help-39

[gentle perch]

75.2?

[vestal pier]

so f = 5/9 c+32

[restive solstice]

Yep.

[vestal pier]

which means you multiply 24 by 5/9

and then add 32

is it not

its the equation given on the sheet

he sent

this

[light helm]

9/5 *
not 5/9 *

[vestal pier]

our curriculum is different in the uk

mostly just algebra

oh it is 9/5

so 9/5 (24) + 32

what about 6.24 litres to ml @gentle perch

[gentle perch]

not sure, am i using the metric prefixes for it?

[vestal pier]

so 1 litre = 1000ml

[gentle perch]

so 6.24 divided by 1000?

is that on my paper?

[vestal pier]

not divided

if you have 6.24 litres

and you want it to be millilitres

and 1 litre represents 1000 millilitres

how many millilitres do you have

[gentle perch]

i guess around 6

[vestal pier]

alright think of it as this

1l=1000ml

6.24l=####ml?

[gentle perch]

6024?

you have to keep in mind im a bit slow when it comes to this, this is a fundamentals class haha

[vestal pier]

close

what is 0.24 multiplied by 1000

think of it as 3 steps of x10

[gentle perch]

240

[vestal pier]

right

so if 6l=6000ml

and 0.24l=240ml

you can add those up and what do you get?

[gentle perch]

6,240?

[vestal pier]

correct

dont forget your units

6240 ml

i used to forget to write my units

lost some marks occasionally

[gentle perch]

yeah i have to remind myself as well

[vestal pier]

alright last one

gallons to pints

[gentle perch]

am i using 2 pints is quart or 4 quarts is 1 gallon

[vestal pier]

2.4 gallons

this is a double step process

was is 4 multiplied by 2.4

[gentle perch]

9.6

[vestal pier]

and what is 9.6x2

because you have 9.6 quarts

[gentle perch]

19.2

[vestal pier]

yes

[gentle perch]

pt

[vestal pier]

good job we completed that row

would you like to do the rest of the other row?

[gentle perch]

yes

[vestal pier]

do you have the question for the other row?

h

alright

[gentle perch]

wym

[vestal pier]

so this is compound shapes

meaning it consists of more than 1 shape

could you draw a line and see where you could seperate it

into 2 shapes

[vestal pier]

correct

[gentle perch]

so is 3 on top of the rectangle and 6 is on top of the square?

[vestal pier]

now what do you think you should do?

we can work out area first

[gentle perch]

Like this

[vestal pier]

yes

ok lets work out perimeter first i guess

so

your missing 1 length

how will you work out x

[gentle perch]

add them together?

[vestal pier]

nope

so on the right side you got 4

you see it?

[gentle perch]

yes

[vestal pier]

if x+4=7

what does x equal

[gentle perch]

3

[vestal pier]

correct

so replace x with 3

and add all those numbers up

[gentle perch]

on the first side or just all of them

[vestal pier]

all of them

perimeter is everything outside

[gentle perch]

32

[vestal pier]

yes

now the area

which is mildly more complicated

[vestal pier]

so here you've split it into 2 rectangles

how do you work out the area of a rectangle?

[gentle perch]

do i use the formula sheet on this

[vestal pier]

not really needed

[gentle perch]

kk

[vestal pier]

but if you need a reminder

maybe

[gentle perch]

do i use x

[vestal pier]

the area of a rectangle is length x width

ok we got rectangle a and b

if the length for a is 7

[gentle perch]

so 3 x 3 x 7 (x 7?)

[vestal pier]

and width is 3

what is the area of a

[restive solstice]

Nope. Just multiply one of the base by one of the height.

[gentle perch]

so 3 x 7?

[vestal pier]

yeah

which is?

[gentle perch]

21

[vestal pier]

you mind taking over after this question?

i gotta revise for my maths exam this friday

[vestal pier]

correct

now you have a length of 6 and a width of 4 for b

whats the area for b?

[honest vigil]

I'll be on standby if necessary.

[gentle perch]

6 x 4 = 24?

[vestal pier]

alright thanks

[vestal pier]

yeah

now you have area a=21
area b=24

what would you do to get the area of this compound shape

with the 2 areas

[gentle perch]

add?

[vestal pier]

yes

so 21+24

[gentle perch]

45

[vestal pier]

correct

so the area is 45 metre square

i'll be on my way now

i have a couple things i have to do unfortunately

[gentle perch]

thank you for the help, it was appreciated!

[gentle perch]

For anyone that wants to take over, this is the page. We just did 2

[honest vigil]

can you attempt q3 using what you now know about perimeters and areas?

[gentle perch]

sure

well i guess i oughta figure out how to draw the right shape itself first

[honest vigil]

the shape of a basketball court is a pretty standard shape.

[gentle perch]

should i just do a square

[honest vigil]

more a rectangle than a square.

[gentle perch]

like this?

[restive solstice]

Yep

But I'd personally label the units too (e.g. 94 ft instead of 94)

[honest vigil]

I'll let Terry take over then.

[gentle perch]

i appreciate the help

so for peri i need to figure out x + 50 = 94?

[restive solstice]

Nope.

You calculate the length of its outer shape. So for rectangle it's 2*(width+height).

[gentle perch]

2 x 94 + 2 x 50?

[restive solstice]

Yep

[gentle perch]

288

and area is 94 + 50?

[restive solstice]

Nope. Do you recall the formula for the area of rectangle?

[gentle perch]

a = lw, so its multiplication?

[restive solstice]

Yep. Juxtaposed multiplication.

[gentle perch]

4,700?

[restive solstice]

If you mean 4 700 instead of 4.7 to 3 decimal places then yes.

[gentle perch]

4,700 is wut i got when calculating 94 x 50 together, is there more to it?

[restive solstice]

Nope, just notation, but I'd personally write 4 700 instead of 4,700 to prevent any confusion as commas is used as decimal separator in some countries.

[gentle perch]

will do

alright question 4

[restive solstice]

Doing part (a)? What do you know about the shape?

[gentle perch]

its a bit of a parallelogram right?

[restive solstice]

Yep. For the question, I'm not sure that if the 3.2 cm given in the question is needed or not.

