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[serene junco]

.close

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[oblique meadow]

Hey, I am having a difficult time understanding a problem. First we represent a 10 by 10 square and a circle. The circle fits perfectly into the square (Figure 1). Let's say we take off a 1 by 1 cube off of each corner of the square.(Figure 2) The second shape has the same perimeter as the original square. Both have a perimeter of 40. Now let's say I repeat the same process of taking off squares a bunch of times.(Figure 3) It should still have the same perimeter as the original square. If we do this and infinite amount of time to get to the exact shape of the circle wouldn't that mean that both the square and the circle have the same perimeter. Obviously this is wrong, the square has a perimeter of 40 and the circle of 10 x pi but I can't figure out how it's wrong.

[mortal trellis]

Well you wouldn't have a circle

It's not smooth

[oblique meadow]

But if you do it an infinite amount of time.

[high wadi]

it's representing a circle

[oblique meadow]

What do you mean?

[mortal trellis]

It looks like a circle

But it's not

[high wadi]

the circle is not perfect

but you can understand

[oblique meadow]

Btw the drawings are not precise. I've simply drawn them for better comprehension

[high wadi]

yes, I understand them

[mortal trellis]

The process isn't continuous. Taking length and then taking the limit is different from first doing the limit and then taking the length

[oblique meadow]

Could you explain it differently, I am not quite sure I understand what you mean.

[mortal trellis]

When doing stuff infinitely often you can't expect that everything works as nicely as you would hope

As another example, if you take a sequence of rational numbers and take their limit, that limit can be irrational even tho all of the numbers in the sequence are rational

[harsh swallow]

this is similar to trying to find the length of a diagonal by taking smaller steps to the side and up.

i don't know a rigorous way to prove it but basically the reason is that every step you take towards your 'limit' has to change in some way

[oblique meadow]

Ok I think I understand.

[harsh swallow]

if every step is constant then you're not 'correctly' looking for the answer

if that makes any sense

[oblique meadow]

.close

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[hazy silo]

if y=-x+4 is tangent to y=(x-a)^(2)+3, what is the variable 'a' equal to

[last ether]

Find the derivative of y=(x-a)^2 + 3

[hazy silo]

ok

.close

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[floral willow]

Why was n = mi+1 used instead of l = (m-1)i+1 ?

This is the rule:

A full m-ary tree with i internal vertices has n = mi+1 vertices, and l = (m-1)i+1 leaves.
Thus, when m is known and the tree is full, we can compute all four of the values e, i, n, and l, given any one of them.  
^ (e)dges, (i)nternal vertices, (n)odes, (l)eaves 

I have this:

(3-1)*(100+1) = 2*101 = 202

[lone heart]

@floral willow Has your question been resolved?

[floral willow]

updated pic

<@&286206848099549185>

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@floral willow Has your question been resolved?

[slate monolith]

#get-advanced-access

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@floral willow Has your question been resolved?

[lone heart]

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[buoyant pulsar]

i need to find the volume of the specified area while it rotates around x=-1

i set up the integral but i think im supposed to include -1 somewhere because of the gap between x=-1 and x=0

idk where im messing up

[remote heron]

yea

can you tell which way you need to be integrating?

i mean dy or dx

sometimes these are easier if you sort the direction first, then try to make radius functions

[buoyant pulsar]

dx

[remote heron]

sorry tea

so, so i'd think of these as inner and outer radius

does that make sense?

your inner radius is where the volume "starts" as you head outwards from x=-1

and the outer radius is where it "stops"

[buoyant pulsar]

yeah, inner being from x=-1 to x=0 and then outer from x=-1 to x=2?

[remote heron]

sort of

first off radius is always positive

if i told you inner radius is 1, does that make sense?

[buoyant pulsar]

yea

[remote heron]

as in, if you rotate this thing

you have to go 1 distance to reach where the solid shape begins

from -1

the outer radius isnt so easy, its a function of x, right

[buoyant pulsar]

right

[remote heron]

if we start at x=-1 but y=0, itll be a different radius, than if we start at x=-1 but y=2

but you know some things about it

like at y=8, its 1

wwhat should it be at y=0?

[buoyant pulsar]

3?

[remote heron]

yea

so we need to make a linear function of x that moves in this way

[buoyant pulsar]

wait so

when we say x is the distance from rectangle

is it from the rectangle to the y axis

or all the way to the axis we rotate upon

cause if its to the y axis then it would be x+1 correct?

[remote heron]

oh wait am i doing the dy way

[buoyant pulsar]

i thi nk thats where i messed up

[remote heron]

sorry im kinda sleepy

[buoyant pulsar]

no youre good haha i think i just solved my own question

[remote heron]

oh

for real?

[buoyant pulsar]

should be like this instead

[remote heron]

you sure this is right?

we can get the answer pretty easy

[buoyant pulsar]

im prety sure

how would you do it?

