help-19

[lime swift]

Ohhhhh

Thanks

Makes sense, that part was ticking me off

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[limpid bough]

just making sure that x = 12 right?

[limpid bough]

from this

180-148 = 32

90-32 = 58

2 x 58 = 116

180 - 116 = 64

116 = 9x+8

108 = 9x

x = 12

can someone help with this too i got a decimal

180-112 = 68

68 x 2 = 136

180-136 = 44

90 - 44 = 46

46 = 10x+14

x = 3.2

[odd edge]

@limpid bough Has your question been resolved?

[limpid bough]

HELP

<@&286206848099549185>

[split gyro]

Yes

[dawn trout]

for clarification is the distance from H to I is 2x-6? and H to J is 4

[split gyro]

Listen IK is 12

[limpid bough]

um how

[split gyro]

Listen from HI the dist is 2x-7

2x-6*

Ok

Ik is 3x

And Jk is 3x-2

If u subtract JK from IK

U will get IJ

[dawn trout]

2x-6 +3x = 3x-2 +4

[split gyro]

That is IK-JK=3x-(3x-2)

[limpid bough]

yes ohh tysmm

[split gyro]

So IJ=2

And HJ= 2x-6+2

Which is equal to 4

So u get x=4 and IK=3x=12

[limpid bough]

wait im so confused

dont we only need to find IK

[split gyro]

Yes

@limpid bough here for help

Listen iK is 3x

[limpid bough]

yes

[split gyro]

So to find IK we need x first

So first we find x

Understood?

[limpid bough]

yes

so we

[split gyro]

Yes

[limpid bough]

do

3x-(3x-2)

[split gyro]

Yes to get IJ

[limpid bough]

which is -2

[split gyro]

No

It's 2

[limpid bough]

ohh yesah

sorry

[split gyro]

Np

We got IJ as 2

[limpid bough]

ohh yeah

[split gyro]

We know HI is 2x-6

[limpid bough]

yess

[split gyro]

So for HJ we do IJ+HI that is 2x-6+2 which is 2x-4

But it is written down that HJ is 4

[limpid bough]

wait but why we need HJ

when we are only find IK

[split gyro]

We need HJ to find x

If we get x we find IK by using 3x

Like listen HJ is 2x-4 and it is given that HJ is equal to 4 so we do
2x-4=4 we get x=4

Now we can find IK
As IK=3x
So IK=3x4=12

[limpid bough]

why is it 2x-4 = 4

[split gyro]

Bro

[limpid bough]

not like 2x- 6

2x-6 -4

[split gyro]

Listen 2x-4=4

Bcz

It's HJ

2x-6 is HI

[limpid bough]

yeah but its not equal

wait what are we finding with this arent we finding IJ

[split gyro]

Bro

IK=3x

We need x first

We are getting x from.HJ

So we need HJ

Wait coming

[limpid bough]

ohhh

i seee

cant we do 2x - 6 = 3x - 2

to find x

[dawn trout]

no

its not the same length

[limpid bough]

then how is 2x - 6 = 4 the same length

[split gyro]

@dawn trout bro help him

[limpid bough]

help

[dawn trout]

look u need to look at it from different angle

try to find similar length

and put it equal to each other

[limpid bough]

so we put 2x - 6 - 4 over the equal sign to find x?

so like 2x - 6 = 4

guys

<@&286206848099549185>

[dawn trout]

take a look at this

[limpid bough]

THANK YOU

that makes sense now

okok tysmm

one last request

can someone check if its actually 3.2

i find it hard to believe x = 3.2

<@&286206848099549185>

[split gyro]

I'm currently not at home

[limpid bough]

:( ok!

its fine guys ty all so much

now i can solve these problems tyyy

[limpid bough]

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[rugged escarp]

haloo

[inland rover]

What is the question?

[mortal trench]

haloo?

[inland rover]

Damn that's deep

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[sage shale]

can someone help me find the perimeter of this

[sand horizon]

What is the perimeter of a semi circle ?

[odd edge]

@sage shale Has your question been resolved?

[sage shale]

9.42?

[mental cairn]

Yes

For one of them

[sage shale]

so you would take away 9.42 4 times?

[mental cairn]

You'd multiply 9.42 by 4

[sage shale]

and then

[mental cairn]

Then for the perimeter of the straight edges just subtract 6 from 12

Then divide by 2

And then multiply what you get with that by 8

[mental cairn]

And add that to what you got from doing this ^

[sage shale]

why do u need todivide by two and multiply by 8?

[mental cairn]

Well you could just multiply by 4

[sage shale]

because there are 4 sides?

[mental cairn]

Yes

[sage shale]

so if i multiply by 4 i dont need to divide by 2 right?

