help-49

[grim vector]

Youre welcome

[lyric scroll]

now i gotta learn tangental lines

on e functions

and then im gonna sleep

[midnight plank]

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[last slate]

hello

[last slate]

im doing some practice tests for an examination tomorrow

The volume of a cube is increasing at a rate of 50cm^2
/sec. Find the rate of increase of the
length of a side when the volume is 125cm^3
(i believe some of these units aren't correct? because volume would be cm^3 for instance, anyway)

im stuck on actually finding ds/dt

[lavish venture]

cm^2/sec

🤔

[last slate]

i have ds/dt = dv/dt x ds/dv

which is, correct, i think

[lavish venture]

just

[last slate]

what im stuck on is finding ds/dv

[lavish venture]

what’s the volume of a cube

in terms of s

[last slate]

s^3?

[lavish venture]

yep

now differentiate dt

[last slate]

What?

[lavish venture]

V = s^3

V and s are both functions of t

[last slate]

Okay

so i'd get 3s^2

that's just dv/dt right?

[lavish venture]

not quite you’re forgetting that s is also a function of t

[last slate]

so what did i just get in terms of labeling

i differentiated v

[lavish venture]

incorrectly

have you done implicit differentiation

it’s like that

[lavish venture]

^^

[last slate]

oh

i did implicit a while ago

[lavish venture]

try differentiating again

[last slate]

differentiating v twice gives v'' which would be 6s

[lavish venture]

nono

i meant

differentiate V again

because your answer was wrong

[last slate]

um okay so

we would differentiate both sides

dv/dt and d/dt (s^3)

[lavish venture]

yep

what does that simplify to

[last slate]

you would have dv/dt = 3s^2?

[lavish venture]

nope

[graceful cradle]

@pokieswar

[lavish venture]

lemme ask you a question

why did you get dv/dt instead of 1

[last slate]

we're differentiating implicitly

[lavish venture]

it’s because V is a function of t so we need to multiply by dv/dt

but so is s

s is a function of t

so it’s not just 3s^2

[last slate]

so its 3s^2 ds/dt?

[lavish venture]

yep

[last slate]

dv/dt = 3s^2 ds/dt

i see

Okay

so from here now

what do we have to do to find ds/dt?

[lavish venture]

so we know dv/dt is 50

and V = 125

what would s be

if V = 125

[last slate]

s would be

3rd root v

3rd rt 125?

[lavish venture]

which is

[last slate]

5

[lavish venture]

then what’s 3s^2

[last slate]

3(5)^2
3 * 25
75?

[lavish venture]

so solve for ds/dt

[lavish venture]

here

[last slate]

uhh

[last slate]

50 = 75 ds/dt

so

50/75 = ds/dt

which is 2/3?

i see

so the rate of increase of a side

is 2/3 cm?

[lavish venture]

/sec

[last slate]

ah yes

Okay thank you let me reread this and write this down

[lavish venture]

you’re welcome

[midnight plank]

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[oak slate]

hello

[oak slate]

$$
\begin{aligned}
&\int_0^1 \mathbf{1}{a + b > p + q \text{ and } ab < pq}(a) , da \
= &\int_0^1 \mathbf{1}{a > p + q - b \text{ and } a < \frac{pq}{b}}(a) , da \
= &\int_0^1 \mathbf{1}{p + q - b < a < \frac{pq}{b}}(a) , da \
= &\int_0^1 \mathbf{1}{a \in (p + q - b, \frac{pq}{b})}(a) , da \
= &\int_0^1
\begin{cases}
1, & \text{ if } a \in (p + q - b, \frac{pq}{b}) \
0, & \text{ otherwise}
\end{cases}
, da
\end{aligned}
$$

[grand pond]

kaue

[oak slate]

all variables are in the interval (0, 1)

unsure how to proceed

i can't just take the length of (p + q - b, pq/b) because p + q - b might go below 0 and pq/b might go above 1

the bold one is the indicator function

[midnight plank]

@oak slate Has your question been resolved?

[oak slate]

<@&286206848099549185> anyone?

[surreal moon]

Yikes. Maybe one of the advanced math channels. #get-advanced-access

What is that? Measure theory?

[oak slate]

it's part of a probability problem i'm trying to solve that have continuous variables

[oak slate]

i will try that, ty

.close

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[last slate]

i kinda

[last slate]

don't know what they want

i thought it would just be

f(b) - f(a)

[tribal temple]

Does this depend on x?

[last slate]

oh

is it just

0

i hate math so much 😭

thank you

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[twilit field]

[
\quad \text{If } n \in \mathbb{N}, \text{ then } \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{4} \right) \left( 1 - \frac{1}{8} \right) \left( 1 - \frac{1}{16} \right) \cdots \left( 1 - \frac{1}{2^n} \right) \geq \frac{1}{4} + \frac{1}{2^{n+1}}.
]

[grand pond]

A dense set(Ping when reply)

[twilit field]

I;m trying to pove this via induction

[spiral osprey]

what have you tried

[twilit field]

The only thing I can think of is multiplying both sides by(1-1/2^{n+1})$

but I doubt that will help

so I eas instead wondering if comparing both to someother function would help

[last slate]

A

[twilit field]

Well, on multiplying both sides by $(1- \frac{1}{2^{n+1}}$ we get $\prod_{i=1}^{n+1} (1- \frac{1}{2^i}} \geq (\frac{1}{4} + \frac{1}{2^{n+1}}))(1- \frac{1}{2^{n+1})$

