....

wait a min

before going to 2nd bracket

[brave edge]

I guess you are writing $2^{1/3}$ as $\frac{p}{q}$

[flat frigate]

Epsilia aka Epe

[solemn willow]

arent it supposed to be $2^{1/3}p^3$

[flat frigate]

mifuyu

[solemn willow]

o3o

[solemn willow]

hmm?

[brave edge]

AH

I just realised

[solemn willow]

or am i wrong

yea thats why i got confused

[brave edge]

[
\frac{p}{q} = 2^{1/3}
]
[
\frac{p^3}{q^3} = 2p^3 = 2q^3
]

[flat frigate]

Epsilia aka Epe

[solemn willow]

kekw

can u explain owo

[brave edge]

That's what's written, it's just that the $2^{1/3}$ and the $p^3/q^3$ are not separated in the shitty writing they showed you

[flat frigate]

Epsilia aka Epe

[solemn willow]

ah i see

[brave edge]

Since $p^3/q^3 = (\frac{p}{q})^3$

[flat frigate]

Epsilia aka Epe

[brave edge]

Then you can use your absurdity hypothesis

About $2^{1/3}$ being rational

[flat frigate]

Epsilia aka Epe

[solemn willow]

ah i see

[solemn willow]

how about

the 2nd bracket

[brave edge]

So this one then, if you see the number you're studying at the power of 3

Then you get that $\frac{p^3}{q^3} = 2$

[flat frigate]

Epsilia aka Epe

[brave edge]

Which means that $p^3 = 2q^3$

[flat frigate]

Epsilia aka Epe

[solemn willow]

OHHHHHHHHHHHHHHHH

[brave edge]

So yeah there is an other part that I didn't write correctly

[solemn willow]

goddamn

now it's clear to me lmao

i got so confused for hours

thank u so much epe ❤️

[brave edge]

They should have written it like that :
[
\frac{p}{q} = 2^{1/3}
]
[
\leftrightarrow \frac{p^3}{q^3} = 2
]
[
\leftrightarrow p^3 = 2q^3
]

[solemn willow]

true.......

but there is no comma or any symbols at all

so i get confused

and i think i get the 2nd bracket as well o3o

[flat frigate]

Epsilia aka Epe

[solemn willow]

thank youuuuuu epe

[crisp latch]

pls don't close 2 min

[brave edge]

Lmao

[solemn willow]

oh ok sure

[brave edge]

Npppppp

[crisp latch]

@brave edge can I say my method ( I didn't really get what u did ) ?

can u check it ?

[brave edge]

Yeah sure

[solemn willow]

sure sure i wanna see it too

[brave edge]

Good luck for writing in LaTeX

[crisp latch]

acc to definition of rational num
it is of form p/q where q not equal to 0 and hcf of p,q=1

lets assume 2^1/3 is rational
here p^3/q^3= 2
so p^3=2q^3
so that means p is a multiple of 2
so p=2k

[brave edge]

It is exactly what's written here so far

[crisp latch]

p^3=2q^3
(2k)^3=2q^3
4k^3=q^3
so this means q should also be a multiple of 2
but we know that for it to be a rational no hcf of (p,q) = 1
this contradicts to our assumption that 2^1/3 is rational

correct ?

ya cool
I didn't see it as it was way ahead

thnx

@solemn willow you can close it and thnx for keeping it open

[solemn willow]

o.o

thanks for ur confusion though xD

.close

[safe radish]

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[crisp latch]

no no I did same only

[safe radish]

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[magic shale]

Hello, I need help with an application of quotient rule for derivatives

[magic shale]

The function C(r) where V is some unknown constant

I cant tell if the derivative of the 4V/r term would be zero over rsquared or if it would be -4V/rsquared

Because if it isnt 0 i really dont know how to solve this question.. i struggle with knowing the little details of the application of these rules

[weak dirge]

Do you have the problem?