[mental cairn]

Correct

[sage shale]

ok thx sm

[mental cairn]

Your welcome

[sage shale]

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[mystic saffron]

As Salaam Alaikum

[mystic saffron]

Hello I need help with refreshing high school grade mathematics

Basically Algebra and need to learn Trigonometry better. In-Shaa'-Allah

[void yew]

!da2a

[odd edge]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[plain ruin]

You know how to solve quadratic equations, inequalities?

To understand trigonometry understand circle and triangle

The main thing is to practice

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

To give some context. I had to leave school due to illness and I am still ill. But Alhamdulillah, I am recovering. So I need to refresh cuz I don't really remember much

I need Arithmetic,Algebra and Trigonometry

I remember Arithmetic but I think I should refresh that too

That's all 🩵

[plain ruin]

Better to ask your teacher about which themes they passed

[mystic saffron]

It's been 5+ years.

I changed schools too

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@mystic saffron Has your question been resolved?

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[tidal yacht]

can someone take me through

i found my answer(2x-y)(2x+y)(4x^2+y^2)

just to clarify

[icy kindle]

apply the difference of squares a couple of times

[tidal yacht]

k

but do i first find the square root of 16?

[icy kindle]

you'd need to, yeah.

[tidal yacht]

ok so ((4x)^2-y^4))

[tulip mica]

This is correct

[tidal yacht]

thanks

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[normal sundial]

isnt it the other way around?

[worn sandal]

Can you show the entire thing

[normal sundial]

soltution also says [-3,2]

but i think its [-2,3]

[normal sundial]

do you know it?

is it wrong?

[worn sandal]

are you just trying to find eigenvectors of A

[normal sundial]

yh

i.e find the vectors in the nulspace

[worn sandal]

not really the same thing

If you want to find eigenvectors, then you want to the find null space of $A-\lambda I$

[clever fjord]

Ari

[normal sundial]

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[sleek gate]

can i ask for help but its in indonesian language

[sleek gate]

In a right triangle, the length of one angle is 30°. If the length of the vertical side is 10 cm, what is the length of the hypotenuse.

this is in translate

[ember oak]

There's insufficient information to answer this question

[sleek gate]

ikr thats why im asking because the website give answers to it but i dont understand

[ember oak]

What's the answer it gives?

[sleek gate]

Length of hypotenuse (hypotenuse) = √(vertex side^2 + base side^2)
= √(10^2 + 10^2)
= √(100 + 100)
= √200
= 10√2 cm (answer)

[hollow quarry]

That angle can't be 30

[ember oak]

Okay they made a typo and meant 45 degrees. That's the only way

[sleek gate]

then how can if we have the 45 degrees to answer the question?

[hollow quarry]

What's the other angle in the triangle if it's 45 degrees

[sleek gate]

90?

[hollow quarry]

Other than that

[sleek gate]

35

[hollow quarry]

Hmm? 180 - 90 - 45 = ...

[sleek gate]

45

sorry

wait so if the degrees is the same its going to give the same cm?

[hollow quarry]

Yes, the sides facing the angles will have the same length

[sleek gate]

ouhhh okayy thankss

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[gentle kelp]

hi

[split gyro]

Hi

[gentle kelp]

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[haughty saddle]

Im stuck in here

[haughty saddle]

i found my width and my height, but now what?

[odd edge]

@haughty saddle Has your question been resolved?

[odd edge]

@haughty saddle Has your question been resolved?

[north badger]

What did you find?

[haughty saddle]

the width is 0.5, the height is f(x) 1/2.5 + 2 for the first rectangle

[north badger]

Ok that's correct for the first rectangle

But what would be the height in general of the i'th rectangle?

[haughty saddle]

thats where i get confused

[north badger]

What is the x position of the right edge of the i'th rectangle?

For the 1st rectangle, it is x=2.5

[haughty saddle]

its 7

[north badger]

No, i want a formula that works for the i'th rectangle

1st rectangle: x=2.5
2nd rectangle: x=3
3rd rectangle: x=3.5
...
i-th rectangle: x=???

[haughty saddle]

0.5 times i?

[north badger]

Almost, but that would give x=0.5 for the first rectangle

And x=1 for rhe 2nd rectangle

And so on

[haughty saddle]

its increasing by 0.5, i got that

[north badger]

You're just missing a +2

x = 0.5 i + 2

[haughty saddle]

why +2?

[north badger]

Because the rectangles start at x=2.5 (right edge)

So if i=1, then x=0.5×1 + 2 = 2.5
For i=2, x=0.5×2 + 2 = 3
Etc...