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

Well, on multiplying both sides by $\left( 1 - \frac{1}{2^{n+1}} \right)$, we get
$[
\prod_{i=1}^{n+1} \left( 1 - \frac{1}{2^i} \right) \geq \left( \frac{1}{4} + \frac{1}{2^{n+1}} \right) \left( 1 - \frac{1}{2^{n+1}} \right).
]$

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

oh

got ti

thanks

.close

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[twilit field]

oops

wrong problem

So I started by adding $\frac{1}{2^{n+1}}$ to both sides

[grand pond]

A dense set(Ping when reply)

[twilit field]

so I now just need to compare 1+n/2 + 1/(2^{n+1}) and 1+{n+1}/2

the right is greater than the left for all n>1

we're done

[modern shore]

im sorry can u like

[modern shore]

do this

[modern shore]

with this

[twilit field]

okay

sure

$\sum_{i=1}^{n+1} \frac{1}{2^i} \geq 1+ \frac{n}{2} + \frac{1}{2^{n+1}}$

[grand pond]

A dense set(Ping when reply)

[spiral osprey]

are you trying to prove this?

[twilit field]

We now compare the right hand side to $1+ \frac{n+1}{2}$

[twilit field]

ye, by induction

[grand pond]

A dense set(Ping when reply)

[spiral osprey]

seems like you're using circular reasoning

[twilit field]

now

*how

[spiral osprey]

you're assuming the statment is true when you have to prove it

[twilit field]

Well, using induction, it's fine until here

[spiral osprey]

oh wwiat i see what you're doing

[twilit field]

So is it fine?

[spiral osprey]

usually you want to work with only one side of the inequality

[spiral osprey]

i'd argue that the next term of this should have a +1 outside of the exponent

[twilit field]

hmm?

[spiral osprey]

$\frac{1}{2^n+1}$

[grand pond]

dingus

[twilit field]

oh

right

oops

😭

makes it much more messy

[twilit field]

$\sum_{i=1}^{2^n+1} \frac{1}{i} \geq 1+ \frac{n}{2} + \frac{1}{2^{n}+1}$

[spiral osprey]

your left hand side changes as well

[twilit field]

ah yes

[grand pond]

A dense set(Ping when reply)

[twilit field]

So we have to compare this and 1+(n+1)/2

We want to determine if $1+ \frac{n}{2} + \frac{1}{2^n+1} \leq 1+ \frac{n+1}{2}$ os true

[grand pond]

A dense set(Ping when reply)

[twilit field]

we can cancel out the 1 and n/2

to get $\frac{1}{2^{n}+1}<1/2$

[grand pond]

A dense set(Ping when reply)

[twilit field]

which is always true

so we're done

[midnight plank]

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[dire thorn]

I'm not completely sure what went wrong, can someone help

[keen bear]

hello

[dire thorn]

hi

[zealous schooner]

that limit isn't 0

it's infinity

[dire thorn]

oh shoot

i put cos(pi/2) = 1 💀

[ancient plank]

Put tan = sin/cos and then take cos = t

[dire thorn]

ight got it

thanks guys

that was pretty dumb 💀

.close

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[oblique prawn]

how to start 10d

[lethal path]

!status

[midnight plank]

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

[oblique prawn]

2, i tried subbing in the result from (c) and i dont think i got anywhere

then i considered z^6+z^3+1 expansion

but i just get costhetas and cos2pi/9 etc which wouldnt help

[lethal path]

try considering the expression $8 \sin(\pi/9) \cos(\pi/9) \cos(2 \pi/9) \cos(4 \pi/9) $

and see what it simplifies to

[grand pond]

south, just south

[midnight plank]

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[burnt torrent]

Hi, I have a question, I started studying math this year and my logic professor is not so good and there are not really many things shared that I can learn from. I have first logic exam next weekend and I need some book or pdf that I can find some exercises with answers. Any suggestions?

[burnt torrent]

The exam will cover everything from basics to set profs like this

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[merry garnet]

If Francine pays Php5000 monthly to her savings account, which is compounded quarterly for 0.25%, then what's the value of p?

[merry garnet]

.close

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[merry garnet]

Leo saves ₱ 7,500 every 3 months in a bank that pays 2 % compounded monthly. How much will be his savings after 12 years?

[peak ridge]

Compound interest foemila

Should be this

I havent done financial math in a while 😭

[merry garnet]

ok thanks

.close

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[last slate]

damn

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[last slate]

.

HI

[last slate]

I HAVE A QUESTION

IN TRIANGLE ABC, MEDIANS BD AND CE ARE PERPENDICULAR TO EACH OTHER AT POINT P. IF BD = 18, CE = 15, THEN FIND AR(AEPD)/10

KINDLY HELP

IS THERE ANY CONSTRUCTION?

<@&286206848099549185>

[spring wave]

you can consruct on ggb

[last slate]

WHAT

?

GGB?

[spring wave]

geogebra

[last slate]

WDYM

CONSTRUCT WHAT?

[spring wave]

u r asking for a construction of this traingle right?

[last slate]

NO

I DREW THIS MYSELF FROM THE QUESTION

[spring wave]

?