[honest vigil]

it is.

[gentle perch]

is 3.2 the base?

[restive solstice]

Nope, it's the height.

[gentle perch]

ah so 12 is the base?

so 12 x 3.2?

[restive solstice]

Yep.

[gentle perch]

38.4

confused about the triangle, whats base and whats height in this circumstance

[restive solstice]

Noticed any patterns between this and the question? (Note that the base is at a right angle to the height labelled).

[gentle perch]

so the height would be 4 ft and the base would be 16?

1/2 x 16 x 4?

32 ft

[restive solstice]

Yep

[gentle perch]

next one is trapezoid?

[restive solstice]

Yep. Can you recall (or derive) the area of a trapezoid?

[gentle perch]

a = 1/2 x 2.4 + 5.6 x 6?

[restive solstice]

Nope. Multiplicative operations like multiplication and division takes precedence over additive operations like addition and subtraction.

So there should be a parenthesis like (2.5 + 5.6).

[gentle perch]

how should i calculate it in my calculator for the test when a similar question pops up

[honest vigil]

[(2.4 + 5.6) x 6] / 2.

[gentle perch]

the calculator is not advanced like that haha

[honest vigil]

that's why the brackets are there to tell you in what order to go.

[gentle perch]

ohh

well 2.4 + 5.6 is 8

[honest vigil]

if that was not clear enough, add 2.4 + 5.6, then multiply the result by 6, then divide that result by 2.
(dividing by 2 first, then multiplying by 6, should result in the same answer however, but I do not like to do division before anything else for a reason related to how calculators work.)

[gentle perch]

x 6 is 48

divided by 2 is 24

so 24 in?

[restive solstice]

Yep.

[gentle perch]

Alrighty. I’ll send the second to last page.

[honest vigil]

also if it was not clear, you should try to always do division last on calculators.

(respecting the order of operations, of course.)

[gentle perch]

yeah ive heard something like that before, and the order of operations obv. but i forgot it could also play into something like this

[honest vigil]

the problem is not explicitly about the order of operations, but rather how calculators store intermediate results.

but otherwise, you can proceed as usual.

sorry for interrupting.

[gentle perch]

all good! i appreciate the information

circumference on circle says (obv not the proper way to type it) c = nd = nr. so is that c = 3.14 x 2 x 8 = 2 x 3.14 x 8?

[restive solstice]

Nope. Because the circumference of a circle is given as pi * diameter = 2 * pi * radius.

And the diameter (not the radius) of the circle is 8 cm.

[gentle perch]

gotcha gotcha

alright let me think on this for a sec

so the diameter is not 2r but instead 8cm?

[restive solstice]

Both 2r (2 * radius) and the diameter is 8 cm. If you look at the diagram given in the question, you'll see that the radius is not 8 cm.

[gentle perch]

ok ok

so how do i figure out the radius

[restive solstice]

Since you're given the diameter, you solve d = 8 = 2r (basically divide the diameter by 2). (I got to go now)

[gentle perch]

gotcha gotcha, thank you for the help! ill try to figure that out

8 divided by 2 is 4

C = 3.14 x 8 = 2 x 3.14 x 4?

[honest vigil]

if you are allowed to take pi = 3.14, yes.

[gentle perch]

is the answer 25.12?

[honest vigil]

,w 8*3.14

yes.

oh I used ,w instead of ,calc, zzz.

[gentle perch]

all good

so i can move onto 6?

wait no area

[honest vigil]

6 is rather simple.

[gentle perch]

a = 3.14 x 4(2) aka 3.14 x 4 x 3.14 x 4?

the answer i get is 157.7536 which doesnt seem right

[fleet ridge]

Close, but there is no need to repeat 3.14

[honest vigil]

but 6 is not asking for the area, right?

[fleet ridge]

He's doing 5

[honest vigil]

oh wait area for 5.

sorry.

[gentle perch]

it has the small 2 at the top so i thought that meant squared

[honest vigil]

only the radius is squared.

[gentle perch]

all good

[civic saddle]

the operation of squaring comes first !

[gentle perch]

gotcha

[honest vigil]

alright I see too many helpers here, I'll step back to prevent confusion again.

[gentle perch]

so 12.56?

[fleet ridge]

No

You still need to square the 4

So its 3.14x4x4

[gentle perch]

ohh ok ok

50.24

[fleet ridge]

Yes

[gentle perch]

alright number 6

[fleet ridge]

Should be easy

[gentle perch]

is it just 2 divided by 5?

[fleet ridge]

No

Try again

[gentle perch]

2 x 5?

[fleet ridge]

Yes

[gentle perch]

10

[fleet ridge]

Because a diameter is twice the radius

[fleet ridge]

Yes

[gentle perch]

i dont notice a specific prism section on my page for 7

[fleet ridge]

Its a cuboid

Also known as a rectangular prism

Do you know the volume of a cuboid?

[gentle perch]

no, i see a rectangular solid thing on the paper though, is that related?

[fleet ridge]

The volume of a cuboid is lbh

[fleet ridge]

Thats a cuboid

Sorry I had to do something

[gentle perch]

all good

[fleet ridge]

So the volume of a cuboid is the volume of the rectangle multiplied by the height

[gentle perch]

is the height 3?

[fleet ridge]

Yes

[gentle perch]

does 3m go on top of the prism as well?

for when im calculating

[sterile python]

For prisms you just take height x length x width for volume

Height not necessarily 3 here

[gentle perch]

i see i see

then how do i figure that out

[sterile python]

?

Figure what out?

[gentle perch]

the height, you said it wasnt 3

[sterile python]

It could be 8

[fleet ridge]

Just assume the height to be 3

[gentle perch]

okey dokey

[fleet ridge]

And find the other dimensions too

[gentle perch]

so 3 x 3 x 8?

[sterile python]

Depends on how do you project the surfaces on different direction

[gentle perch]

72

[fleet ridge]

Yes

[sterile python]

This works similarly to the cylinder (your next example)

[gentle perch]

and i use sphere on the paper for it?