These all correspond to the bottom right edge of the rectangles

Then the height is easy to find because it's just f(x) with x being those bottom right positions

[haughty saddle]

i still dont understand the +2

you said because the rectangle starts at 2.5. But what if the rectangle starts at 1.5?

[north badger]

For example for the 7th rectangle, the bottomr right corner is at x=0.5×7 + 2

[haughty saddle]

ooohh

[north badger]

If the integration/sum starts earlier, then you would have +another number

[haughty saddle]

Okay i understand that now

[north badger]

Now, using x=0.5 i + 2, the height of the i-th rectangle is simply f(x) = f(0.5 i + 2) = ...

[haughty saddle]

then i put that inside the 1/2 + 2?

the f(x)?

[north badger]

Yep

If you can write it out here that would be great

[haughty saddle]

im using pen and paper first, then ill write it here

[north badger]

Yeah np

[haughty saddle]

i have this so far.... (1/0.5i + 2 ) +2

[north badger]

Yeahhh but those parentheses dont look too good

You should have $f(x)=\frac{1}{0.5i+2}+2$

[clever fjord]

π=√g

[haughty saddle]

yes thats what i have

but i dont see that in the answer

[north badger]

Now, all there's left to do is multiply the height of the i-th rectangle with it's width, and add them all together

[haughty saddle]

so.. 0.5 i + 2 * 0.5?

[north badger]

Not really, it's $0.5 \times f(x) = 0.5 \qty(\frac{1}{0.5 i +2} + 2)$

[clever fjord]

π=√g

[north badger]

width × height

Distribute the 0.5, and simplify the fraction, and you'll see your answer

[haughty saddle]

i dont know what is it with me, but i cant seem to simply it lol

[north badger]

First distribute the 0.5 to get $$0.5\times\frac{1}{0.5i+2} + 0.5\times 2 = \frac{0.5}{0.5 i+2} + 1$$

[clever fjord]

π=√g

[north badger]

Now multiply the numerator and denominator of the fraction by 2 to simplify it

(i.e. dividing by 0.5)

[haughty saddle]

1/0.5 i + 4 + 1?

[north badger]

Almost, it's $$\frac{0.5\times2}{(0.5i+2)\times2} + 1 = \frac{1}{i+4}+1$$ Which gives you as final answer the sum of all rectangles $$\sum_{i=1}^{10} \frac{1}{i+4}+1$$

[clever fjord]

π=√g

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[ivory raven]

taking the usual example of the power set of naturals, the idea is that if the set were countable, then you could biject the elements of the set to the naturals: i.e. list them out

then you generate a new element of the set (via diagonalisation) that is not equal to any element in the set, showing that this listing is not a bijection since it is not surjective

the basic setup is just a proof by contradiction, and diagonalisation is used to show that your new element is distinct from any listed ones

yes, we are assuming that it is countable for a contradiction

assume that the set is countable

so you can biject to the naturals and list them out

because this is a bijection, every real number is in this list (a_i)

now construct a new real number b by taking the first digit of a_1 and adding 1 (roll to 0 if this digit is 9)

that is the first digit of b

now do the same thing, and add 1 to the second digit of a_2

this is the second digit of b

continue for the whole list

this generates a decimal number in the set

but it cannot be on the list, because it is distinct from the nth number in the nth decimal place

so the listing cannot be a bijection

[odd edge]

@dreamy wasp Has your question been resolved?

[ivory raven]

suppose it's on the list, at position n

so b = a_n

but by construction, we picked the nth digit to be the nth digit of a_n, plus 1

so they cannot be equal at that decimal place

so b is not equal to a_n for any n

[odd edge]

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[green finch]

I need help with this trigonometry functions question.

[green finch]

<@&286206848099549185>

[warped grove]

Please don't ping Helpers until 15 minutes have passed with no response

[green finch]

sorry

<@&286206848099549185>

/close

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[fresh sphinx]

can someone verify this

wait lol li did wrong

[cold swift]

was gonna say

[fresh sphinx]

lol idk why i wrote 1 to 0

[cold swift]

when -p+1>0 it doesnt diverge

[fresh sphinx]

oh

[cold swift]

cuz the u would be in the numerator

so itll just be 0/(-p+1)

also for when p=0

you didnt do any substitution there so the bounds shouldnt change

it should still be from 1 to 2

[fresh sphinx]

yeah lol that was a mistake

when p=0 it should be always convergent right

[cold swift]

yeah

[fresh sphinx]

wait wdym by this

youre saying that the integral converges when p<1 right

[cold swift]

yeah

[fresh sphinx]

ah right

[cold swift]

im just saying that when you have a positive exponent there

[fresh sphinx]

so including 0

[cold swift]

the 0 is in the numerator

yeah

p<1 and including 0

so basically just p<1

[fresh sphinx]

so is the answer just all real numbers but 1

[cold swift]

yeah

wait no

if -p+1<0 for say

the 0 would be in the denominator

so itll diverge

[fresh sphinx]

right

[cold swift]

so just all real numbers less than 1

[fresh sphinx]

so the integral converges when p(1, inf)

[cold swift]

uhh p<1 not p>1

[fresh sphinx]

oh '

[cold swift]

remember -p+1>0

yeah you just swapped the direction

[fresh sphinx]

alright tysm

[cold swift]

yw!