[last slate]

I AM ASKING WHETHER THIS QUESTION REQUIRES ADDITIONAL CONSTRUCTIONS FOR IT TO BE SOLVED

[spring wave]

oh

[last slate]

I'LL COME BACK 10 MINUTES LATER

YOU CAN DROP MESSAGES

[spring wave]

if you make the third median the problem becomes easy

[midnight plank]

@last slate Has your question been resolved?

[last slate]

.CLOSE

.close

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[keen shadow]

what do i do when i am arranging terms, and theres a term without a + or - ? (that sounds really confusing, i know)

[last slate]

do you mean as in how 54 doesnt have a plus or minus at the front

[keen shadow]

yup

[last slate]

ok so

its still a positive number

so its a plus

[junior flower]

hi rosy ily

[last slate]

hii laylaaa

ilyt

[keen shadow]

yeah, but like do i add or subtract 54?

D:

[last slate]

add

[keen shadow]

ah.

kay then :3

.close

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[last slate]

its easy if a value doesnt have a sign before it, its a positive, if it has a negative value, then when you move it you move it with the negative sign

ehoops channel closed

nvm

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[naive flint]

why would this not be

25 x 1mol/180.16

= to 0.1387

like if they want it in moles thats the only conversion i can think of tho

unless u add

ill try something actually brb

ya ok it didnt work

[sage helm]

Where's that coming from?

I think you should try and find the weight of the oxygen in 25 g of glucose

[naive flint]

well

what i think now is

I should take the 25 grams X 1mol/180.16

the molar mass

THen we have 6 carbon atoms

I show my work

or should i do this

one sec

[naive flint]

i feel like this set up makes more sense tho cuz i can cancel out

[plain osprey]

any idea on how to find the number of oxygen atoms in 25 grams of glucose?

[naive flint]

not rlly

I just kind of do dimesional analysis

dont rlly understand the concept

[plain osprey]

i mean tbf

are u allowed to look for the chemical formula of glucose

[naive flint]

o

yes

[plain osprey]

then find its molar mass

[naive flint]

ok 180.16

[plain osprey]

oh u already did

u know how that the molar mass × the moles = the mass?

or are u not familiar with this formula

[naive flint]

oh

no

= to mass of what

glucose?

[plain osprey]

yes

[naive flint]

oh

ok

[plain osprey]

well u can use it to find the moles of glucose

ur given 25 grams

as the mass

and the molar mass

[plain osprey]

is this

[naive flint]

Yep

[plain osprey]

now

you know the formula moles = number of molecules/avogadros number?

[naive flint]

yep

[plain osprey]

u can use it to find number of molecules

ur given avogadros number

which is 6.022 × 10²³

maybe 6.03, depending where u live

and moles of glucose u have it

[naive flint]

Yes

[plain osprey]

now each molecule of glucose

has 6 oxygens

[plain osprey]

given this chemical formula

[naive flint]

Yeah

[plain osprey]

so 6 × number of molecules shluld be the total number of oxygens right?

[naive flint]

yeah

times by 6 at the end

[plain osprey]

yes

thats how u get number of oxygens in 25 grams of glucose

now whats the amount of moles?

[naive flint]

is

0.1387?

[plain osprey]

no i meant like

formula speaking

what would u do after finding number of oxygen atoms in 25 grams of glucose

u would divide it by avogadros number, bud

to get the amount of moles

[naive flint]

oh

[plain osprey]

i didnt do the numeric activity for this

didnt do calculations

so i cant say if im correct or not

[naive flint]

oh Ok

[plain osprey]

but by formulas, i think its valid

thats all i can show

[midnight plank]

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[dense bay]

i determined that the determinant J(u,v)=-2u. but i couldnt figure out how to prove c):

[dense bay]

afaik
$$f_{XY}(x,y) = f_{U,V}(u,v) \abs{J(u,v)}$$
$$f_{XY}(x,y) = f_{U,V}(u,v) \cdot 2u$$

[grand pond]

jack

[dense bay]

but the question stated that
$$f_{XY}(x,y) = f_{U,V}(u,v) / 2u$$???

[grand pond]

jack

[patent fractal]

i think the jacobian should be on the other side

because $x = g_1(u,v)$, and $y=g_2(u,v)$

[grand pond]

rain

[patent fractal]

so they are the change of coordinate

[dense bay]

ah ty, were you referring to this formula?

(im having a hard time finding the formula i'm supposed to be using)

[patent fractal]

basically yeah

but i abstract away the integral

[midnight plank]

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[merry garnet]

Pi would like to deposit ₱2,000 every 6 months in a fund that earns 2% compounded semi-annually for his college studies. Is this condition a simple annuity or a general annuity? Why?

It is a simple annuity because m = 2 and p = 2

It is a simple annuity because m = 6 and p = 6

It is a general annuity because m = 6 and p = 2

It is a general annuity because m = 2 and p = 6

[merry garnet]

<@&286206848099549185>

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[merry garnet]

is it m=2 p=2

[midnight plank]

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[merry garnet]

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[jolly roost]

im a bit confused

[jolly roost]

the answer key for e) says

can someone explain how this is true?

[fallow scarab]

Multiply both sides by -1

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[last slate]

I’m confused is this C?

[oak slate]

what are you confused about

[last slate]

I’m confused on how I’m supposed to do to find which fence needs more fencing

[gaunt nimbus]

perimeter

[oak slate]

the fence will be placed on the boundaries of the shapes, so you need to find the perimeter of each one.

whoever has the biggest perimeter needs the most fence

[last slate]

So A needs more fencing?