[sterile python]

Suppose you have a cylinder, then it is composed of layers of circles with same size

Have you learned about sphere volume formula?

[gentle perch]

this is a fundamentals class and whats on the paper is basically what weve learned aside from whats also on this practice test

[fleet ridge]

So what do you know related to areas and volumes

[gentle perch]

barely anything aside from whats been chatted about here tonight

[fleet ridge]

Oh

Well you'll have to learn about areas and volumes of common shapes

[gentle perch]

isnt that whats on the formula paper im allowed to use?

[fleet ridge]

I think so

[gentle perch]

here

[fleet ridge]

Yes

A cuboid is a rectangular solid

[gentle perch]

so ill be using sphere?

[fleet ridge]

No q8 is a cylinder

[gentle perch]

ohh

so 3.14 x 3 x 3 x 8?

[fleet ridge]

Yes

[gentle perch]

i got 226.08

[fleet ridge]

Yes

[gentle perch]

alright well do 9 and 10 then ill go to sleep and work on the last page tomorrow i think

[fleet ridge]

K

[gentle perch]

so itll be 3.14 x 2 x 2 x 3.2 divided by 3?

[fleet ridge]

Yes

But you also have to round

[gentle perch]

ya so i got 13.3973333333, i forget am i going from left to right or right to left

[fleet ridge]

What do you mean

[gentle perch]

nvm

13.40

[fleet ridge]

Yes

[gentle perch]

how do i find the radius for 10

[fleet ridge]

The diameter is given

[gentle perch]

4/3 x 3.14 x 23.6 x 23.6 x 23.6?

[fleet ridge]

No

You can find the radius form the diameter

[gentle perch]

im confused

[fleet ridge]

What do you mean

Someone already said how to find radius form diameter

[grave parcel]

Radius = Diameter divided by 2

[gentle perch]

i remember now, thank you

[grave parcel]

👍

[gentle perch]

11.8

4/3 x 3.14 x 11.8 x 11.8 x 11.8

[grave parcel]

which question are u doing?

[gentle perch]

6878.80?

[gentle perch]

10

[fleet ridge]

Yes

Wait

[grave parcel]

uhhh

hold on let me see

[fleet ridge]

It should be 6878.63 right.

[grave parcel]

@gentle perch are u using pi or 3.14?

[grave parcel]

.80 and .63 are both acceptable ig

[gentle perch]

it should be 83 i think

[grave parcel]

if you plug it all into ur calculator yes

[fleet ridge]

I meant .83

Mb

[gentle perch]

all good

alright, im gonna get some rest now, thank you for helping, everyone! very appreciated 💗

[grave parcel]

ye you have had this channel up for hours lol

get some rest good job

[gentle perch]

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[pearl pond]

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[rustic gale]

Vowels never come together.
Or

No two vowels comes together.

Both phrases same hai or different????

[cinder drum]

BITSAT ki prep kr rha hai kya??

[fleet ridge]

This is a math server

[smoky gull]

logical reasoning is a part of maths

[honest oyster]

What does hai mean

[fleet ridge]

I guess

[smoky gull]

Both phrases are same or different is what the last sentence means

[cinder drum]

hai is like... u can ignore that it would still mean same

[fleet ridge]

I think it means 'is'

[smoky gull]

-# ffs not translatable

[cinder drum]

literally na, but not in this context

[fleet ridge]

Oh

Na?

[cinder drum]

nvm 😭

[smoky gull]

just focus on the Q

[rustic gale]

Yo ki hunda c

[cinder drum]

Entrence exam hunda c

[rustic gale]

Pilani jane ka koi mood nahin mera

.close

[pearl pond]

Channel closed

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Use .reopen if this was a mistake.

[cinder drum]

🥀

[pearl pond]

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[ocean hazel]

Why is the velocity time graph a pair of horizontal lines when drawn using a displacement time graph? The values on my graph are correct, but the shape is not.
-# First image is the question and answer, Second Image is my working.

[bright apex]

ru converting v/t into s/t or s/t to v/t

[ocean hazel]

is s is displacement then yes, they're asking us to convert displacment time graph into a velocity time graph

[bright apex]

and. ur question is?

[ocean hazel]

^

[bright apex]

so u want to convert s/t to v/t

[ocean hazel]

no

[bright apex]

then

[ocean hazel]

I'm asking why is the velocity time graph a pair of horizontal lines when drawn using a displacement time graph?

[bright apex]

do u know differentiation

[ebon skiff]

velocity is just the slope of the displacement-time graph

[ocean hazel]

oh my days I can't believe I overlooked this 🤦‍♂️
Slope is constant so velocity will also be constant right hence why its horizontal since a horizontal line is neither increasing or decraesing in velocity its constant.

[ebon skiff]

exactly

[ocean hazel]

Got it, thanks for your help.

.close

[pearl pond]

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[ebon skiff]

np

[pearl pond]

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[valid nova]

this $\log_{10}(5) \cdot \log_{10}(20)$

[jolly parrot]

TheAstorPastor

[valid nova]

or

$\log_{10}(5 \cdot \log_{10}(20))$

[jolly parrot]

TheAstorPastor

[fleet ridge]

I think its this one

[fresh solar]

This one

[valid nova]

hmm ok

.close

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[north talon]

i noticed that if an = bn, then am = bm for all m >= n

[slim vigil]

Holy handwriting

[north talon]

so it suffices to show that the sequence cn = | an - bn| converges to 0

[slim vigil]

<@&268886789983436800>

[slim vigil]

For i=0,1,...?

Can you check the problem statement?

[north talon]

index starts from a0

wdym?

[slim vigil]

Oh nvm

[north talon]

oh wait

if a0=b0 we are done so assume a0 > b0

[slim vigil]

Can't you just let a_0 and b_0 be 2 different numbers then define the sequence as such?

Can you check the problem statement?