[magic shale]

This is from lecture so its the best ive got unfortunately

Ive realised now the derivative ive got is wrong

[weak dirge]

you're asking $\frac d {dr} (2 \pi r^2 )$?

[flat frigate]

KurtDee

[weak dirge]

I'm not sure I see it

can you define c(r)?

[weak dirge]

okay

[magic shale]

Just need help for the highlighted term getting the first derivative

[weak dirge]

$C(r) = 2 \pi r^2 + 4V/r$ ?

[flat frigate]

KurtDee

[magic shale]

Yeah where V is some constant

[weak dirge]

Sure

So you can do $\frac d {dr} 2 \pi r^2$ right?

[magic shale]

Assuming domain is (0,infinity)

[flat frigate]

KurtDee

[magic shale]

Yea

[weak dirge]

So lets do two ways

The way I would solve it and quotient rule

I see that as $4Vr^{-1}$, so $\frac d {dr} 4Vr^{-1} = -1*4V r^{-2}$

[flat frigate]

KurtDee

[weak dirge]

alternatively using quotient rule

$f(r) = 4V, g(r) = r, \frac{f'g - fg'}{g^2}$

[flat frigate]

KurtDee

[magic shale]

Hm thats what i feared because now i dont know how to optimize this

[weak dirge]

$f' = 0, g' = 1, \frac{f'g - fg'}{g^2} = \frac{0-4V}{r^2} = -4vr^{-2}$

make sense?

[flat frigate]

KurtDee

[magic shale]

Thank you Kurt that makes sense

[weak dirge]

wonderful

[magic shale]

I will try to work on this some more

.close

[safe radish]

Channel closed

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[weak dirge]

good luck

[magic shale]

.reopen

[safe radish]

✅

[magic shale]

One more question @weak dirge if youre still there

[weak dirge]

sup

[magic shale]

If one term in a two term derivative at a point is undefined and the other is 4 lets say

Do we ignore the undefined term or is the whole thing unusable

[weak dirge]

We pretty much ignore the point

Hold up

let me get an example

It really dependsssss on the limit at that point

[magic shale]

Ill show u one sec

[weak dirge]

https://en.wikipedia.org/wiki/Dirac_delta_function

So the derivative of this function is a constant 0

because the limit at f(0) = 0

[weak dirge]

C''(r)?

[magic shale]

Yeah

Like what can i do with that if im trying to find global max and min

[weak dirge]

Global max and min

Hm

[magic shale]

Global min specifically

[weak dirge]

Do you have points at C' == 0?

[magic shale]

The question is basically asking what is the lowest cost C(r) possible for a cynlinder manufacturer

[weak dirge]

if so, it would be either those points or the bounds

otherwise it would be the bounds

[magic shale]

C’0 is undefined as well

[weak dirge]

what's the range of C

[magic shale]

Open

[weak dirge]

[0,\infty) ?

or open at 0

[magic shale]

Yeah positive range

To infinity

I dont think it goes below a certain point anyway

[weak dirge]

So you can't remove points that are not contained in the derivative

I'm trying to think of an example

[magic shale]

Ill type the question

[weak dirge]

basically yeah lets say we have 1/x shifted left one unit so like 1/(x+1) with range [0,infty)

So because we have no point on the left we dont have an f', but f(0) is still the max

[magic shale]

A closed cylindrical tank of constant volume V is manufactured where material for the sides costs twice as much per square unit than the ends what dimensions should be used to minimize total cost

[weak dirge]

Really it comes down to a test

consider all C' = 0, and the bounds of the function

obviously if C(\infty) is a solution there is no minimum

[magic shale]

Global min is between r=0 and r=infinity at some point since both ends tend to infinity

[weak dirge]

Yeah so just plug those points in

You can do a refined approach and consider C''

[magic shale]

Which points though

[weak dirge]

x where C'(x)=0

those are critical points

[magic shale]

Ohhhh

[weak dirge]

min max saddles

[magic shale]

Thanks ill try that

[weak dirge]

yeah

if there aren't too many you don't really need to consider C''

unless it asks you to classify those points

in that case yeah you do

@ me if you need me

[magic shale]

Thanks Kurt!!