[remote heron]

take two cylinders

height 8

one radius 1 and one radius 3

concentric

less the volume of the 1 radius from the 3

then divide it by 2

thats your answer

[chrome salmon]

yes be sure

[remote heron]

which is uhh

[buoyant pulsar]

80pi/3 is what i got for volume

[remote heron]

$\frac{8(\pi \cdot3^2-\pi\cdot 1^2)}{2}$

[chrome salmon]

nice answer

[remote heron]

o man discord dont freeze

[ocean seal]

jan Niku

[remote heron]

but

who knows

i think 32 pi

[ocean seal]

fäf

[chrome salmon]

getting 80pi/3 this way

but need someone to confirm

[lone heart]

@buoyant pulsar Has your question been resolved?

[remote heron]

hmm i get 8/3 pi

but i am bad at math

[chrome salmon]

and here me trying to be half as good as you

[remote heron]

i figured its $\pi\int _0 ^8 \qty(1-\frac y4 )^2 \dd y$

[ocean seal]

jan Niku

[chrome salmon]

i wonder'

[remote heron]

same

i dont like all these different answers

[chrome salmon]

I got the same answer as Dariush tho

[remote heron]

ah

then just im wrong

[chrome salmon]

although I'd like for someone to confirm cuz we both may have done the same mistake

[lone heart]

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[ivory gull]

Could someone please show me how to integrate ln x, this is what I have so far and I think its wrong

[ivory gull]

I think its supposed to be x ln x - x

but I am not sure how to get there

[maiden arrow]

uv-vdu

you accidentally did uv-dvdu

(sorry, i don't know how to use the rich text bot here)

it would be xln(x) - int(1)dx

= xln(x) - x

ach no

sorry

[ivory gull]

It's okay

[maiden arrow]

you just forgot that du is 1/x (dx)

which multiplied by x is just int(1)dx

[ivory gull]

Ohhh

So

Integral x dx/x

makes it integral of 1

[maiden arrow]

yep

[ivory gull]

which makes it x

Ahhh

[maiden arrow]

i got confused there because you wrote v = int(dx)

i normally write dv instead of v for that step

[ivory gull]

I understand my mistake now, but could you please tell me how you would solve it?

Like

I feel like I get so messy

because really this is only one part of a bigger question

[maiden arrow]

I would write it the same as you did

you just forgot to multiply du into v

[ivory gull]

Ohh okay

[maiden arrow]

instead you just wrote x

[ivory gull]

Do I need to do by parts twice here?

[maiden arrow]

what is the upper bound there?

[ivory gull]

e^2

[maiden arrow]

gotcha. let me do this one myself real quick

[ivory gull]

Thank you!

[maiden arrow]

also, is that the natural log cubed of sqrtx or the natural log of the cube root of x

[ivory gull]

x^(1/3)

[maiden arrow]

gotcha

okay. let me see if i can work this rich text bot real quick

forgive me if i mess this up

$1/3intxlnx$

[ivory gull]

This is what I have so far

[maiden arrow]

well that didn't work

[maiden arrow]

you should be using different u and dv values

do you know LIATE?

[ivory gull]

I did integration by parts twice, but not sure if it was necessary

Yes I learned that today, but tbh I don't know what is really considered what

Like

What is x?

From liate

[maiden arrow]

alright. LIATE is the order for which you choose u for integration by parts

[ivory gull]

Yes

[maiden arrow]

x is algebraic

[ivory gull]

Ohh

[maiden arrow]

for this, u would be ln(x)

[ivory gull]

and not x

[maiden arrow]

so du = 1/x

dv = x

v = (x^2)/2

that should get you on the right track

[ivory gull]

Would u be ln x^(1/3)?

[maiden arrow]

no

ln(x^(1/3)) can be simplified

[ivory gull]

Ohh

[maiden arrow]

to just 1/3lnx

[ivory gull]

yeah

[maiden arrow]

log rules

pull the 1/3 out

and you're just integrating xlnx

[ivory gull]

Thank you so much

[maiden arrow]

np

[ivory gull]

Im going to restart

Really appreciate the help!

[maiden arrow]

no problem. should be fairly quick now that the u's and dv's are good.

[ivory gull]

Yes, thank you again

[lone heart]

@ivory gull Has your question been resolved?

[ivory gull]

How can I make this

Look like this?

I know I have to factor out a 1/4

But because there is that e^4

It makes me confused a bit

Nvm I think if I just get the common denominator it will be clear to me

lmao im so stupid

.close

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[spare moss]

Can someone give me the answer to all the questions with solution...

[spare moss]

Find the area of the shaded region

[plucky token]

they're all just finding the area of the separate shapes and either adding or subtracting

[spare moss]

Yo can you show the solutions? From number 1 to number 3 with answers im pretty confused with our topic on math rn...

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

@spare moss Has your question been resolved?

[lone heart]

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[limber vapor]

can I please get some help on this question
(this question is in the topic of proofs- math ext 2 year 12-australia)

also this one pls

[desert crane]

okay so are you generally familiar with the structure of an induction proof?