How do I find the perimeter

[oak slate]

the perimeters are the lengths of these lines

for the first one you might want to use the pythegorean theorem to find the length of the curved lines

[last slate]

So

15^2+5^2=c^2?

[oak slate]

where

[last slate]

I counted the units

For A

[last slate]

Oh I flipped

[soft cave]

Yes

Similarly for the other side

[last slate]

Wdym

[oak slate]

you want to find the length of the sides

you can count the units of height and base of these triangles and use pythagoras to find the length of the side of the shape

[last slate]

Is a^2 15?

[last slate]

I’m sorry I’m still confused

[oak slate]

how about you start with the second one

how would you find the perimeter of B

[last slate]

I would add up the sides

10 and 6

[oak slate]

how did you get 10 and 6?

[soft cave]

They are 11 and 7 but you got them from the right place

[last slate]

Oh I didn’t count the right angles

Yea it’s 11 and 7

[oak slate]

and if you know the sides are 11 and 7, how do you find the perimeter?

[last slate]

Wouldn’t you add each side so like 11+11+7+7=36

[soft cave]

Exactly

[oak slate]

yes

[last slate]

So now what do I do with it

[soft cave]

That’s your answer for the perimeter of B

Now we need the perimeter of A, we were working on

[last slate]

Oh

[oak slate]

how many units are there?

assume its a straight line

[last slate]

15

[soft cave]

Yes, what about that bottom part

[oak slate]

what about here?

[last slate]

4

[oak slate]

sorry bad drawing

[last slate]

3

[soft cave]

Yes

[last slate]

So do I just add now?

152+32=36

Woah

[soft cave]

It is also important to know the bottom left angle is right

[oak slate]

where did you find 152?

[last slate]

I tried to do this sign *

But it made it weird

I was trying to multiply 15 and the 3

By 2

[oak slate]

remember neither 15 nor 3 are sides of the shape

[last slate]

Oh

[oak slate]

the green side is a side of the shape

so how would you find the green part if you know those lengths are 15 and 3?

[last slate]

Subtract?

[oak slate]

do you know the pythegean theorem?

[last slate]

Yea

[oak slate]

what does it say

[last slate]

A^2+B^2=C^2

[oak slate]

and what do A,B,C represent?

[last slate]

A is the 15 B is the 3 and C is the green line I’m assuming

[oak slate]

yep

so you can sub the values in the equation and solve for C

[last slate]

So 234=C^2?

[oak slate]

yeah

but you need an extra step

[last slate]

Divide?

[oak slate]

you need to √ both sides

[last slate]

3 radical 26

[oak slate]

yes

you found one side

now you need to find the other

[last slate]

How would do it for the variable

[oak slate]

wym

[last slate]

For c^2

[oak slate]

√ is the inverse of squaring, they undo each other

√(x²) = x

(assuming x positive)

so you just get C

which is what you're looking for

that's why you took the square root

[last slate]

??

So

Would it just equal C

[last slate]

Oh

So 3 radical 26=C

[oak slate]

yeah

now you find the other side

[oak slate]

the one on the green triangle

[last slate]

.close

[midnight plank]

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[viral dagger]

find all real (x,y,z) such that $\max(x^2+yz, y^2+zx, z^2+xy)=2\left(\frac{x+y+z}{3}\right)^2$

[grand pond]

Skissue ping4response

[viral dagger]

(1,1,1) works

[zealous schooner]

WLOG assume x>=y>=z, then which of the terms is the largest on the left?

hmm actually wait

you'll need to consider a few cases I think

[feral sedge]

LOG? as you wish…

log(x) >= log(y) >= log(z)

[zealous schooner]

thanks Mqnic

very helpful

[feral sedge]

:D

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

i asked someone at the comp and they found (a,a,a) works for all a which is wtf??

[spring wave]

RHS = 2/9 * (x+y+z)^2
= 2/9 * ((x^2 +yz)+(y^2+zx)+(z^2+xy) + (xy+yz+zx))

[viral dagger]

uh

[spring wave]

<= 2/9 * ((x^2 +yz)+(y^2+zx)+(z^2+xy) + x^2 + y^2 + z^2)

[spring wave]

<= 2/9 * ((x^2 +yz)+(y^2+zx)+(z^2+xy) + (x^2 + xy)/2 + (y^2+ zx)/2 + (z^2+xy)/2)

equality condition being x = y = z

[spring wave]

= ((x^2+yz)+(y^2+zx)+(z^2+xy) )/ 3

let
x^2 + yz = a
y^2 + zx = b
z^2 + xy = c

that means maximum of (a,b,c) is smaller than or equal to its mean

that only happens when a=b=c, and it meets equality condition

so the set of solutions is x = y = z

This was a nice qn

am using xy + yz + zx <= x^2 + y^2 + z^2

we can do this as xy <= (x^2+y^2)/2

@viral dagger does this make sense?

[viral dagger]

how is x^2+y^2+z^2<=(x^2+xy)/2+(y^2+zx)/2+(z^2+xy)/2

oh wait

oo thats clever

thanks! that made alot of sense

.close

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[feral sedge]

Note: x = y = z satisfies

Claim: No other solutions

Suppose for contradiction, at least two of x, y, z differ.