[north talon]

consider ci = ai - bi, we know c0 > 0
expand c{i+1} = ci + floor{√bi} - floor{√ai}

[north talon]

its correct

[slim vigil]

I can just let a=0, b=2 or sth then calculate the rest

[north talon]

maybe just seems weird translated, lol

[slim vigil]

Same with any 2 other numbers

There's no conditions on a_0 and b_0

[north talon]

i mean the problem probably means it works for any arbitrary a0,b0

[slim vigil]

Which it obviously doesn't

Try picking 2 random different values for a_0 and b_0

You can calculate a_1 and b_1 from those, then so on

Check the original problem

[north talon]

seems correct, xd

oh, wait

[split void]

is it all n?

[north talon]

in the solution he uses ai+1 = ai - floor √ bi

????

its subtraction ig

my bad

[bronze surge]

hi

[split void]

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[slim vigil]

So what's the question?

[north talon]

$a_{i+1} = a_i - \lfloor \sqrt{b_i} \rfloor$ and $b_{i+1} = b_i - \lfloor \sqrt{a_i} \rfloor$

[jolly parrot]

Copter

[tropic saddle]

I'm not seeing an obvious reason why it shouldnt work. the first few examples I did do work

[north talon]

-w-

[rough forge]

-w-

[tropic saddle]

although it does seem like eventually we have a_i = b_i+(something small) and then the likelihood that floor(sqrt(a_i)) and floor(sqrt(b_i)) are different gets smaller and smaller so you might be stuck

[north talon]

well i think i solved this either way

if you wlog a0 > b0 you get that the sequence an - bn is eventually constant

then for large enough n we must have floor √an = floor √bn

suppose that an - bn isnt constant to zero, say c >0

set floor √an = floor √bn = x ⇒ an = x^2 + a, bn = x^2 + b for some constants a,b

0 <= b < a <= 2x

if a >= x+1, you can simplify a{i+1} = (x+1)^2 + (a-(x+1)) so it suffices to look at 0 <= b < a <= x

then you can bound to get a contradiction :D

[tropic saddle]

is your x supposed to be independent of n?

and those a,b?

[north talon]

x,a,b are constants

[tropic saddle]

thats obviously bs

[north talon]

-w-

[tropic saddle]

a_n and b_n keep growing

[north talon]

oh i didnt mean it for all n

should i index it differently

[north talon]

n here is the minimum n such that an - bn is constant

[tropic saddle]

well ok you didnt post where you got your contradiction so who knows whether that part checks out

but ok rest does

[slim vigil]

Well they said all n, if a_0≠b_0 then it's false?

[tropic saddle]

they didnt

[slim vigil]

Is this the problem statement?

Oh nvm

I'm blind

Sorry

[north talon]

$a_{N+2a} = (x+a)^2$ and $b_{N+2a} = (x+a)^2 + b-a < (x+a)^2$ $\$ which $\lfloor \sqrt b_{N+2a} \rfloor < \lfloor \sqrt a_{N+2a} \rfloor$ which is a contradiction.

[jolly parrot]

Copter

[tropic saddle]

ok i'll believe you that those equalities are true

[north talon]

baka latex

[tropic saddle]

then checks out

[north talon]

nice :3

.close

[pearl pond]

Channel closed

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[fringe geyser]

Heya, can anyone help me with a mechanics question in the DMs?

[fringe geyser]

I’m not actually allowed to post the question anywhere but I would like to see if what I did was correct?

[plush bramble]

Physics server in #old-network

[plush bramble]

Please don't abuse help channels. They're for math help

[fringe geyser]

It’s to do with maths

[autumn fossil]

is it some kind of competition?

[fringe geyser]

Mechanics in maths

[autumn fossil]

if youre not supposed to share the question, are you supposed to get helped with them?

[fringe geyser]

No of course not

[plush bramble]

No. Not everything is math

If you have a math question, ask it

If you don't, please go to
https://discord.gg/physics

[flint basalt]

please post your question here if you cant post it anywhere try asking a classmate or professor for help

[autumn fossil]

Hm, didnt we allow physics and related subjects in the past?

[flint basalt]

its allowed, but harder to get help

[fringe geyser]

Bro if you do maths it’s inevitable that you do mechanics too

Right my bad sorry for asking for help

[hoary relic]

Please just post the question so we can help you.

[flint basalt]

it might be better to ask in the physics server mentioned above. otherwise we cant do anything if you cant post the question

[fringe geyser]

If you’re gonna be a smart ass I genuinely don’t want help

[flint basalt]

if you want help post the question...

[plush bramble]

<@&268886789983436800>

[plush bramble]

Wrong

[fluid oar]

What do you mean by that

Is this academic dishonesty

[fringe geyser]

No?

[plush bramble]

Lmao

[fringe geyser]

Bro u guys are actually neeks

I’m taking my ass out this server

[spiral pivot]

Do you mean neet?

[flint basalt]

if you cant explain why you can't post the question thats a big pointer that you are trying to get help on something you shouldnt be getting help on

[spiral pivot]

Anyway, typically, we request that you actually ask your question to get help.

[warped violet]

they aren't here

[flint basalt]

they gone

[autumn fossil]

too late

[spiral pivot]

.close

[pearl pond]

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[flint basalt]

<@&268886789983436800>

[pearl pond]

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[flint basalt]

mrbeast has returned

[unborn abyss]

he never left

[modest tartan]

lol

[hoary relic]

just make a bot bait channel atp

[pearl pond]

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[eternal veldt]

Could someone help me with question letter B pls

[fresh solar]

find an expression for h in terms of m (mass) and substitute that in "rubener's law"

[eternal veldt]

like this?

[fresh solar]

yes

but E is proportional to the square of h so better to introduce a constant too

now you can just proceed with forming the equation using the given values

[eternal veldt]

is this what you mean?

[fresh solar]

where did you get this from?

[eternal veldt]

i thought because the cheeta weighing 220 is k time the E of the cheeta weighing 64

I put it as K * E

and substitute the mass as 64

[junior loom]

do you know what it means for things to be in proportion

[eternal veldt]

nope not really

[junior loom]

i would start there

[eternal veldt]

When two values change at the same rate?

[fresh solar]

I will step back for you to explain then

[junior loom]

two variables x,y are in proportion iff there's a nonzero constant k such that x=ky

[junior loom]

start at the beginning

m is directly proportional to h³, what does that mean

as an equation

[eternal veldt]

When h^3 increase m increases?