[odd edge]

@fresh sphinx Has your question been resolved?

[fresh sphinx]

@cold swift sorry for the ping, im trying to show as much work as i can so im showing the function behaviour as x approaches infinity and why it diverges

so i have this limit and obviously it doesnt exist but like is there any work to do to prove that or not really

cause symbolab doesnt say anything and idk either

[cold swift]

i mean limits are intuitive so its not really necessary

[nocturne brook]

ahh ln(2) is negative isn't it?

[cold swift]

ig you can move the (ln(2))^x to the numerator and do lhopitals

ln(2) is a decimal between 0 and 1

not negative

[nocturne brook]

i'm silly my b

[fresh sphinx]

thats true

i mean i really cba

i dont think thats the point of this question anyway

ty again

[odd edge]

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[split gyro]

Yo

Yo

Yea

They reach home at 5

They used to reach home at 5

It means we can take that

Zuleica school gets over at 4

I feel

Cz it got one hour early

So it ended at 3

She was at station

She walked home

She would have walked for 1 hour 12 minutes

Yea

It's a ratio/proportion

Hints 33

Is the answer 72?

Wait I get it

Wilma

Left the house

And takes I say x time to reach

So she will take x time to return back

Let's say Wilma's school gets over at y

so they are saying y+2x=5 pm

Right

Now they say that once Wilma's school got over one hour early

That is y-1

Oh no

Too many distances

Does zuleica takes time to reach train station

Should I add that also

Ok see

Wilma goes to train station

Assuming it takes x time

Returns back I take it also takes x time

Zuleica school gets over at y

So we can say that

Y+2X=5 pm

Yes

Cz there is no time lag in it

It does not affect

Whether her school gets 1 hour early ober

Over* or she reaches 1 hour early

Ok if you say so

But what about that 1 hour shift

Ok if I take it as that

Ok

Sure

But it says Wilma leaves at ususal time

Usual*

Yes I said that only earlier

Listen let's say they take x time reach home

But today they reach at 4:48

That is x-12

Assuming they reach home at 5 pm

I am saying that

Let's say if Wilma takes 1 hour to get home

4-5

Now zuleica

Started walking at 3

Her mother went out at 4

3-4 she walked

This means

Aah

[odd edge]

@dusky goblet Has your question been resolved?

[odd edge]

@dusky goblet Has your question been resolved?

[copper ore]

@dusky goblet still need help?

[north pivot]

some 1 help>
?

[north pivot]

I need help:)

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[visual geyser]

guys, I need help

[visual geyser]

Examine the convergence of improper integral integrated over 1 to infinity integrating the function 1/(x(1+x^2)) wrt dx

[nocturne brook]

what convergence tests have you tried

[visual geyser]

the comparison test?

[nocturne brook]

ok

[visual geyser]

with x^3

in denominatir

[nocturne brook]

sure

did that give joy?

[visual geyser]

nope

[nocturne brook]

how come?

,w graph 1/x^3 and 1/[x(x^2+1)]

[clever fjord]

Results provided by WolframAlpha

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Requested by i_am_flip

[visual geyser]

so it converges

i guess im stuck in limits

[nocturne brook]

that's not what I had in mind, but alright

[visual geyser]

oops

solved

[nocturne brook]

what I thought of was this

[visual geyser]

]

[visual geyser]

sorry 😦

[nocturne brook]

both 1/x^3 and the target function are positive on [1, infty), with 1/x^3 being greater

[visual geyser]

yes

[nocturne brook]

and we can see that the corresponding integral of 1/x^3 converges

[visual geyser]

and i guess i have one more question

can i share?

[nocturne brook]

so the smaller one also does

sure

[visual geyser]

integrating from sinx to cosx the function e^(-t^2)dt. finding the value of derivative of function at pi/4

not quiet sure about the approach

leibnitz rule?

[nocturne brook]

isn't that like FTC part 1 and/or part 2?

[visual geyser]

what is FTC?