Let S = (x^2 + yz) + (y^2 + zx) + (z^2 + xy)
Let s = x + y + z, m = s/3
Let a = x - m, b = y - m, c = z - m
(By definition, a + b + c = 0)

Note:
(a + b + c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ca)
(ab + bc + ca) = -1/2 * (a^2 + b^2 + c^2)

Note:
(x^2 + yz) = 2m^2 + am + a^2 + bc
(y^2 + zx) = 2m^2 + bm + b^2 + ca
(z^2 + xy) = 2m^2 + cm + c^2 + ab

Summing:
S = 6m^2 + (a + b + c)m + (a^2 + b^2 + c^2) + (ab + bc + ca)
= 6m^2 + 1/2 * (a^2 + b^2 + c^2)

Thus:
S/3 = 2m^2 + 1/6 * (a^2 + b^2 + c^2)

Since maximum must be as large as average:
max(x^2 + yz, y^2 + zx, z^2 + xy) >= 2m^2 + 1/6 * (a^2 + b^2 + c^2)
> 2m^2 = 2((x + y + z)/3)^2

Hence:
LHS > RHS. Contradiction.
x = y = z.

(alternative but longer way, i think this is correct)

[viral dagger]

uhh

.reopen

[midnight plank]

✅

[viral dagger]

hoq did you get the x^2+yz=2m^2+am+a^2+bc

[spring wave]

he substituted x = a + m y = b + m and z= c + m

[feral sedge]

(x^2 + yz) = (m + a)^2 + (m + b)(m + c)
= (m^2 + 2am + a^2) + (m^2 + (b + c)m + bc)
= 2m^2 + am + (a + b + c)m + a^2 + bc
= 2m^2 + am + a^2 + bc

[viral dagger]

oo thanks

.close

[midnight plank]

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[viral dagger]

personally i feel like the first solution was more possible for me to find in a competition, but thanks both of you guys!

[nova pike]

and also mind I ask what competition is this?

[viral dagger]

lmao

its small0

[nova pike]

Oh alright thanks anyway!

[spring wave]

welcome

[nova pike]

mb, I was replying to skissue but your solution is good!

[midnight plank]

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[last slate]

if medians BD and CE are perpendicular to each other, find the area of the triangle ABC if BD = x and CE = y

[midnight plank]

@last slate Has your question been resolved?

[last slate]

bro what

Not solved yet

<@&286206848099549185>

[spring wave]

did you try constructing the third median?

the answer should be 6*(x/3)*(2y/3)*1/2

or 2xy/3

here is a diagram, sry that the labels are different

you see why each of these subtriangle have the same area?

[spring wave]

@last slate

[last slate]

y?

[spring wave]

call the centroid M

yk area of trinagle is 1/2 * base * height ?

[last slate]

yes

Ok?

[spring wave]

so in this diagram GMK = KMI ?

[last slate]

oh yeah

[spring wave]

And now GHK = KHI

so GMH = HMI

[last slate]

yes

[spring wave]

and so HMJ = JMI = HML = LMG

[last slate]

oh got it

[spring wave]

yk that medians divide each other in 2:1 ratio?

[last slate]

yup

[spring wave]

so you can figure out the area of GMK

and then multiply it by 6

[last slate]

oh yes

then we get the area

great

gg

thx

.cloase

.close

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[twilit field]

I 'm trying to prove $\sqrt[3] {2} \notin \Q$

[grand pond]

A dense set(Ping when reply)

[last slate]

Q = rational?

[twilit field]

yes

[last slate]

ok

[twilit field]

We proceed by contradiction, Let $\frac{p}{q} = \sqrt[3]{2}$
\
Then $p^3=2q^3$,so $p$ is even.
\
We then have $8k^3=2q^3$, which tells us $q$ is even too, which make p,q not co=prime, thus our assumuption must be wrong and $\sqrt[3] {2}$ isn't rational

[last slate]

do we just have to show that (p,q) ≠ 1?

[twilit field]

(p,q)?

[last slate]

GCD

[twilit field]

We assume they are co-prime , so yeah

[last slate]

then it's easy

[grand pond]

A dense set(Ping when reply)

[last slate]

but what does p,q not being coprime have to do with that?

[slow sand]

seems correct

@twilit field is that all?

[last slate]

we may say every rational number has a simplest form, but here it isn't possible so it must be irrational (since it can be either rational or irrational, not complex)

[slow sand]

Ig you can prove why x³ being even implies x is even

which is also fairly easy

[last slate]

yeah

[twilit field]

yeah, contrapositive would work like a charm

[slow sand]

uhh

you wanna close the channel now?

[last slate]

!DONE

[midnight plank]

If you are done with this channel, please mark your problem as solved by typing .close

[last slate]

as if you don't know

[twilit field]

I know

give me a few minutes to breathe

.close

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[last slate]

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[north hound]

The region bounded by the lines x = 1, y = 1, y = -1 and the curve x + y^2= 0 is rotated about the line x= 4 to form a solid. When the region is rotated, the line segment S at height y sweeps out an annulus.

a) Show that the area of the annulus at height y is equal to π (y^2+ 8y^2+ 7) (u^3).

b) Hence, find the volume of the solid.

[north hound]

can someone please help

.

[midnight plank]

@north hound Has your question been resolved?