[limber vapor]

yes

[desert crane]

so what will your base case be?

[limber vapor]

1?

[limber vapor]

wait, can we please look at this question first please?

[desert crane]

yep, since we're showing this for positive integers n. can you verify that the base case works when n = 1?

ok sure

[limber vapor]

thanks

[desert crane]

what are you not sure about on that one?

[limber vapor]

I am not quite sure where to start
is it possible for you to write (hand write in fine) me a working out so I can ask questions to you within the working out?

[desert crane]

sure

so what inequality could be useful here, when comparing two sides of a triangle to the third?

[limber vapor]

the one with absolute values?

💀 sorry if I sound stupid

[desert crane]

"triangle inequality" should ring a bell

[limber vapor]

yeah yeah, that one
the one with absolute values

oh right, so we start with this one?

[desert crane]

yeah, this means that the sum of two sides of a triangle is always greater than the third side (if the three points aren't collinear)

here, what are the candidates for longest side?

[limber vapor]

not sure

[desert crane]

well it definitely can't be $log_{10}(19)$ right

[ocean seal]

lyra

[desert crane]

depending on $n$, would you agree that the longest side has length either $log_{10}67$ or $log_{10}n$

[ocean seal]

lyra

[desert crane]

?

[limber vapor]

$log{10}n$?

[ocean seal]

Mastermind_879

[limber vapor]

wouldn't this be just the question

[limber vapor]

😶 is it possible for you to this for me?

😭 sorry for my lack of knowledge at my own school topic

[desert crane]

no need to apologize

[limber vapor]

can you write a handwritten/typed working out for me?

[desert crane]

yeah working on it

[limber vapor]

thank you

[limber vapor]

thanks, I will try it out

[desert crane]

sounds good, let me know what you get

[lone heart]

@limber vapor Has your question been resolved?

[limber vapor]

yeah, I was able to find it 🥹

but since N is an integer but N>67/19

but 67/19 is not an integer

[limber vapor]

is the possible values just all the integers inbetween them

thank you for helping me with this question btw

[limber vapor]

can you maybe help me with this one now, if you're not busy

[desert crane]

yes, you just want to find the integers n that satisfy this - it's fine if the bounds aren't integral, just take the least/greatest integers that do satisfy them and everything in between

sure i can help w that

so we said base case is n = 1. can you check that this works?

[limber vapor]

in theory it wouldn't work, and log 1 is 0

[desert crane]

why's log coming into the picture? we just want to check if the formula proposed in the question works for n = 1, i.e. take the first derivative of sin ax and check that it is equal to the right-hand side

[limber vapor]

oh, the new question

right

my bad

[desert crane]

OH sorry yeah oops

should've specified mb

[limber vapor]

😅 no no, I am slow

okay

let me see

[limber vapor]

how would I try it out without knowing what constant "a" is?
(general question, not meaning to say it isn't possible)

[desert crane]

well it doesn't matter what a is to find the derivative of sin ax (it will depend on a, but that's fine)

[limber vapor]

oh

I don't think it is
the answer is negative to the original thing... I might have made a mistake

as far as I know cos(x+90) doesn't equal to sin(x)

did I make a mistake?

[desert crane]

well on the right-hand side you'll get a*sin(ax + 90), which simplifies to what?

[limber vapor]

a*cos( ax)

[desert crane]

yep, and that's also the form we want from the derivative right

[limber vapor]

OH
but it is still neg to what we want tho

[desert crane]

how? the derivative of sin ax is a*cos(ax), which lines up with what you got

[limber vapor]

brain fart. yes. you're right

[desert crane]

lmao all good

okay so now we can carry out the induction step: how would you set this up?

[quiet tinsel]

in this why are we not considering log 67 + log n> log 19

sorry to disturb you guys but I had that doubt

[limber vapor]

assume true for n=k
then prove true for n=k+1

or do you mean for me to write the actual thing

[desert crane]

yeah thats the setup

how would you apply it here

[limber vapor]

not sure how true your statement is
but even if it is true, this lovely helper already presented enough to answer the question

[limber vapor]

as in you want me to write it?

[desert crane]

log 67 is already greater than log 19, so this is true for any n

[desert crane]

i think this would help yeah

[limber vapor]

okay

[quiet tinsel]

oh thnx man

[quiet tinsel]

really sorry to disturb you guys again but I just wanna check the final answer of this question. Mine is coming out to be 1270

[limber vapor]

sorry for the wait btw, just working on it

@desert crane I am stuck on proving k+1 is true
I tried differentiating f^k(x) to get to it to prove it
but am having trouble, am I going towards the right direction?

[lone heart]

@limber vapor Has your question been resolved?

[limber vapor]

resolved

[floral kiln]

hello

can someone help me with 3659 x 26 with a screenshot of working out

[gusty gorge]

do you want to learn multiplication with the fast fourier transform

[floral kiln]

yes pls

[floral kiln]

yesss

[lone heart]

@limber vapor Has your question been resolved?