[nocturne brook]

fundamental theorem of calculus

[visual geyser]

f(x) is result of the integration and f'(pi/4) is asked

[visual geyser]

yup

[nocturne brook]

if $F(t)$ is an antiderivative for $f(t)$, then $\frac d{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac d{dx}\left[F(b(x))-F(a(x))\right]$

and this is attackable

uhhhhh

[clever fjord]

Flip

[visual geyser]

is it? i got some different formula here 😦

[nocturne brook]

that's right; I didn't compute the actual derivative on purpose

[visual geyser]

ohh! i didnt pay attention

Im integrating the question wrt dt

and limit is in x

[nocturne brook]

yup

[visual geyser]

so we go with this thing?

will we?

[nocturne brook]

yeah it works

[visual geyser]

oh great!

thank you so much bro 🙂

[nocturne brook]

thankfully it doesn't require information about the antiderivative of f

[visual geyser]

yup

[odd edge]

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[short scaffold]

is my solution correct for (b)?

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<@&286206848099549185>

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[sand horizon]

I found delta is 0.0446, but its probably just a rounding difference, seems good !

[short scaffold]

thankiess!!

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[short scaffold]

I'm having trouble on the last part

[sterile blaze]

a^2-b^2=(a+b)(a-b)

[short scaffold]

oh shit

sorryyy

thankss!!

.close

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[sour grove]

9|(6x+1) = -9

but what do i do with that 9

the first 9

[tulip mica]

is this solve for x?

[sour grove]

yes

[tulip mica]

divide both sides by 9

[sour grove]

oh ok

so x = 0

[odd magnet]

divide

[sour grove]

-9/9 = -1

[tulip mica]

yes

[sour grove]

So |(6x+1) = -1

[tulip mica]

yes

now square sides

[sour grove]

and then i use square

so that would be 6x+1 = 1

6x = 0

[tulip mica]

yes

[sour grove]

so x = 0?

[tulip mica]

wait

sorry

My bad

[tulip mica]

this is not possible

no real roots exist

[sour grove]

Oh so the question isnt possible?

oh i can see the asnwer

its x=0

yeah

means = no sollution

[tulip mica]

Kind of. the question is possible, but the answer to it doesnt exist. Deep

😂

[sour grove]

xD. well thnx man. I didnt knew what to do with that 9. But now i do. Simple have to divide

.closed

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[deft stump]

whats the answer

[rough flint]

A

[deft stump]

thank you

.close

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[sturdy galleon]

alr bet

[sturdy galleon]

oh my pc

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[orchid willow]

I know its not math but i dont have any other place to ask

I've tried everything I can think of but I cant seem to find the way to solve this. The exercise says:
The cubic thermal expansion coefficient of mercury varies with temperature (centigrade) as per the formula you see in the picture, where V_0 is the volume at t = 0°C. The mercury thermometer is calibrated in such a way that 0°C and 100°C coincide with those indicated by a perfect gas thermometer. At the temperature t = 60°C of a perfect gas thermometer, what apparent temperature can be read on the mercury thermometer?

[orchid willow]

For context, i tried to use the perfect gas law to find the expansion coefficient for a perfect gas, find the average for both the coefficients and then relate them to have the same volume change at 100°C, but somehow i always get absurd results

And the right answer should be 59.7°C

[odd edge]

@orchid willow Has your question been resolved?

[odd edge]

@orchid willow Has your question been resolved?

[odd edge]

@orchid willow Has your question been resolved?

[warped grove]

Can you show your work? I don't think the expansion coefficient for a gas matters at all, the gas thermometer is just there to tell us what temperature it is.

I think you'd just want to look at the mercury thermometer at 60°C, using the cubic thermal coefficient

Vs if you just linearly approximated between 0°C and 100°C

So basically find the volume of the mercury at 0°C, 60°C, and 100°C, and compare

[orchid willow]

I'd like to but i erased everything, what i did was found an average thermal coefficient and try to equate the "expansion" using an imaginary V0 for the perfect gas

[orchid willow]

I thought something like this but i struggle to see how to find a proper volume, or a proper average thermal expansion coefficient

I know expansion is not the right word but english is not my first language, just to clarify

[warped grove]

Expansion is the correct word

Basically you will have to integrate the thermal expansion coefficient

(These equations are from Wikipedia)

It comes from solving this differential equation

Do you know how to solve differential equations

[orchid willow]

Yup i do know how to, but i dont know where to start with the volume

[warped grove]

It tells you V0 is the volume

So use that

Even if you don't know the numerical value

[orchid willow]

Im sorry but Im struggling to understand something

[warped grove]

What is it?