[junior loom]

nope, its the same thing as saying m and h³ are in proportion

[junior loom]

^

[eternal veldt]

m = k*h^3

[junior loom]

for some non zero constant k yes

k can be negative and they would be in proportion. they can both increase like m=k² but m is not proportional to k in such a case

[junior loom]

so this is neither sufficient nor necessary

[junior loom]

(for m>=0)

[eternal veldt]

so m beign directionally proportional to h^3 means when height increases the mass is cubed?

[junior loom]

so "Rubner's law" says there is a constant c such that E=ch²

[junior loom]

it means ^

[junior loom]

^

[eternal veldt]

So in Rubners law as the h increases the E increases faster and bigger because it is cubed?

[junior loom]

"faster and bigger" not sure what you mean, comparing what to what?

[eternal veldt]

If H doubles then E becomes 4 times bigger

[junior loom]

no, it becomes 4c times bigger

[eternal veldt]

yea

E = x*h^2

[junior loom]

dont recommend using x for a constant

its not wrong, but its not the convention

[eternal veldt]

E = c*h^2

[junior loom]

yeah

c or k or a greek letter usually is used

[eternal veldt]

and then like the other guys said do i substitue the h i got from h = cuberoot(1.76) into E = c*h^2?

[junior loom]

well you need to find the energy needs of the cheetah and lion

but theyre in proportion

E_[lion,220kg]=kE_[cheeta,64kg]

and then you use the equations from before

going now, gl

[fresh solar]

<@&268886789983436800>

[eternal veldt]

I think i get it now

(Cube root(1.76))^2 = E for Lion

and (0.8)^2 = Engery for cheeta

and lions energy is k times more than the cheeta so the equation becoems like this right?

Then i jsut do algera to sovle for k which will then become 2.28?

.close

[pearl pond]

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[pearl pond]

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[brisk steeple]

How to prove that the arithmetic mean of root(6) and root(2) is equal to the square root of 2 + root(3)?

[autumn fossil]

start by expressing the arithmetic mean of root(6) and root(2)

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

try squaring both sides

[brisk steeple]

yes

I can prove it by squaring both the sides, but I want to move from root(2 + root(3)) to (root(2) + root(6))/2.

[autumn fossil]

youll have to write the inside as a square

2 + sqrt(3) = (a + bsqrt3)^2

[autumn fossil]

expand it

hmm

might be better to write it as sqrt(a) + sqrt(b) instead

2 + sqrt(3) = (sqrt(a) + sqrt(b))^2

[pearl pond]

@brisk steeple Has your question been resolved?

[brisk steeple]

Got it

Thank you

.close

[pearl pond]

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[north talon]

Find all monic polynomials $P(x) $ such that $P(x+1) $ divides $P(x)^2 -1$

[jolly parrot]

Copter

[north talon]

write $P(x) = (x-r_1)(x-r_2)...(x-r_k)$ then for any $x + 1 =r_i$ we know $P(x)^2 = 1$

[jolly parrot]

Copter

[north talon]

i have that $P(r_i-1) = \pm 1$

[jolly parrot]

Copter

[north talon]

but other than that i dont really know what to do since i dont know the signs

because if P(r_1 -1) = 1, then P(r2 -1) = -1 by continuity?

i think thats right?

nvm, that doesnt work

i guess if like P(r1 -1) = 1 = P(r2 -1) would that mean r1 must be a mutiple root?

is that true

by ivt

[viscid shale]

I havent read much and im going away in like a sec

But cant you define P^2 - 1 as a diff of squares Where
P^2(x) - 1 = (P(x)-1)(P(x)+1)

And since the condition of divisility can be read as a logical conjunction
you can search all polynomials that satisfy P(x+1) = P(x)+1 or -1

[north talon]

arrange the roots so theres m many such that P(r-1) = 1 and k-m for the other?

[north talon]

well i dont really know what id do with that

also we dont have equality right

or must it be equal because of the degree?

[viscid shale]

About what in particular?

[north talon]

P(x+1) = P(x) +1 or P(x) -1

is that true?

[viscid shale]

For some polynomial ig.

[pearl pond]

@north talon Has your question been resolved?

[pearl pond]

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[plush bramble]

<@&268886789983436800>

[smoky gull]

<@&268886789983436800>

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

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[pearl pond]

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[spring whale]

What 10+9

[dense jasper]

<@&268886789983436800> troll

[spring whale]

Ok sorry

[radiant terrace]

Don't troll here

It distracts the helpers from helping and the helpees from getting help

.close

[pearl pond]

Channel closed

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[spring whale]

What (-190)-(968)×770:197

[radiant terrace]

A day long mute for trolling

[pearl pond]

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[rotund narwhal]

Hi this isn’t due in 10 minute and this is my kitchen table

I am completely lost

My teacher in class told me to do chain rule and product rule

This is numerical methods btw

How can I start like how do I setup the problem?

[ashen ivy]

i don't think your professor likes you \
what is your first instinct when seeing $\dv{t} \left( \pdv{f}{t} + f \pdv{f}{y} \right)$?

[jolly parrot]

haseeb ♥

[rotund narwhal]

Differentiate

But differentiate with respect to f and f’/y?

[ashen ivy]

well, with respect to t, but using the chain rule

[rotund narwhal]

but what is the original function thats what throws me off?

[ashen ivy]

let's get the easy part out of the way first though: $$= \dv{t} \left( \pdv{f}{t} \right) + \dv{t} \left( f \pdv{f}{y} \right)$$

[jolly parrot]

haseeb ♥

[ashen ivy]

my bad

but let's focus on one term at a time

what does f depend on?

[rotund narwhal]

y(t)

[ashen ivy]

yes, and what else?

[rotund narwhal]

sorry

t

[ashen ivy]

? that's ok

no harm in guessing

yes, t and y(t), because we have f(t, y(t))

do you know how to draw the "dependency tree" for a function like this?