.close

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[sand hornet]

find the area of the shaded region

[lean otter]

answer is in help 12

you might wanna close this channel

[sand hornet]

.close

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[graceful patrol]

How do I make v the subject?

Equations with derivatives confuse me

[safe radish]

@graceful patrol Has your question been resolved?

[brave edge]

Integrating with respect to t

[graceful patrol]

No clue how to do that 😬

[brave edge]

You have $\frac{dv}{dt} = \frac{Kt}{m}$

[flat frigate]

Epsilia aka Epe

[brave edge]

So you have the expression of the derivative

And you want the function itself

What function would have $\frac{K}{m}t$ as its derivative with respect to t ?

[flat frigate]

Epsilia aka Epe

[graceful patrol]

KT^2/2m

[brave edge]

Gg

[graceful patrol]

Sorry lower case t

Thanks a lot

[brave edge]

Yeah but you forgot something

When you differentiate a constant, you get 0

Since you can write $\frac{K}{m}t$ as $\frac{K}{m}t + 0$

[flat frigate]

Epsilia aka Epe

[brave edge]

Don't forget a constant

[graceful patrol]

But wouldn’t the 0 be negligible?

[brave edge]

Why would it be ? If you were to differentiate your function, okay it would get the derivative you had before

But who said it was the only function for which it worked ?

[graceful patrol]

Ah i see

Thanks a lotttt

[brave edge]

You can close

[graceful patrol]

.close

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[cerulean sparrow]

https://i.imgur.com/tkVOhTe.png

[cerulean sparrow]

How did they get the value of a?

[weak dirge]

Probably zscore * std let me check

yep

[cerulean sparrow]

what is zscore again @_@

[weak dirge]

lol

hold up

zscore = number of standard deviations for the confidence interval

so you have 0.9 two sided confidence interval

that's 0.05 on both sides

[cerulean sparrow]

how did they get 0.9 in the first place?

[weak dirge]

uhhh looks arbitrary

let me checl

Yeah it looks like to be given by the problem

you want 90% confidence it seems like

is that correct?

[cerulean sparrow]

oh

Yeah

[weak dirge]

so you leave out 5% on both sides

zscore transforms % confidence to the number of standard deviations

https://www.ztable.net/

if you check out 1.64, you get ~5%

so you need 1.64 standard deviations, multiply by std

which is 0.49

make sense?

[cerulean sparrow]

im looking at the column and row but how does 5% become 1.64?

[weak dirge]

So lets look at the image embedded

at z=0, you get a straight half

with z = - whatever, you're getting the area of the tail

it's not a linear relationship, in fact there is no equation that can calculate this exactly

So we use charts to change it

so at 1.64 units, you're left with 5% of the graph

but like at 2 units, you have 0.0227 left

but all that this chart is is telling you how many standard deviations correlate to the % of data captured

does that make sense?

[cerulean sparrow]

Yeah

from the middle, z=0
the left tail is -2 and the right is 2?

[weak dirge]

Well okay

we want 90% of the data, that means we leave out 10%

and we're doing this on both sides as you can see

by [mu-a, mu+a]

so a needs to capture 5% of the data

a = (number of stds for 5%) * (standard deviation size)

[cerulean sparrow]

I see, how did you get 1.64 again? i couldnt figure out how to use that chart

[weak dirge]

so within the chart we have the area left over

we're looking for 10%/2 = 0.05

the sides represent the 1's and the 10th's place, and the top represents the 100th's place

[cerulean sparrow]

wait nvm I think i got it

[cerulean sparrow]

By doing this, the ans is still off by a bit, why is that?

[weak dirge]

how off?