[lethal path]

cause the question mentions annulus, it wants you to use the disk method

there must be some typos in π (y^4 + 8y^2+ 7) (u^3)

[north hound]

oh wait yeah its y^4 not y^2

[lethal path]

but essentially you are integrating $\pi (r(y))^2$

[grand pond]

southlander!

[lethal path]

note that you have to first find the volume of revolution of x + y^2 = 0
and then find the volume of revolution of x = 1

[north hound]

uh im really confused

[lethal path]

so subtract the two integrals of revolution

[lethal path]

how so?

[north hound]

like im unsure what the bounds for the integration should be

[lethal path]

oh they are y = -1 to y = 1

[north hound]

wait so i need to find the integration of x=1 and then subtract the integration of x+y^2=0

right

[lethal path]

it's the other way around

[lethal path]

and also you're forgetting the formula of the volume of revolution

so it would look like $\pi \int (r_2 (y))^2 \ dy - \pi \int (r_1 (y))^2 \ dy$

[grand pond]

southlander!

[north hound]

um sorry im still confused

which one is r1 and which one is r2

also i have to subtract k=4 right

from both integrations

[lethal path]

yes, you need to do that for both

[lethal path]

$r_2$ is the horizontal distance between $x + y^2 = 0$ and $x = 4$

$r_1$ is the horizontal distance between $x = 1$ and $x = 4$

[grand pond]

southlander!

[north hound]

wait i think i get it

thank you so much

[lethal path]

no worries!

you should get the exact same thing without the u^3

[north hound]

yep

thanks a lot

.close

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[odd lagoon]

I need help, I am really stuck trying to figure out what test to use

[odd lagoon]

it feels like every test is a dead end

I don't have the intuition for this stuff

The series is written like this:

sum x=1 to infinity 1/(f(x) + g(x))

and the only info I am given is that

f(x) is weaker than g(x) as x approaches infinity

so i thought it looks similar to a harmonic series

because i can ignore f(x) when we go to infinity

i basically only have to evaluate 1/(g(x))

but g(x) is not equal to x

so normally i would go the integral test

[twilit jetty]

do you have any info on g(x)

[odd lagoon]

no that's the thing, isnt this too little information?

i would just integrate it otherwise, right?

like swap x for something else

integrate it

if the integral of 1/g(x) was infinity it would diverge, or if the integral of 1/g(x) was a finite number it would converge

im just gonna answer that

.close

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[formal cloak]

is the working out for this correct? I though volume for a cone was 1/3 x l x r2

[muted panther]

No it is incorrect, they didn't divide by 3

[lethal path]

LMAOOOO

they calculated the volume of the cylinder by accident

[muted panther]

And by the way, you forgot to multiply by pi in your expression (in your message just above), also you should use ^ to write powers : (1/3)*pi*(r^2)*h

[midnight plank]

@formal cloak Has your question been resolved?

[formal cloak]

thank you guys

[midnight plank]

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[twilit field]

[
[-1,1]
]

[
[-1,1] \subseteq \bigcap_{i \in \mathbb{N}} A_i \quad \text{as} \quad [-1,1] \subseteq [-n,n] \quad \forall n \in \mathbb{N}.
]

[
\bigcap_{i \in \mathbb{N}} A_i \subseteq [-1,1] \quad \text{since we want } \quad [-i,i] \in \bigcap_{i \in \mathbb{N}} A_i \quad \forall i \in \mathbb{N}.
]

For (i = 1), we find ([-1,1] \in \bigcap_{i \in \mathbb{N}} A_i), thus (\bigcap_{i \in \mathbb{N}} A_i \subseteq [-1,1]). Therefore,

[
\bigcap_{i \in \mathbb{N}} A_i = [-1,1].
]

How does this look?

[exotic pelican]

that thing after the since makes no sense

[twilit field]

hmm

[grand pond]

A dense set(Ping when reply)

[exotic pelican]

$A_1 \cap \bigcap_{i \in \mathbb{N} \setminus {1}} A_i \subseteq A_1$

[grand pond]

rbit

[twilit field]

hmm?

why is what I said wrong

I'm saying all elements in the interection need to belong to all A_i

[exotic pelican]

because like, [-i, i] is not an element of what comes next

[runic hamlet]

take the proof I posted in the other channel and adapt it to this problem

[exotic pelican]

nor a subset

you probably meant $[-i, i] \in {A_1, A_2, ...}$

[twilit field]

yes

[grand pond]

rbit

[twilit field]

hmm

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

Yes

[twilit field]

So how do I prove reverse includion

We need [-i,i] \subseteq A_i \forall i \in \N, [-1,1] is one such set, thus $[-1,1] \subseteq \bigcap A_i$

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

I imagined more something along the lines of:

if $x\in\bigcap_{n\in\bN} A_n$, then it follows that $1/n\leq x\leq 2-1/n$ for all $n\in\bN$. In particular for $n=1$ we get $1\leq x\leq 1$, meaning that $x\in{1}$. This shows $\bigcap_{n\in\bN} A_n \subseteq {1}$.

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

ooh

If $x \in \bigcap_{i \in \N}A_i $, it then follows $-i \leq x \leq i, \froall i \i \N$. For n=1, $-1 \leq x \leq 1$,,thus $x\ in [-1,1]$ so $\bigcap_{i\in \N} A_i \subseteq [-1,1]$

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

$If ( x \in \bigcap_{i \in \mathbb{N}} A_i ), it then follows that ( -i \leq x \leq i, \ \forall i \in \mathbb{N} ). For ( n = 1 ), ( -1 \leq x \leq 1 ), thus ( x \in [-1,1] ). Therefore, ( \bigcap_{i \in \mathbb{N}} A_i \subseteq [-1,1] ).