[rotund narwhal]

no i would love to learn how

[ashen ivy]

let's say you have a function f(x(t),y(t)) that depends on x(t) and y(t). then f depends on x and y (but not t!), whereas x depends on t and y depends on t

we represent this with a tree diagram:

$\begin{tikzcd}
& f & \
x && y \
t && t
\arrow[from=1-2, to=2-1]
\arrow[from=1-2, to=2-3]
\arrow[from=2-1, to=3-1]
\arrow[from=2-3, to=3-3]
\end{tikzcd}$

[jolly parrot]

haseeb ♥

[rotund narwhal]

sure

[ashen ivy]

the key is, we can 'read off' the chain rule from here

$f \to x \to t$ gives $\pdv{f}{x} \cdot \dv{x}{t}$

[jolly parrot]

haseeb ♥

[ashen ivy]

$f \to y \to t$ gives $\pdv{f}{y} \cdot \dv{y}{t}$

[jolly parrot]

haseeb ♥

[ashen ivy]

then we add up all possible branches to get $$\dv{f}{t} = \pdv{f}{x} \dv{x}{t} + \pdv{f}{y} \dv{y}{t}$$

[jolly parrot]

haseeb ♥

[ashen ivy]

anyway, you don't have to do it like that, but i think it's a little easier to work with the chain rule this way

you can see each "level" of dependence

[rotund narwhal]

ik we dont have to see it that way but i prefer to see it this way

[ashen ivy]

yeah i like to think this is easier :)

back to the problem, we have $\dv{t} \left( \pdv{f}{x} \right)$

[jolly parrot]

haseeb ♥

[ashen ivy]

treat $\pdv{f}{x}$ as a function, write out what it depends on, then get the chain rule for it

[jolly parrot]

haseeb ♥

[rotund narwhal]

oh this gets long lol

given f
/
x. y

wouldnt i need to do two sets of chain rule

[ashen ivy]

welcome to numerical methods 💔

[ashen ivy]

yes, which is why we get two terms (one for x and one for y)

but note, you have t and y(t), so your tree looks a little different

[rotund narwhal]

so. ƒ
/
/
(∂y/∂t)*(∂y(t)/dt) (∂t/∂y) * (∂y/∂t)

no

wait

[ashen ivy]

yes, in the case when you have f(x,y)

[rotund narwhal]

in place of x i should put t

[ashen ivy]

but now we're considering f(t, y(t))

[ashen ivy]

does t depend on anything, though?

[rotund narwhal]

y

[ashen ivy]

hmm i think i posed this wrong

we want d/dt so we want to terminate each branch at t

[rotund narwhal]

like this?

ima stop being lazy one sec

re writing my dependency tree

[ashen ivy]

not quite, let me show you the tree for this because its pretty different

ooh ok ill let you write it

[rotund narwhal]

okay no i wrote it out wrong

and im back to being lost

[ashen ivy]

ok i feel like you're almost there

[ashen ivy]

if f depends on t by itself, the branch is gonna be f -> t

and we're finding $\dv{t}$ of the function $\pdv{f}{x} (t, y(t))$

[jolly parrot]

haseeb ♥

[ashen ivy]

which depends on (t, y(t))*

[ashen ivy]

$\begin{tikzcd}
& {\pdv{f}{t}} & \
t && {y(t)} \
&& t
\arrow[from=1-2, to=2-1]
\arrow[from=1-2, to=2-3]
\arrow[from=2-3, to=3-3]
\end{tikzcd}$

in fairness i gave you a really bad example for what we were doing

[rotund narwhal]

[ashen ivy]

so we get this asymmetrical tree => our chain rule will involve dt/dt = 1

can you write the chain rule for $\dv{t} \left( \pdv{f}{x} \right)$ based off this tree?

[jolly parrot]

haseeb ♥

[rotund narwhal]

∂y/dt *t ?

no

i cant

[ashen ivy]

start with some derivative of f

[rotund narwhal]

oh

∂(f)/dt

[ashen ivy]

ok that's one branch done

oh wait actually, the function isn't f

it's $\pdv{f}{x}$

[jolly parrot]

haseeb ♥

[ashen ivy]

so the t-partial of that function

[rotund narwhal]

(∂∂(f)) /∂t

?

no wait

shouldnt it be product rule?

[ashen ivy]

you can write stuff down and take a picture, i recognize typing is not the easiest

especially with all these damn ∂

[rotund narwhal]

lol okay sorry

[ashen ivy]

nah whatevers easiest for u

[ashen ivy]

it'll be a second partial, $\pdv{t} \left( \pdv{f}{t} \right)$

oh shoot im hallucinating

call me chatgpt

[jolly parrot]

haseeb ♥

[rotund narwhal]

thank u sorry i threw u offf

[jolly parrot]

haseeb ♥

[ashen ivy]

ok fully corrected, sorry

so!

we have one branch, now we need ∂f -> y -> t

also this is NOT a 10min question

[rotund narwhal]

i completely agree

i wrote down the dependency tree

[ashen ivy]

almost, but there's no x

(my bad on that)

so u can just erase those

[rotund narwhal]

sorry no no

wrong pgoto

[ashen ivy]

oh lmao

[ashen ivy]

so, we have $\pdv[2]{f}{t} + \square$, and that second term depends on $\partial f \to y,$ then $y \to t$

aa

[jolly parrot]

haseeb ♥

[rotund narwhal]

One sec

[ashen ivy]

we want a + between the two branches, and then we need one in terms of f

[rotund narwhal]

Uhh

[ashen ivy]

so $\pdv[2]{f}{t} + \pdv{y}\left( \pdv{f}{t}\right)\pdv{y}{t}$

[rotund narwhal]

Ohhhh

[ashen ivy]

we had a branch $f_t \to y$ which gave us $\pdv{f_t}{y}$, then multiplied with $y \to t$ which gives us $\pdv{y}{t}$

[jolly parrot]

haseeb ♥

haseeb ♥

[ashen ivy]

forgot what we were differentiating

[rotund narwhal]

Like

Sorry d(f)/d(t)

[ashen ivy]

so we end up with $\dv{t} \left(\pdv{f}{t} \right) = \pdv[2]{f}{t} + \pdv{f}{y}{t} \cdot \pdv{y}{t}$

[jolly parrot]

haseeb ♥

[ashen ivy]

does that make sense? we can think of the chain rule however you usually think about it if that is easier

[rotund narwhal]

My laptop just died

[ashen ivy]

natural reaction to the multivariate chain rule

[rotund narwhal]

So where there is a split in the branches product rule and then chain rule

[ashen ivy]

i would say, each branch is multiplication, then you're adding them together

the whole thing is the chain rule

$\overset{\partial f \to t}{\pdv[2]{f}{t}} + \overset{\partial f \to y}{\pdv{f}{y}{t}} \cdot \overset{y \to t}{\pdv{y}{t}}$

[jolly parrot]

haseeb ♥

[rotund narwhal]

How is it just chain rule if ur adding?