[cerulean sparrow]

ans is apparently .8092

instead of .4902 I also tried 3.1/ sqr(40) to see if its a rounding problem but its quite off

[weak dirge]

Hold on

let me get the exact answer

I'm not sure where they're rounding

Yeah I think it's a rounding error

I'm not exactly sure

in any case you're about 1%

which is generally fine?

[cerulean sparrow]

Also , I checked their intervals and seems like the a value is widely different

https://i.imgur.com/KYALytF.png

So strange

[weak dirge]

I dont understand this

oh

yeah that's really weird

I have no idea

[cerulean sparrow]

Hm, maybe typo

i'll check it with my teacher. .

[weak dirge]

they could be rounding? that's my only idea

I think you should email/talk to him/her

[safe radish]

@cerulean sparrow Has your question been resolved?

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[timid goblet]

<@&286206848099549185>

[timid goblet]

Help

[queen parcel]

Brayden - don't ping Helpers right away

Or everyone or here

Read #❓how-to-get-help

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@timid goblet Has your question been resolved?

[ember bough]

hint: alternate interior angles

[safe radish]

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[lean otter]

That is an interesting problem. I would think that you need to find the integrals for the volume of both rotations of the section and then equate them and solve for mu

Makes sense?

And since you want to keep them in terms of x, you can use two different methods for volume

Shell and disc method for example, I think you'll need both

[last halo]

Okay will try soon

[safe radish]

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[long minnow]

im not a helper but i know how to do that question if you still need help

have you attempted a)?

[velvet quartz]

i already finish

[long minnow]

good job lol

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[velvet quartz]

@long minnow low

[long minnow]

?

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[velvet quartz]

how about this ?

how about 11 - 14

[vestal valley]

You need help with 11 and 14?

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[strong pawn]

how would i find the area in purple?
i know the coordinates of all the points, and i know the distances between the points surrounding the area

[inland ivy]

I would start with the area of triangle PCR

Then find the radius of circle C

Find the arclength from P to R

Find the area of the segment

[strong pawn]

i got the arclength, but how would i from that find the area of the segment?

[safe radish]

@strong pawn Has your question been resolved?

[dreamy glen]

I think it is called a "sector". Try to figure out what fraction of the area of the circle is covered by the sector.

Like, the central angle out of 360 degrees. Or the arclength out of the circumference.

[strong pawn]

so like find the area between the chord of QR and the arclength?

[dreamy glen]

Oh. You are looking at the triangle PQR, not PCR?

[strong pawn]

yeah

the area of the purple segment

[dreamy glen]

Bread was suggesting that you use the triangle PCR. QCR is a "slice" of the circle, so its area might be easier to figure out.

Essentially comes down to the same thing though. The segment is the sector without a triangle.

[strong pawn]

ooh like find the area of PCR and then subtract the area of QCR?

[dreamy glen]

Yes. The sector QCR, not the triangle QCR.

[frail glacier]

um can someone help me

[strong pawn]

aah i see

[dreamy glen]

Might need to clear about that in your answers.

[strong pawn]

thanks a lot

.close

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[safe radish]

@full badger Has your question been resolved?

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[vale bronze]

5(b) please

[thin bridge]

what have you tried so far?

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[vale bronze]

.reopen

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✅

[vale bronze]

<@&286206848099549185>

I know that angle DAC is 18.4

[thin bridge]

to deteremine <CAB consider getting more information about the sides of triangle CAB

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[slate mango]

b^2 -18<0
How can I write this in interval form pls help

[slate mango]

I got this but it is wrong (-sqrt(18), sqrt(18))

[quasi bison]

when you say "wrong" do you mean "the homework system rejects it"

[thin bridge]

technically the interval is correct,
but isn't simplified

[safe radish]

@slate mango Has your question been resolved?

[slate mango]

Yes

[slate mango]

How can I simplify it, could you please help?