[orchid willow]

The book gives me the expansion coefficient function for mercury and then talks about a perfect gas, which has 1/T as the expansion coefficient function

[warped grove]

I'm not sure it's relevant how the perfect gas thermometer works, we can just assume its measurements are accurate

[odd edge]

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[warped grove]

.reopen

[odd edge]

✅

[warped grove]

(You forgot to reply to the bot message asking if your question was resolved)

[orchid willow]

Thanks for reopening that!

So i was saying, you said to consider the mercury at 0, 60 and 100

[warped grove]

Yeah

[orchid willow]

Okay now

It says that the thermometer is set such that 0 and 100 are the same for mercury and for the perfect gas

So id assume that the volume inside the thermometer must be equal at that temperatures, is that so?

[warped grove]

No, I don't think the volume of the perfect gas is relevant

[orchid willow]

Okay

[warped grove]

I think this is just a way of saying that the 0°C and 100°C marks are accurate

[orchid willow]

And where should i get the result (i.e. 59.7) out of?

[warped grove]

At 0°C, the mercury is at the 0°C mark, and at 100°C, the mercury is at the 100°C mark

Well did you calculate the volume of the mercury at the three temperature I said yet

At 0°C, you can assume it's some baseline volume V0

What about at 60°C?

(I haven't worked out this problem fully btw, but this seems like the most logical direction to go to me)

[orchid willow]

I would need to linearise the coefficient between 0 and 100 using the mean integral theorem, the problem in doing so is that at 60°C the volume is somewhere around 2*10^-4 the initial volume

Which is all but logical

[warped grove]

What is the mean integral theorem?

[orchid willow]

It's just a way of finding average values of a function by integrating it

[warped grove]

I don't understand, can you show me what you're doing

[warped grove]

Why not just use the formula I gave you

[orchid willow]

Because the book is giving me a different function, which is not necessarily realistic but its the thing i should use in the exercise

[warped grove]

What

This is the same formula as in the book

1/V0 dV/dT

isn't it

[orchid willow]

Yup but i already have the expression of it

The point of the exercise is to consider the real behaviour of mercury

Therefore the expression is given

[warped grove]

Yes

I know

Did you try solving the differential equation yet

[orchid willow]

Yes i did

[warped grove]

What did you get, can you show me

[orchid willow]

a b and c are the values in the book i just didnt want to copy them

[warped grove]

can you explain why you're taking the value of the cubic expansion coefficient and dividing by 100?

[orchid willow]

Wait a second

I get what you say now maybe

The reason was to find an average value between 0°C and 100°C

But i guess you wanted me to solve the differential equation for T = 60°C

[warped grove]

Yes, I think you should calculate the value of V when t = 60°C

[orchid willow]

Okay then let me try quickly

Okay so

The result is that the volume is 0.011 the initial volume

And thats why i said before that this is all but logical

[warped grove]

btw are you sure the answer is 59.7°C? I ended up getting 59.5°C

[orchid willow]

I'll show you what i did

[warped grove]

okay, show me

I think you're probably just interpreting that number wrong

[orchid willow]

The book says so, might be an approximation issue

I wouldnt bother for a .2 difference

Thats what i did

Wait a second

What i got is the difference between the initial volume and the volume at 60°C

But i missed a point

So at 60°C i should have 1.011*V0

[warped grove]

Yeah exactly, I think instead of linearizing you should do the integration though

[orchid willow]

That's what i did tho, did i miss something?

[warped grove]

If you linearize, then your approximation will be off enough, that you'll get exactly 60°C I think

[warped grove]

After solving the differential equation, you should get this formula

[warped grove]

^

(But instead of alpha_V, the letter you're using is gamma)

[orchid willow]

But isnt that what i just integrated?

[warped grove]

Hm wait I'm not exactly sure I understand what you did, what did you plug in for t here?

[orchid willow]

I mean, i managed to isolate dV, integrate both sides and find a value of V

[orchid willow]

It was from 60 to 0, because it varies with the centigrade scale

So i just used 60 in this case

[warped grove]

oh you integrated it from 0 to 60?

[orchid willow]

From 0 to 60 sorry

[orchid willow]

Nono i just miswrote

[warped grove]

Oh so you just plugged in t=60?

[orchid willow]

Yes

[warped grove]

But t varies from 0 to 60 as the mercury heats up from 0 to 60, no?

[orchid willow]

Yes

Cause the function varies in centigrade, not kelvin

So 0 is just 0 here

[warped grove]

Why is that relevant

Nvm I see what you mean

[orchid willow]

Because when you take the square and the cube in the integral it varies

So now that i solved the differential and got a value for the volume at 60 i should also find a value for 100

[warped grove]

Okay yes I agree this is correct

Yeah

[orchid willow]

Then linearise the volume from 0 to 100

[warped grove]

Plug in t=100 next

And then divide the two values

[orchid willow]

And find at which t it equals 1.011V

Is that right?