[grand pond]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[midnight plank]

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[midnight plank]

@white prawn Has your question been resolved?

[midnight plank]

@white prawn Has your question been resolved?

[spring wave]

Idk energy flow diagram, but work energy thrm is just "initial energy of a system + work done on the system = final energy of the system"

[midnight plank]

@white prawn Has your question been resolved?

[midnight plank]

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[bitter fox]

i am getting this wrong and i dont get how

[bitter fox]

usually i use chat gpt to tutor me but i dont thinnk its working this time

[smoky solar]

what does 0 bar represent?

[bitter fox]

i have no idea what bar means in this context

srry

i posted all the info of the question

im pretty sure the referennce angle of something like -120 is 60 right?

you do 180-|angle| if its greater than 90 but less that 180

[smoky solar]

must be the supplementary sign

try submitting the answers in radians instead of degrees

cuz the main angle is in radians

[bitter fox]

OK

[bitter fox]

Thank you you are the best

-close

[smoky solar]

no problem

[bitter fox]

/close

[smoky solar]

dot

[bitter fox]

.close

[midnight plank]

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[midnight plank]

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[last slate]

Can someone explain?

[fallow scarab]

Have you heard of tip to tail

Just means B - A component wise

6-5 = 1

-1 - 1 = -2

[last slate]

0-2=-2

?

@fallow scarab so doesnt that mean i am right tho....?

[fallow scarab]

What

Where is your calculation contradicting anything else

[last slate]

nthn, just wanted to confirm if this is true

because if it is, then i am confused because i am struggling with

i decided that A is the point where the 90 degree angle is

so that

[fallow scarab]

Then what's there to explain

[last slate]

$\overrightarrow{AB}=\begin{pmatrix}1\-2\-2\end{pmatrix}$ and $\overrightarrow{AC}=\begin{pmatrix}-2\1\-2\end{pmatrix}$

[grand pond]

forMaths

[last slate]

but like this $\overrightarrow{AB}\cdot\overrightarrow{AC}\neq0$

[grand pond]

forMaths

[last slate]

when it is supposed to be considering we are trying to prove that it is a right angled triangle

wait no

it is

lol

ignore everything i said

.close

[midnight plank]

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[jolly roost]

Is the momentum conserved if an object is in projectile motion?

[midnight plank]

@jolly roost Has your question been resolved?

[peak herald]

what do you mean by this?

"is momentum conserved if-" yes.

[midnight plank]

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[keen harbor]

I wanted to ask as question regarding some symbols that contain a ' prime marking. I realized I don't know the terminology to use.

[keen harbor]

How would you refer to those vectors with the prime marking. "Primed vectors", "vectors with the prime symbol", etc.

I wanted to ask something like:
Why is it that the vectors in the third bullet point <are primed?>, just because their basis B' has a prime. Since they represent the same coefficients, shouldn't they be displayed as the same vectors -- meaning, like in bullet point 2, without primes?

[humble torrent]

A prime here is to denote something possibly different but in some sense related

It’s more of a physiological thing

[humble torrent]

If you read out loud the name of say the vector v, then v’ is simply said “v prime”, etc

[keen harbor]

Thanks for answering the first question.

Regarding the question itself,

[humble torrent]

How come?

They say B, B’ are different bases

[keen harbor]

I understand what the prime normally means. My point is that they are exactly the same; no different, and don't even need to be though of as something different. Having the prime is potentially more confusing that including the prime, since the coefficients are exactly the name. It is the resulting vector that is different.

[humble torrent]

Also those are scalars not vectors

In the coordinates

[humble torrent]

Nope

The coefficients need not be the same

If you have a different basis

[sharp coral]

we have that $v_1 \ne v_1'$ because they are different numbers. however they are similar in that they are both the first component of the vector $\vec v$ (just with respect to a different basis)

[grand pond]

cloud

[humble torrent]

Sure the linear combination is the same vector, but that doesn’t mean the scalars and basis vectors are the same, well since by assumption the basis vectors are different

[keen harbor]

The image says "to express the same vector". Further, the first bullet point says "v is a vector". B and B' are different, but the vector coefficients $(v_1, ..., v_n)$ are the same for bullet points #2 and #3.

[humble torrent]

Where are you getting that the scalars are the same?

[grand pond]

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

[keen harbor]

Because they are both from the same vector. It is the basis that is different, and thus the resulting vector.

[humble torrent]

Do you know what the notation means in the image?

I.e [v]_B?

[keen harbor]

A linear combination between the scalars/coefficients in vector v, with basis B.

[humble torrent]

Do u e.g. know what it means to express a vector in a basis?