[lone heart]

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[true sandal]

is my solution wrong?

[true sandal]

https://tutorial.math.lamar.edu/classes/calcII/MoreSequences.aspx

I used the method here at the last example

[worn fox]

Why do you think its wrong

[true sandal]

I am struggling with this topic alot

with sequences and series

[worn fox]

It looks fine to me

[alpine sable]

they give you this checklist.

You could re-state the boundary over which the sequence decreases.

But It also increases over second boundary, which you should add.

[lone heart]

@true sandal Has your question been resolved?

[true sandal]

where do you get this

[alpine sable]

from the webpage you linked. paul's online notes

[true sandal]

why is it blue

[alpine sable]

it might be because of my addon that forces dark-mode on sites

[lone heart]

@true sandal Has your question been resolved?

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[tulip hound]

On the first trip, the car spent 0.375 parts of the gasoline that was in the tank and another 5 liters. On the second trip, 80% of the remaining gasoline and the last 4 liters. How many liters of gasoline were in the tank initially

[flint pecan]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[tulip hound]

1

I mean if I take total gasoline amount as x, would first day be x-0.375x-5?

[gray isle]

yes

[flint pecan]

yeah so that would be the remaining gasoline after the first trip

20% of that minus 4 liters would be 0

[gray isle]

80% of that minus 4 liters would be 0
no to that

[tulip hound]

So whole solution is

x-0.375x-5-(x-0.375x-5/2)-4=0

[gray isle]

no

wheres
-(x-0.375x-5/2)
coming from

[tulip hound]

I mean not /2

4/5

[gray isle]

can you rewrite it

[tulip hound]

one sec

x-0.375x-5-(x-0.375x-5×4/5)-4=0

[gray isle]

misplaced (),

don't use cross for multiplication

you want something representing 4/5 being multiplied to the whole x-0.375x-5
not just the -5

[tulip hound]

Oh yes

But

[gray isle]

x-0.375x-5-(x-0.375x-5) * 4/5-4=0

[tulip hound]

You know Ickwant that

Meant*

[gray isle]

no i do not

[tulip hound]

Well I meant u know now then

[gray isle]

and solve that equation

[tulip hound]

Yeah thanks

.close

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[fierce delta]

show that;

[fierce delta]

ignore the fraction line in the n-1 over k part

i tried expanding everything first,

n!/(n-k)! x k! = (n-1)!/(n-k-1)! x (k-1)! + (n-1)!/(n-1-k)! x k!

does this look correct?

[lone heart]

@fierce delta Has your question been resolved?

[fierce delta]

.close

[lone heart]

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[silver zephyr]

How do i find the x intercepts and domain of a sin graph

[chrome salmon]

When is y=0?

First find domain I'd say

[silver zephyr]

alright

how do i know when y=0?

when y=0 x=0 too?

oh like that

and to find the domain what do i do

find the derivative?

[flint pecan]

why would x=0 when a denominator can't be 0

[silver zephyr]

no clue, just a guess

so, y=0, making it 0=sin(1/x)

and now what do i do to find a domain

a domain is the ~"height" right?

[wide oxide]

No that’s the range

Domain is the horizontal

[silver zephyr]

ohh okay

[wide oxide]

Domain is basically all the values of x for which the graph is not undefined if it correct

[silver zephyr]

how do i find all the values for x

[wide oxide]

Hmm ok so what value of x would leave the graph undefined or the equation undefined

As u can see it’s 1/x

What would x need to be for 1/x=undefined

[upper hull]

y = sin(1/x) oscillates infinitely many times between -1 and 1 as x approaches zero, but it never touches or crosses the x-axis.

The amplitude of the oscillations also increases without bound as x approaches zero.

[silver zephyr]

at x=0 ?

[upper hull]

and domain is intuitively all real numbers except for x=0

[upper hull]

which means it doesn't have x intercept

[wide oxide]

The domain would either be in set notation set builder notation if I’m correct

[flint pecan]

it does have x intercepts

[silver zephyr]

so essentially its all real numbers excluding 0, written sa (-∞, 0) U (0, ∞) ? (using interval notation.)

[wide oxide]

Uhh yes I would think so

[chrome salmon]

ye

[silver zephyr]

Alright, thanks guys

.close

[lone heart]

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[solar trail]

what do i do

[solar trail]

should i take y common or smt?

or should i consider the lcm of their denominators?

y is supposed to be 0 i dont exactly know how

[alpine sable]

what are you trying to do

[solar trail]

find y

[gray isle]

factorisation helps

[upper hull]

can u see why 0 is a solution?

[gray isle]

should i take y common
yeh

[solar trail]

its the equation given?

[solar trail]

and then what

[alpine sable]

try writing root{8}y with root{2} denominator

[gray isle]

what do you have after taking y common

[alpine sable]

then you get the answer

[upper hull]

just time sqrt 2 on both sides

[alpine sable]

y = 0 !!!