[quasi bison]

can you show the problem as it appears in the homework system and your input exactly as it appears

we cannot at this stage rule out an input formatting error

but another possibility is that there were some instructions you missed

[slate mango]

Sure one moment
Pic removed

[quasi bison]

...this is weird

(-sqrt(28), sqrt(28)) seems like it should be the correct answer

[slate mango]

Oh

Okay, I'll try something else, thanks for trying!

Got it, after trying multiple stuff

Answer is

(0, sqrt(28))

For the first one it is (sqrt(28), inf)

Thanks!

.close

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[subtle void]

How many 6-digit odd numbers can be formed from the digits 1,2,3,4,5 and 6 (with repetition) such that they are divisible by 3? I know that the sum of the digits should be divisible by 3 too but other than that I'm lost as to how to approach this problem

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[lean otter]

i have a triangle drawn out

but what happens next I only have 2 sides.

[ancient escarp]

you have the angle needed for law of cosines

[lean otter]

how

[ancient escarp]

northeast is a straight 45 degree angle up and to the right

west is a straight line to the left

[lean otter]

True thank you. Never looked at it that way.

[ancient escarp]

👍

[lean otter]

wbout this

[ancient escarp]

for a parallelogram, side lengths across from each other are congruent

so you have all side lengths, use law of cosines and solve for angle

[lean otter]

💀

I'm so braindead.

ty

.close

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[limpid shell]

hey I am stumped on how I can apply Z-scores to a probability tables

[limpid shell]

This is the problem I am on

\

I need to find the probability score of this, using this table

Actually Using these tables I got this answer

Hey I got it figured out, .close

.close

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[drifting relic]

Yo

[ancient escarp]

[drifting relic]

In a study of the accuracy of fast food drive- through orders, Restaurant A had 268 accurate orders and 63 that were not accurate.
a. Construct a 95% confidence interval estimate of the percentage of orders that are not accurate.
b. Compare the results from part (a) to this 95% confidence interval for the percentage of orders that are not accurate at Restaurant B: 0.173 < p < 0.254. What do
you conclude?

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[drifting relic]

@ancient escarp

<@&286206848099549185>

[pastel verge]

What have you tried

[drifting relic]

I’ve been trying to construct the interval but I can’t

[pastel verge]

They gave you 63 orders are not accurate out of total (268+63) orders

So you know the proportion

[drifting relic]

I still don’t understand

[pastel verge]

You have to find the interval for proportion in percent

What's your proportion?

[drifting relic]

Idk

[pastel verge]

63/(268+63)

You have to make confidence interval around this

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[drifting relic]

can u just give me the answer

[cyan vale]

this not homework help server

[pastel verge]

The question is not can , the question is will

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[random trench]

How do I find the point of a discontinuity?

[random trench]

Lets say I have y= x^2 -16/x^2-7x +12

So after I factored it the equation it becomes (x+4)(x-4)/(x-4)(x-3)

but I dont know what to do after

[ember bough]

uh

[opal surge]

The denominator cannot equal zero, so the x-values where the denominator equals zero are points of discontinuity.

[ember bough]

ok

[ember bough]

that's exactly what i wanted to say

[random trench]

Sorry can you guys walk me through it, im so confused with the point of discontinuity thing

[opal surge]

In your case your denominator is (x-4)(x-3). Set it equal to zero.

What x-values get your left hand side equal to zero? I think it’s called finding the roots.

[random trench]

Wait so (x-4)(x-3) = 0?

Then isolate for x?

[opal surge]

You don’t have to

Look at the first one (x-4)

[random trench]

yea

[opal surge]

What x values makes that statement 0?

[random trench]

+4?

[opal surge]

Correct.

Now you would have 0(x-3) which basically makes it 0. So 0=0 holds true. Therefore x=4 is one of the points of discontinuity.

Then look at (x-3) to find the second point.

[random trench]

x=3?

[opal surge]

Yup in the end you got x=3 and x=4

So you have points of discontinuity when x= 3 and x=4.