[warped grove]

I think you just divide the additional volume at 60 by the additional volume at 100

you should get 0.597

in other words the 59.7°C mark

[orchid willow]

I actually got 60.1

But i see how it should work

[warped grove]

Huh

How did that happen

What number did you get for the volume at 100

[orchid willow]

I got 0.0183 as additional volume

Or 1.0183V0 ad total volume

[warped grove]

oh okay I got the same, I would try dividing just the additional volumes, not the total volumes; I got 0.010937/0.018315 = 0.597...

[orchid willow]

Oh wait

Thats why

I approximated the value at 60 to 0.011

Okay i got 59.7°C

Thanks a lot!

[odd edge]

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[halcyon walrus]

You wanna rationalise that?

[dawn tiger]

hi, i've tried explaining in that channel, don't post in multiple channels and don't advertise your channel in other places

#help-35

.close

yes

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[chilly jacinth]

how do i prove iv.

i did the rest

[proper moth]

couldn't you just graph it?

or is that not allowed

[chilly jacinth]

not allowed

[orchid torrent]

Think of an inequality you can use to relate these two quantities

You should think of ||the power mean inequality||

[orchid torrent]

Resource on this if you’re unfamiliar: ||https://brilliant.org/wiki/power-mean-qagh/||

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[blissful depot]

Let ABCD be a convex quadrilateral and M, N means the sides AB respectively CD. Let it be shown that the relation 4MN^2 = AD^2 + BC^2 + AC^2 + BD^2 - AB^2 - CD^2 takes place.

[blissful depot]

I applied law of cosines

in AMB, BMN, CDM, ADN

but I can't seem to get the 4MN^2 out

[odd edge]

@blissful depot Has your question been resolved?

[blissful depot]

<@&286206848099549185>

[mystic saffron]

help

[odd edge]

@blissful depot Has your question been resolved?

[odd edge]

@blissful depot Has your question been resolved?

[blissful depot]

<@&286206848099549185>

[blissful depot]

uh

who pinged me

[odd edge]

@blissful depot Has your question been resolved?

[blissful depot]

<@&286206848099549185>

[lusty barn]

Still need help? @blissful depot

[blissful depot]

Yes, please.

Ive gotten down to this with law of cosi es

Cosines

[supple mural]

I think you can spam stewarts on MDC, ACB, ADB

[blissful depot]

Ill try that

[blissful depot]

Do I add them up then?

[supple mural]

Just manipulate the equations into the desired form

[blissful depot]

We can pull out MN² from MDC

Yet I dont know about the other terms

[supple mural]

try it

[blissful depot]

Ive gotten this

Thats not so good

I mean i can square it on rhs but it doesnt help

[supple mural]

keep going

stewarts just works

[blissful depot]

with what

i dont know what to replace where

lol

[supple mural]

this

[blissful depot]

Yeah i did it on them

[supple mural]

ok then just substitute and win

first simplify your equations

divide by AB

then its just algebra

[blissful depot]

Should i divide by cd in the first one?

[supple mural]

yeah

[blissful depot]

Oh my god i just saw it

The solve

[supple mural]

good

[blissful depot]

Nevermind, yet we're close

Simplifying makes it worse...

[supple mural]

you simplified wrong

try again

[blissful depot]

This is the best form i could get it to

[supple mural]

thats wrong

why do you have stuff in parnthesis squared

[blissful depot]

Ah great, i replaced with CM instead of CM²

I see

Now i just need the AB² and CD² and we're done

[supple mural]

this is still false

check your stewarts

[blissful depot]

alr

No, ive made a mistake at common factors

Yep, it works now

Thank you @supple mural

Have a great day

.close

[odd edge]

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[supple mural]

np

[odd edge]

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[tame harness]

need help with this thats 19 questions long please @ me

[dawn tiger]

asking for help on quizzes/tests is against the #rules

[tame harness]

ohh kk

.close

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[spring nacelle]

At first glance this looks like something that you would do with mean value theorem

[odd edge]

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[weary pelican]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[heady plover]

1

I know power series btw

[weary pelican]

ok

[weary pelican]

can you write this in sum notation?

(as a power series)

[heady plover]

$\sum_{k=1}^{\infty}{kx^{k-1}}$

[dawn tiger]

hmm that looks like d/dx of something

[weary pelican]

um some stuff wrong

what's n doing in the sum?

and are you sure that there is an end to the sum?

[green elm]

what's n doing anywhere, for that matter 😁

[weary pelican]

.