[upper hull]

-7y=0

y is 0

-3y-4y

[alpine sable]

damn i didn't see that immediately.

[solar trail]

OH

tysm yall got it

.close

[lone heart]

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[alpine sable]

how do i solve this? the answer to b) is "no"

[whole shell]

did you get

the answer for a

what is it

[alpine sable]

i have the answer key, though

brb

[upper hull]

ig introduce coordinate, (0,8) is vertex, and one of the point on the graph is (6,0)

[whole shell]

@alpine sable now

that you have the answer for a

what you do is

since the truck is 3.9 wide

it means its

[upper hull]

0=a(6-0)+8

[whole shell]

1.95

from either end

[upper hull]

and find a and expand to get A

[whole shell]

so sub in x = 1.95 or x = -1.95 (doesnt matter which)

and see if f(x) is >= 4.8

[upper hull]

sub in 4.8 to the quadratic to find the width of the tunnel at the height

[whole shell]

also im pretty sure

[carmine reef]

As usual, you're not helping

just ignore him he's a troll

[upper hull]

[whole shell]

you said (0,8) is the vertex, so you know that the quadratic starts from the middle

yet you said (6,0)

and not (3,0)

[upper hull]

(3,0)

is the point, mb

[whole shell]

also, making f(x) = 4.8 is unnecessary

its easier to find the height at the width you have, than finding the width at the height you have

[upper hull]

both works

u should prob test both, but one being not fitting is a sufficient condition

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[potent ferry]

can someone help me with this question?? a ball has a radius that is 5.2mm what is the volume

[gray isle]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[upper hull]

V = (4/3) * π * r^3

is the formula

just plug radius in

,w (4/3) * π * (5.2 mm)^3

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by )( < > )(#1583

[carmine reef]

did Mr ou need to help with that one

[vale wigeon]

OP is nowhere to be seen 👻

[gray isle]

also nfi why wolfram rounded

does it hate units or something

[carmine reef]

maybe pi is just rational after all

[gray isle]

,calc 590/(5.2^3 * 4/3)

[ocean seal]

Result:

3.147047109695

[astral briar]

588.977413115

[gray isle]

back to being serious

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

@potent ferry Has your question been resolved?

[lone heart]

@potent ferry Has your question been resolved?

[lone heart]

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[turbid root]

Hi

[turbid root]

I need help with last question. The answers for the first few questions are

A)The median height: 35cm
B) the first and third quartile are 30cm and 39 cm respectively

C) the 30th percentile is 31cm

D) 4 plants.

I need help with 4 ii)

[lone heart]

@turbid root Has your question been resolved?

[lone heart]

@turbid root Has your question been resolved?

[lone heart]

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[tacit arch]

might want to translate

[alpine sable]

السلام عليكم

عرب

هلا والله

engleezy no

no englez

arbic

[worn fox]

There are not many Arabic speaking helpers

[alpine sable]

unfortunately😞

dw im just missinf arounf i dont need actual help thank you though

[worn fox]

? Why waste time

[alpine sable]

trying to find arabic helpers

so we can discuss something

also if ur curious

the problem here is

x^2+4x+4😂😂

[tacit arch]

we're not if you don't need help. just close

[alpine sable]

ok👍🏻

close

[tacit arch]

.close

[lone heart]

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[high wolf]

could someone help me understand how they got final answer?

[high wolf]

Like,.I know they plugged asin(pheta) in for x..

But somehow they got (1-sin^2(pheta)) afterwards

[vague coral]

theta* 💀

[vague coral]

factorization

[high wolf]

Shouldn’t it be (1-cos^2(theta))

[vague coral]

a²-a²sin²(theta) = a²(1-sin²(theta))

cos²(theta)+sin²(theta) = 1

cos²(theta) = 1-sin²(theta)

so a²-a²sin²(theta) = a²cos²(theta)

[high wolf]

God I’m confused on this..

It’s just factoring trig functions are a pain..

Hold on

[high wolf]

How we get 1?

[vague coral]

I said you factor

[lone heart]

@high wolf Has your question been resolved?

[lone heart]

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[tacit arch]

you really shouldn't be mean

[twin haven]

sorry

i thought nobody would see it

[echo socket]

!status

[lone heart]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[sly mantle]

people say you are how you act when no one’s looking. just because you think we can’t see it doesn’t make it ok. consider this a warning

[twin haven]

2

[alpine sable]

provide what you have tried

[echo socket]

Could you show what you've got so far?

[twin haven]

i like cat

[alpine sable]

and what your confusion is

[twin haven]

i got none

i didnt know what to do

[alpine sable]

why choose number 2 then...?

[echo socket]

You said you got stick midway

[echo socket]

Anyways, start by calculating how much area do the walls have in total

[twin haven]

what do i do to times it

[echo socket]

Perhaps it would be nice to know how much area does just one wall have

Any idea on how to get that given each wall is 2.7m high and 4.2m long?