Sorry didn’t mean to use hyperbole

Of saying just

Because ur mostly multiplying

But there is addition too

[ashen ivy]

the essential idea is, the chain rule can be thought of as a dot product between two vectors

[rotund narwhal]

wtf i didnt know that

when do u learn that lol

[ashen ivy]

if you take multivariate calc on its own, these ideas are explored more thoroughly

i think for your course, they want you to be comfortable with the formula

and this tree system is one way of remembering it

[rotund narwhal]

yes i love the tree system

[ashen ivy]

but i'll state it: for $\vx(t) = \cv{x(t) \ y(t)},$ we have $\nabla (f \circ \vx) = \nabla f_{\vx} \cdot \nabla \vx$, where $\nabla g = \cv{\pdv{g}{x} ; \pdv{g}{y}}^T$ is the gradient vector

[jolly parrot]

haseeb ♥

[ashen ivy]

if you haven't seen this before, the gist is that you have a dot product of two vectors, and the vectors are derivatives of a bunch of different functions

$\nabla f(\vx)$ the partials of $f$ evaluated at a specific $\vx$, and $\nabla x$ the 'partials' of $\vx$ (which are just normal derivatives, bc single variable)

[jolly parrot]

haseeb ♥

[ashen ivy]

im just throwing word salad around, i think

[rotund narwhal]

a little bit

[rotund narwhal]

and i know u dont want to hear this

the only pattern within the tree that dignifies addition is the split correct?

i have calculated a gradient once i havent mess with partial gradients though

[ashen ivy]

every full path from the parent function to the variable

[ashen ivy]

the gradient = partial gradient

so if you have
$\begin{tikzcd}
&& {f(u,t)} & \
& {u(t,x)} && t \
x & t
\arrow[from=1-3, to=2-2]
\arrow[from=1-3, to=2-4]
\arrow[from=2-2, to=3-1]
\arrow[from=2-2, to=3-2]
\end{tikzcd}$

[jolly parrot]

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

[ashen ivy]

then you will have three paths, not two

$\pdv{f}{u} \pdv{u}{t} + \pdv{f}{u} \pdv{u}{x} + \pdv{f}{t}$

[jolly parrot]

haseeb ♥

[ashen ivy]

also, note here that $\pdv{y}{t} = \dv{y}{t}$, because $y(t)$ is single-variable

[jolly parrot]

haseeb ♥

[rotund narwhal]

okay okay

its starting to click

[ashen ivy]

ok fire

can you try the chain rule for this diagram?

$\begin{tikzcd}
& {f(x,y)} && \
{x(t)} && {y(t,s)} \
t && t & s \
&&& t
\arrow[from=1-2, to=2-1]
\arrow[from=1-2, to=2-3]
\arrow[from=2-1, to=3-1]
\arrow[from=2-3, to=3-3]
\arrow[from=2-3, to=3-4]
\arrow[from=3-4, to=4-4]
\end{tikzcd}$

[jolly parrot]

haseeb ♥

[rotund narwhal]

ngl i was gonna do the other ∂/∂t(f(∂f/∂y))

but im gonna do that because i need practice

[ashen ivy]

honestly just go for it

if it clicked then it clicked

[rotund narwhal]

lemme try ur example practice doesnt hurt

[rotund narwhal]

this is assuming f(x,y) isnt a already a partial dervative like the last one correct?

oh wait im wrong

no im right

man i wish u were here just to double check me

but im pretty sure im right

[pearl pond]

@rotund narwhal Has your question been resolved?

[pearl pond]

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[pearl pond]

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[ocean hazel]

<@&268886789983436800>

[pearl pond]

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[bronze glen]

hello i came across this statement "Partial substitution of the variable in the expression of the limits should be avoided" can anyone give an example of what partial substitution exactly means

[tropic saddle]

you only replace some occurences of that variable and leave others in

[bronze glen]

wait so 1s

ive solved a q

did i make that error there

1 min ill send an image

cuz i have no idea how i can solve it

otherwise

[safe prairie]

lim(x tends to 0) x/x,
and hypothetically ur next step only made the denominator 0, which made u falsely believe that everything would be +inf?

[bronze glen]

you would have to simplify it cuz its an indeterminate form right

[bronze glen]

sorry for the trash image quality btw

[light helm]

here product law
and both limits converge so there isn't an issue

[bronze glen]

didnt get it, because like i substituted value for sinx/x but i left out the sinx part in the denominator tho

[light helm]

lim as x to 0 of f(x)g(x)
= (lim as x to 0 of f(x)) * ( lim as x to 0 of g(x))
if both limits exist

[safe prairie]

yes so there’s a risk if u dont avoid ur concerns

[bronze glen]

oh

so we assumed that the limit of the remaining part also exists?

or like we knew it exists

so here substituting was okay?

[light helm]

you can see if it exists from evaluating

[bronze glen]

okay so if it exists then partial substituting is allowed?