[clever fjord]

nino

[weary pelican]

much better

now

does kx^k-1 ring a bell?

[sinful grove]

times x and then divide by x

[heady plover]

$1/(1-x^2)$

[clever fjord]

nino

[weary pelican]

woah woah we're just getting started

no that's not it

[heady plover]

ok so 1/(1-x)

[weary pelican]

still not it

listen for a sec

get out of "power series" mode

and just try to recall where you've seen kx^k-1 before

[heady plover]

(x^k)'

[weary pelican]

yes

exactly

[weary pelican]

but the derivative is additive

(let's just say the derivative of an infinite sum is the sum of the derivatives within the radius of convergence)

so

[weary pelican]

this sum

[glass dawn]

I don't know what type of math this is.

[weary pelican]

is the derivative of what?

[weary pelican]

are you here to help or? this kind of comment isn't helping :/

[heady plover]

$\sum_{k=0}^{\infty}{x^{k}}$ i guess

[clever fjord]

nino

[weary pelican]

starting from both k=0 or k=1 works

[glass dawn]

You can use double $ to format it

[weary pelican]

but good choice of starting from k=0

[heady plover]

ok

[weary pelican]

so, now back into power series mode.

What is this sum you wrote equal to?

[heady plover]

1/(1-x)

[weary pelican]

yes

btw valid for which x?

[heady plover]

|x|<1

[weary pelican]

exactly

[weary pelican]

this is 1/(1-x)

[weary pelican]

this is its derivative

[heady plover]

ooh so

it should be 1/(1-x)^2?

[weary pelican]

:(((( almost

yes ok

you got it

[heady plover]

aah!

Tysm

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[fallen sinew]

How do i find the average depth (question c)?

[sand night]

average depth would be the midline

[fallen sinew]

oh so would it be 4

[sand night]

ye

[fallen sinew]

ok

tyy

[sand night]

yup

[fallen sinew]

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[reef lily]

Just a question about statistics

[reef lily]

Im not sure how to do b ii)

i get the rest

but why is the variance the same? is there some formula to use?

E(x) = 2.1

Var (x) = 1.29

E(Y) = 1.9

[lethal pond]

Y is the number of times he drops the ball

so it's 4-X right?

[reef lily]

no Y is dropped

[reef lily]

yeah thats what E(Y) is

[lethal pond]

yeah mb typo

[lethal pond]

no that's what Y is

E[Y] = 4-E[X] by linearity of expectation

also Var[Y] = Var[X] because generally Var[aX+b] = a^2 Var[X]

you can verify this if you havent seen it

[reef lily]

ohhh

[lethal pond]

by just writing out the square and opening it up

[reef lily]

we skipped that section of the book

[lethal pond]

ok it's straightforward

let's just do it

[reef lily]

nah its ok we are supposed to learn it for a later topic

its later in the ciriculim statement

[lethal pond]

ok

[reef lily]

thanks tho

[lethal pond]

i mean for the special case of 4-X

you can still compute it explicitly

but yeah

[reef lily]

i see

thanks

[lethal pond]

sure

[reef lily]

close.

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[mystic saffron]

hello, what is the method for this question?

[spring nacelle]

Let the original number of students from first year be x

Write out what those statements say in terms of algebraic equations

Solve for x

[mystic saffron]

sorry but i still dont understand

[spring nacelle]

Which part are you having trouble translating into equations

Go sentence by sentence

[mystic saffron]

ik what the question wants

but idk how to get the total

the ratio was originally 3:10 then became 6:5 after adding 270 and 120

how am i supposed to know the original

[spring nacelle]

Can you set up the equation for this?

In terms of x

[mystic saffron]

3/13 x + 10/13 x = x?

[spring nacelle]

This tells us nothing about x

That's just x=x

Let's go step by step

[mystic saffron]

3/13x + 270 = 5/11y

no?

[spring nacelle]

What is y?

[mystic saffron]

sure

[mystic saffron]

the new total when 120 and 270 is added

akhh this is hurting my brain

[spring nacelle]

So first sentence

How many second year students are there originally in terms of x

[mystic saffron]

10/13 x

[spring nacelle]

No

[mystic saffron]

oh

[spring nacelle]

This is how we defined x

[mystic saffron]

10/3 x

right?

[spring nacelle]

Correct

[mystic saffron]

alright so 10/3 x is the second year and x is the first

[spring nacelle]

Then what is the number of first and second years after more enter

[mystic saffron]

10/3 x + 270 for the second year

and x + 120 for the first

[spring nacelle]

Nope

You switched the years for the entering students

[mystic saffron]

oh my bad

so 10/3 x + 120