[twin haven]

2.7 times 4.2m ????

[echo socket]

Yeah

So four of those walls will take 4 * 2.7 * 4.2

[alpine sable]

also this question is weird lmao. Did they include the door part to exclude it or include it with the walls? since I don't see many people painting the doors themselves

[echo socket]

But we need to account for the door

[twin haven]

so i times

[echo socket]

Would be nonsensical to mention the door only to include it

[echo socket]

How much area does the door take?

[twin haven]

1.6

[echo socket]

Right, so we will have to subtract that from this

[twin haven]

so 45.36 subtract 1.6=43.76

[echo socket]

You should get 43.76

Yeah

So we need to have an area of 43.76m^2

Let's now evaluate how much litres of paint that is

[twin haven]

87.52

[echo socket]

Each litre of paint can cover 15m^2 of wall, so we will have to divide 43.76 by 15

And, yeah, not multiply

[twin haven]

2.197333333333333333

[twin haven]

2.197333333333?

[echo socket]

Yeah

[twin haven]

then whs

what

[echo socket]

And since 1.4litres of paint costs 12pounds

Each litre of paint should cost 12/1.4 pounds

[twin haven]

okay

[echo socket]

Or 60/7 pounds

Meaning we have to multiply what we got by 60/7

And that should be the answer

[twin haven]

420

[echo socket]

You got 420?

[twin haven]

yes

times

or is it divide

[echo socket]

Times 60/7

[twin haven]

mmm

im confused

[twin haven]

what is it

[echo socket]

Just multiply that by 60/7

You should get about 25

[twin haven]

ohh

[twin haven]

is it 2.19333333 times 60/70 which is fraction

[echo socket]

Yeah, 60/7 though

You can just multiply by 60 and then divide by 7 if you are using a calculator

[twin haven]

18.8??

[echo socket]

You should get 25

[twin haven]

so is 25

[echo socket]

Yes

[twin haven]

ok

[twin haven]

its wrong

its 30

[echo socket]

Does it show the solution? Cause I don't see what's wrong with my reasoning

[twin haven]

no

it just shows answer

[echo socket]

Hm, so it's supposed to be 2.5*1.4 litres of paint

Or 52.5m^2 of area painted

[twin haven]

so what is answer

@echo socket so its 52.5m times 2

[twin haven]

its wrong

[twin haven]

how do i do it

[twin haven]

its not 105

@echo socket what is it

Tom is applying for a visa.
He has to pay £1,500 for the visa as well as a 8% charge on the visa price for an on-the-day service.
If Tom has 1 year to save, how much money should he save per month to be able to cover his costs?
Round your answer to the nearest £10.

.close

[lone heart]

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[twin haven]

.reopen

[lone heart]

✅

[lone heart]

@twin haven Has your question been resolved?

[sullen dome]

So how much area needs to be painted?

[twin haven]

i dont know

[twin haven]

i tried to find out

[sullen dome]

Each wall is 2.4×5.2

In area

[twin haven]

okay

[sullen dome]

But we don't want to paint the door

So we remove the area of the door from the total

[twin haven]

ok

[sullen dome]

So 2.4×5.2-2×0.8

[twin haven]

s0 8.384?

[sullen dome]

That's for the one with the door

Then we have 3 normal walls

Which will just be 3(2.4×5.2)

[twin haven]

ok

[sullen dome]

So sum it all up

We get the total area we need to paint

I get that to be 48.32

[twin haven]

okay

i got 37.44

okay#

[twin haven]

what i do next

[sullen dome]

So we know one tin is 2 liters and that will cover 24m^2

[twin haven]

okay

[twin haven]

so 48

or is it 24 squared

if it is so 576

[sullen dome]

No

m^2 just means area

So know we know 2 tins won't be enough

Right

[twin haven]

okay

[sullen dome]

Because it's going to be that 0.32 left

So we need 3 tins

[twin haven]

okay

[sullen dome]

But the 3rd is 1/2 price

So 12+12+1/2×12

Which is 30

[twin haven]

okay

[sullen dome]

Any questions

[twin haven]

so the answer is 30

[sullen dome]

I believe so

[twin haven]

oi

i understand

is it 2.1 times 4.5-0.9

to be 8.55

[sullen dome]

Where did you get any of these numbers from

[twin haven]

2.1 and 4.5

and 0.9

[sullen dome]

Area of the wall is 2.1×4.5

Area of door =?

[twin haven]

9.45

[sullen dome]

1.8×0.9

B×h

[twin haven]

1.62

[twin haven]

right?

[lone heart]

@twin haven Has your question been resolved?

[lone heart]

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[celest stump]

is the derivative of ln(x) 1/x ? I'm only asking because desmos doesn't show 1/x when getting the derivative

[worn fox]

ln(x) is only defined for x > 0, so the derivative of ln(x) is 1/x for x > 0

[celest stump]

but when solving should i just consider it 1/x ?