[light helm]

an example where partial eval stuffs up. You should recognise this to be a definition of $e$
$$\lim_{n \to \infty} \br{1+\red{\frac 1n}}^n$
if you first eval the 1/n to 0, that'll lead to
$$\lim_{n \to \infty} 1^n = 1$$
(which is the incorrect)

[jolly parrot]

ραμOmeganato5
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[safe prairie]

product yes division no as shown by how 0/0 is invalid

and within product law you check for whether thats ok

[bronze glen]

ohh

okay

1 little q more

Can you just tell me where I can start

in this q

[static quartz]

L'Hopital rule

[bronze glen]

we havent done it yet so i cant use it in qs

[static quartz]

then use trigonometric identities in the numerator

[light helm]

harmonic trig / aux form

[static quartz]

to simplify it

[bronze glen]

whats aux form

[bronze glen]

okay

[light helm]

auxillary angle

[bronze glen]

idk that either

ig ill try trigonometric identities

[bronze glen]

i reached a dead end

So what next

i dont think it can be simplified ahead

[light helm]

sub u = alpha - pi/4

[bronze glen]

didnt understand

[light helm]

make a substitution

[bronze glen]

oh

and u tends to 0?

[light helm]

Yes

[bronze glen]

ohh

so its root 2

thanks so much man, really appreciate your help

ill close this

.close

[pearl pond]

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[pearl pond]

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[winged pulsar]

Apparently it’s “1”… and I’m getting negative square root of 3

[fresh solar]

You are correct

[split void]

its defo not 1 so the answer key is wrong.

[maiden kernel]

it is very much negative sqrt(3)

[pearl pond]

@winged pulsar Has your question been resolved?

[winged pulsar]

And is the writing correct?would I get full pts on an exam?

[fresh solar]

Eh that depends on your teacher and marking scheme although I think it's sufficient lol

[pearl pond]

@winged pulsar Has your question been resolved?

[winged pulsar]

But it makes sense?

[split void]

yes.

[pearl pond]

@winged pulsar Has your question been resolved?

[pearl pond]

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[jolly parrot]

Renato

[honest oyster]

So what have you tried?

[stoic imp]

I am not sure if I need to use weak or strong induction

what I did was to prove that the base case n = 2 holds

a2 = 2! = 2

[honest oyster]

Uh weak is fine for this case

[stoic imp]

why

[honest oyster]

Cause it only depends on the one before it

Either induction is fine

[stoic imp]

when to use strong induction vs when to use weak induction

@honest oyster

[honest oyster]

You can always use strong induction
You can only use weak induction for problems where the k+1 term only depends on the k term

[stoic imp]

why not use strong induction all the time then and ditch weak induction forever

[honest oyster]

No one's stopping you from doing that

[stoic imp]

can you help me prove this shit using strong induction

@honest oyster

[honest oyster]

The proof is pretty much the same for strong or weak induction

Just substitute a(n)=S into the recurrence formula and do some algebra manipulation

S here being what we're proving it equals cause I'm too lazy to write that out

[stoic imp]

can you help me with the proof structure aswell

[summer imp]

(Remark: weak and strong induction are logically equivalent. In a case like this, where your "next step" depends solely on the previous one and no other, it's just neater in the induction step to use weak induction and write

Suppose that a_n = n! ..... We will show that a_(n+1) = (n+1)! ...
rather than
Suppose that for all 2 <= m <= n, a_m = m! .... We will show that a_(n+1) = (n+1)!...

The latter is overkill since you're never going to need to call for other than the previous statement. It's just a bunch of words and variables added in for nothing. The argument is the exact same.)

You need a base case either way, so you should start with that.

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

can you help me with proving this using strong induction\

[summer imp]

Is there a specific reason why you absolutely want to use strong induction? Again it's just adding variables and words that obscure what you're doing since you only need to refer to the previous step in your problem.

But yes you can do it with strong induction if you so wish.
What is your base case?

[stoic imp]

because I cant distinguish when to use something or the other

as far as I am concerned everytime they ask me to prove something using induction and I see a recurrence relation or subsequences I just do strong induction

@summer imp

[summer imp]

The recursive definition of a_(n+1) refers only to the previous case a_n. If there was a_{n-1}, a_{n-2} or something else in there yes, strong induction would make arguing easier probably.

You use strong induction to facilitate problems which you can define recursively in terms of more than just the previous step.

But ok use strong induction

[stoic imp]

can we use strong induction here aswell?

[summer imp]

They're equivalent, you can always use either.

[stoic imp]

why use weak induction if you can just use strong induction

[summer imp]

Because it's just a neater statement.

[stoic imp]

care to elaborate?

[summer imp]

It's just simpler and clearer to use weak induction when you don't need strong induction since you're not introducing unnecessary variables.

[stoic imp]

can you give a concrete example that is easier

because I am the opposite I think I would prefer to do strong induction everytime forever and ever

@summer imp

[summer imp]

It's a matter of simplicity and elegance. It also makes it easier in the sense that there's only one way you can apply your induction hypothesis (as opposed to strong induction where there's multiple ways you may be able to split the problem in smaller parts).\

If you take the statement that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$ for instance, you have :\
\textbf{Base case}\$\sum_{i=1}^1 i = 1 = \frac{1\cdot 2}{2}$. \
\textbf{Induction step (for weak induction)}\ Suppose that $\sum_{i=1}^k i = \frac{k(k+1)}{2}$. Then $\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^k i\right ) + k+1 = \frac{k(k+1)}{2} + k+1 = \frac{k(k+1)}{2} + \frac{2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$ \qed.\

\textbf{Induction step (for strong induction)}\ Suppose that for every $1 \le m \le k$, $\sum_{i=1}^m i = \frac{m(m+1)}{2}$. Then $\sum_{i=1}^{k+1} i = \left(\sum_{i=1}^k i \right) + k+1 = \frac{k(k+1)}{2} + k+1 = \frac{k(k+1)}{2} + \frac{2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$ \qed.

[jolly parrot]

Azyrashacorki

[summer imp]

The argument is the same, you're just introducing extra stuff you don't need.

[stoic imp]

seems more elegant to me the strong induction case

[summer imp]

It's the exact same argument.

The only difference is adding more stuff at the beginning.

It's less concise with strong induction, and if you can do it with less assumptions (even though it turns out they're equivalent), then why bother?

[stoic imp]

can you use weak induction with recurrence sequences or subsequences?

[summer imp]

It depends on whether they are defined recursively in terms of the previous case only or in terms of other cases.

[stoic imp]

if you have a recurrence sequence that depends on the previous two terms