Taking your factored rational. Trying plugging x for 4 or 3in your calculator. It will give you an error or undefined.

[random trench]

ahhh okay, thank you

.close

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[fleet folio]

can someone explain to me how to solve this

[fleet folio]

and explain the steps of how to do it

i got 3048 but idk if its right

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[pastel verge]

How?

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[fleet folio]

i did it

is it right

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[glacial moat]

From the input $602+a = 605+b = abc$, could someone tell me how did Wolfram Alpha straightaway got a = 14, c=4 and b =11 ? Did it just guess? I cannot find a way to solve the system.

[flat frigate]

Yamni_Ista

[prisma wren]

A=3+B
605+b=(3+b)bc
605+b=3b+b^2*c

I believe you can then write c in terms of b and then b in terms of a, to solve for c

[strange quest]

you just have the same equation twice here tho

[lean otter]

it doesn't work

I think calculator interpreted them integer then tried for some values but I have found a way to determine that b is odd

b+3=a, b(b+3)c=605+b, then b times b+3 makes an even number, that means opposite must also be even. If 605+b is even, b is odd

then a is even

[glacial moat]

Yeah, but I don't think that is useful, is it? It's probably impossible to solve the problem using this system, hmm.

[prisma wren]

Yeah I tried to solve it by hand, looks like you just need to brute force it

[strange quest]

You can for sure get a parametric solution I think

[glacial moat]

The problem: Cuboid – consists out of 1 cm^3 cubes. When you take away one row, you get a cuboid with a volume of 602 cm^3, when you take away one column, you get a cuboid with a volume of 605 cm^3. What were the lengths of the individual sides of the cuboid?

[strange quest]

say x is the width, y is the length, z is the height

y is constant

right

crap i gtg.

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@glacial moat Has your question been resolved?

[strange quest]

Ok so you have (x-1)yz=602 and (z-1)xy=605

[glacial moat]

Yeah, I got that as well, but can you do anything with it?

[strange quest]

I think youd need a third equation somehow

can you take a screenshot of

https://cdn.discordapp.com/attachments/903481106819612714/912403654626140160/unknown.png

what you plugged in

here

[glacial moat]

Just 602+a = 605+b = abc

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[glacial moat]

<@&286206848099549185>

[lean otter]

"roughly", can you provide full statement?

[glacial moat]

Okay, I've deleted the word roughly. There is no more information in the statement and this is almost a word for word translation.

[strange quest]

Well

Plugging it into wolfram alpha

You get this

So technically there are infinite solutions, but only one works for you cause they have to be positive integers

Idk how you’d find this without either a calculator or just plugging stuff in

Also

[strange quest]

Where did you even get this from

[lean otter]

@glacial moat you still here?

anyway, lets assume that a,b,c are three positive integers

then (602+a)/a must be an integer so 602/a must be integer and 602=2 * 7 * 43, so (2 * 7 * 43)/a must be an interger
then (605+b)/b must be an integer so 605/b must be an integer and 605 = 5 * 11 * 11, so (5 * 11 * 11)/b must be an integer
then you know 602+a=605+b so a=3+b
so (2 * 7 * 43)/(3+b) must be an integer and (5 * 11 * 11)/(b) must be an integer you can write all possibilities so:
b+3 = 2 or = 7 or =43 or = 14 or = 86 or = 301
b = 5 or b = 11 or b=121 or b=55
you can see that only numbers whose difference is 3 is 11 and 14 so a=14 and b=11, then volume will be 602+14=616, and c = 616 /(11 * 14)=4

@glacial moat

[fierce basin]

Can anyone help me find the range please 👉👈

[worthy hemlock]

Don't ask in an occupied channel
Read #❓how-to-get-help

[glacial moat]

Thanks so much for the help.

.close

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[sand palm]

Help

Don’t get it at all

[inland ivy]

Do you know what a negative in a power means

[sand palm]

Yes

[inland ivy]

Where do you get stuck then

[sand palm]

The question