[worn fox]

solving what?

[celest stump]

general equations

or should i write that x > 0

[worn fox]

you need to be more specific really

[celest stump]

like this

[weary wyvern]

My guy, write down what's correct

[celest stump]

finding the derivative

[worn fox]

well that function is defined for all $x \neq 0$

[ocean seal]

ΣAC

[celest stump]

well obviously there x can be less than 0

[weary wyvern]

Yes this is the so called natural domain of defintion

The domain and codomain are part of the definition of a function

Which is not specified here but that's fine

[celest stump]

thank you

.close

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[vapid steppe]

yo

[vapid steppe]

can someone explain 9

,rotate

[vapid steppe]

why did they multiply 12 and 80

[fallen verge]

12(lb)*80(dollars/lb)=960dollars

Since we are looking at cost, the unit is dollars as expected

[vapid steppe]

how did you know to multiply 12 and 80

and not the other numbers

[lone heart]

@vapid steppe Has your question been resolved?

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[alpine sable]

i dont get how to do this inductive proof

[remote heron]

were you able to do the base case

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Yes sorry

[alpine sable]

base case is n = 1

right

so 3 > 1

[lone heart]

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[simple gull]

Hi, can anyone please give me a hint why this proof is wrong? Personally, I think m/pn is just a specific case and therefore not representative for other cases.

[potent garnet]

to me the last sentence looks a bit sketchy

actually, yea i agree with u

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[brazen osprey]

So far I've just tested basic points like -3, -1, -2, 0, 1, 2. I know that 0, 1 and 2 are all =. However when I tried [-1, 1] it said I was wrong. Don't understand why and how to fix it

[gray isle]

well the interval you just wrote includes
-1, 0, 1
which you've just stated are values where |x| = x^2

[brazen osprey]

yeah

[gray isle]

where f(x) isn't greater than g(x)

which is why its wrong

[brazen osprey]

but negative and positive numbers wouldn't work because of the exponent

[gray isle]

it would be helpful to graph these two functions

[brazen osprey]

ok

So just 1 and -1?

[gray isle]

no

[brazen osprey]

Oh wait nevermind

x<-1 and 1<x

[gray isle]

no

can you indicate the locations where the red graph of |x|
is above the blue graph of x^2

[brazen osprey]

x>-1 and x<1?

(-1,0)U(0,1)

[gray isle]

x>-1 and x<1?
not quite, you forgot about exclusion of zero there

(-1,0)U(0,1)
yes

[brazen osprey]

Thanks : )

.close

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[simple zealot]

I dont understand why this is incorrect

Am I missing something here?

[gray isle]

where is x^4 in the denominator for your planned decomposition coming from?
and also the degree of 4x^4 is still higher than that of x^2 - 4
so you should continue with your long division

[simple zealot]

i was adding the x^4 and getting the lcd for the equation i had but i now see what your talking about

so i keep going until theyre arent equal ?

So like this ?

.close

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[thorny moon]

how

[thorny moon]

this is what i did

but its wrong

[spice tide]

the formula has (1/2)at iirc

so you should have -4.9m/s^2

wait a second that's wrong I think xD

[scarlet oriole]

displacement is given by Vot+1/2(a*t^2)

you divide by two after you square

@thorny moon

note that it says 30degrees from the vertical

[spice tide]

yeah they got the 30 degrees part right

he using 5.5m/s

[scarlet oriole]

that's wrong

should be around 9 m/s

9.5

11*cos30

[spice tide]

oh from the vertical lol

[spice tide]

they can't use this formula because they don't have delta x right

[scarlet oriole]

that's what is being asked

[spice tide]

no it says how long till max height

v^2=(v_0)^2+2a(x) do you solve that for x

[scarlet oriole]

max height is when derivative is 0

[spice tide]

and then use the formula you posted?

[scarlet oriole]

they all are deriving the general displacement equation

the only problem they made is velocity reference

[lone heart]

@thorny moon Has your question been resolved?

[thorny moon]

whaatt the heeel is vot

[scarlet oriole]

this

[thorny moon]

im only using 5 equation of motion

im noob

[scarlet oriole]

what are those equations?

post them

you are working with acceleration the equation i posted is something you should be familiar with

[thorny moon]

bro whats the difference if its launched vertically vs horizontally

ok wait a sec

[scarlet oriole]

@thorny moon

if we divide 2 axes Vertical y and horizontal x

[thorny moon]

its posting

my wifi

[scarlet oriole]

yeah

i posted equation 5

[thorny moon]

oh

why didnt it work with my equation?

equation2?

[scarlet oriole]

all equation you posted are deriving 1 equation

they are put in a way to help you find what you need faster

last column explains that

[thorny moon]

yes

[scarlet oriole]

"variables not in equation"

[spice tide]

you can use equation 2

[scarlet oriole]

literally means this variable is missing so probably you are looking for it

[thorny moon]

i mean we basically have vi

[scarlet oriole]

wait

that's the mistake you did

[scarlet oriole]

this is the reference

[thorny moon]

thats what i used but i got the wrong answerrr

idk what being launched from the vertical is

😭😭

[spice tide]

you did used the wrong value for v

[scarlet oriole]

you got the wrong answer because you calculated with wrong Vi

[thorny moon]

no

[scarlet oriole]

30 degrees from vertical

[thorny moon]

what

help

[spice tide]

the angle is from the vertical not the horizontal for some reason

[scarlet oriole]

meaning you aim up and tilt either right or left with 30degrees

[thorny moon]

whats the difference

[scarlet oriole]

that's the meaning of referece from vertical

[spice tide]

so use cos(30)

[spice tide]

this

[thorny moon]

what if i said from the horizontal

[scarlet oriole]

just referencing to trick students

than you either aim right or left doesnt matter

and from there you tilt 30 degrees up

cause down is the ground

[spice tide]

you can think of it as 60 degrees from the horizontal if you want

[thorny moon]

so basically it doesnt matter if u launch from vertical or horizontal it would give u the same?

[thorny moon]

OH

acc?

thats legal?

[scarlet oriole]

bro

💀

[thorny moon]

oh it worked

[scarlet oriole]

people learning physics only to pass

[thorny moon]

💀💀

[scarlet oriole]

ofc it did

[thorny moon]

yessir

[thorny moon]

U NEVER MENTIONED IT WAS THE OPPOSITE 😭😭

[spice tide]

:{

[scarlet oriole]

basically dina

acceleration -9.8

is only vertical

[thorny moon]

mauskateer op

[thorny moon]

what mmama😭😭

[spice tide]

;-; lance noticed the vertical thing

[thorny moon]

acceleration is in vertical plain yes

[spice tide]

i assumed it was horizontal when i read question xD

[scarlet oriole]

yes that means people never deccelerate in X plain

that's something to keep in mind when you get into X,Y physics

when problem asks to find displacement vertically and horizontally

[thorny moon]

wait

[scarlet oriole]

basically the equations you have only work when there is acceleration

whatever ask again when you get to them

but start learning physics with more practicality it will get much easier

[thorny moon]

IF HORIZONTAL PLAIN IS THE OPPOSITE THEN I GOT THIS QUESTION WRONG TOO

[thorny moon]

well😭😭

[thorny moon]

but one of the equations doesnt have acceleration

[scarlet oriole]

it does, you are just solving for it

what i mean no acceleration i mean that a = 0

[thorny moon]

i did this

i think im confused on what launching vertically means

cuz what the hell

[scarlet oriole]

dina do you VC?

[thorny moon]

im so laggy but i can try

[scarlet oriole]

go in the call

[lone heart]

@thorny moon Has your question been resolved?

[lone heart]

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[celest stump]

how do you do this?

[celest stump]

I started with doing nPr(26,6)*nPr(6,2) but apparantly thats wrong

[lone heart]

@celest stump Has your question been resolved?

[thorny moon]

im pretty sure u use combinations bc its a pattern

the order doesnt really matter

[celest stump]

but then why multiply by 8!

[thorny moon]

im tryna figure that out

[celest stump]

wait if igured it out

it doesnt matter what order you picked out the red/black cards

but it does matter what order they are on your hand

i think thats why

[thorny moon]

maybe

tbh i just learned the intro to combinations tdy ask someone else

😭

[lone heart]

@celest stump Has your question been resolved?

[celest stump]

.close

[lone heart]

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Closed by @celest stump

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Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ember glacier]

I am confused about this, because I do not know where to go from aftter finding the second derivative of e^5x(25x^2+20x+2)

[chrome salmon]

what do you check for concave up?

[ember glacier]

open intervals

[chrome salmon]

I mean what's the condition for f' and f'', do you know them?

[ember glacier]

no, i am confused about it

[chrome salmon]

did you do something where you try to find points where f'(x)=0?

[ember glacier]

no this is for f''(x) > 0 and< 0

[chrome salmon]

yes that's what I'm asking for, what did you try?

[ember glacier]

e^5x(25x^2+20x+2) > 0 and e^5x(25x^2+20x+2) < 0

im confused on how to find that tho

im stuck there

[chrome salmon]

you got f'' but you still need to find values for which it's positive and negative

[ember glacier]

i dont understand

[chrome salmon]

first you have to do this

check the procedure if you have a solved example

[ember glacier]

i dont

theres no example for that problem

[chrome salmon]

ok let me see if i can find any

[ember glacier]

okay thank you

[chrome salmon]

https://www.youtube.com/watch?v=iG7t3bl3qNY&t=334s&ab_channel=larryschmidt

found this video

[ember glacier]

yes

thank you for finding me a resource that helps me with the process

[chrome salmon]

you can ping me in case of doubt

[ember glacier]

yes

i am im doubt still

once i set it equal to 0

what do i do

to factor out the possible critical points?

for concavity

[brazen dew]

hi