[humble torrent]

No

The scalars in v depends on ur choice of basis

[sharp coral]

le'ts take a concrete example.
let's take our first basis to be the standard basis:
[ B = {(1,0), (0,1)} ]
Then we have the vector $\vec v = (3,4)$. then in the standard basis it is:
[ [\vec v]{B} = (v_1, v_2)B = (3,4)B = 3(1,0) + 4(0,1) ]
Then we have another basis:
[ B' = { (1,0), (-1,1) } ]
The vector wrt the second basis is:
[ [\vec v]{B'} = (v_1', v_2'){B'} = (7,4){B'} = 7(1,0) + 4(-1,1) ]

[humble torrent]

Thats what it means to talk about the coordinates of a vector

Since it by definition depends on ur choice of basis/coordinate system

[keen harbor]

That doesn't make sense to me

Why wouldn't it be:

3(1, 0) + 4(-1, 1)

[humble torrent]

You’re probably implicitly assuming the standard basis is somehow what determines the coordinates of ur vector

[sharp coral]

that was a typo

[grand pond]

cloud

[humble torrent]

Have u seen the geometrical interpretation of bases when describing vectors? How they’re essentially a coordinate system?

[keen harbor]

No, I don't agree with the last line: 7(1, 0) + 4(-1, 1) -- You are changing both the basis and the coefficients, when only the basis should change.

[humble torrent]

Why?

What is your definition

Consult your book if you don’t believe us

[sharp coral]

the components of a vector with respect to a basis are by definition the coefficients of a linear combination of the basis vectors

[keen harbor]

Let me look for some support material only. Give me a moment.

[humble torrent]

The whole point is to able to describe the same vector in different “coordinate systems”

So if you use the same coordinates, in every basis, then you’re basically blindly assuming that the same coordinates always points to the same destination even in different “measuring units”

[robust isle]

to give a more hands-on example, let's say my thumb is 1 inch long, what you're essentially saying is that units never matter, so ok my thumb is also 1 centimeter long now, it doesn't work like that
thumb = vector, inch and centimeters = 2 bases

[keen harbor]

I didn't find the simply equation I was looking for, by maybe this:

So, that's the value of the vector. With a1, ..., an as coefficients of v, and b1, ..., bn as basis.

[humble torrent]

The sum on the RHS should be equal to the LHS

Now is this true for your example?

[humble torrent]

This one

[sharp coral]

we would also write $\vb v = (a_1, ..., a_n)_B$ where $B$ is the given basis essentially as shorthand for that linear combination

[keen harbor]

In this case, we can change the basis B to be B'. Changing to B' shouldn't also change the values of a

[grand pond]

cloud

[humble torrent]

Are you aware of the implications of this?

[sharp coral]

but B' is a different set of vectors. why would the same coefficients applied to different vectors create the same result?

[keen harbor]

Yes, you can express the same set of coefficients as different coordinates using different basis vectors.

[humble torrent]

Read the practically of this, here and then above that aswell

[humble torrent]

What is the coordinates of a vector in a given basis?

[humble torrent]

In any case, you can check this won’t work by simply computing what you proposed

[keen harbor]

It is this: #help-49 message

[humble torrent]

This doesn’t explicitly answer my question

It is infact not defined in that image

[keen harbor]

That is precisely the answer to your question.

[humble torrent]

So tell me it

[keen harbor]

the resulting coordinate is v

[humble torrent]

?

[keen harbor]

It is a set of basis scaled by a set of coefficients.

[humble torrent]

What are the coordinates of a vector to you?

[keen harbor]

The coordinates of u, or v, or any vector, is the result of scaling a set of basis vectors by a set of scalars; as shown in the image I linked.

[humble torrent]

Possible confusion by using u for you there lmao

Oh god

Why does this read like a GPT response

[keen harbor]

What is it that you dissagree with the image that I linked? Are you saying that's not the coordinates of the vector?

[humble torrent]

Okay so you’re agreeing that the coefficients are the coordinates?

[keen harbor]

This is the page that contains that: https://en.wikipedia.org/wiki/Vector_space

No, they arent

[humble torrent]

Okay so you’re saying the “coordinates” is the vector?

As in the object

There’s literally a definition for coordinates btw

[keen harbor]

I stand corrected. The coefficients are the coordinates.

Now that I have been educated on the correct meaning of coordinates, I still don't understand why in the example show way earlier, there is:

[humble torrent]

Great that we have that agreed on atleast

Oh I see

[keen harbor]

#help-49 message

That these coordinates change from (3, 4) to (7, 4), just because there is a change of basis.

[sharp coral]

yes, coordinates are inherently dependent on a choice of basis

[humble torrent]

Well for starters compute them

And see what happens

As in compute the sum and see if they’re equal to the vector

[keen harbor]

Let's try avoiding rehashing the discussing we just had:

Are the "coordinates" the coefficients before being transformed by the basis or after?

I just agreed with you that coords are the coefficients before the transform.

[humble torrent]

The coordinates is what describes your vector as a tuple, because of the choice of basis

Before choosing a basis, there isn’t inherently coordinates for the vector

[keen harbor]

Lets go with what you're saying. What are the numbers in the tuple called before being transformed by the basis, such that the result is the vector?

[humble torrent]

A vector is a bit more abstract than just a list of numbers

Vectors are elements of a vector space

[sharp coral]

the coordinates are the coefficients of the linear combination in both cases.
so in the first case, the coordinates are (3,4)
because 3b1 + 4b2 = v
and in the second case, the coordinates are (7,4) because 7b1' + 4b2' = v
so in each case the coordinates come directly from the linear combination

[humble torrent]

Which satisfy a bunch of “axioms” or conditions

A vector could be a function, matrix, etc

Since they happen to satisfy the definition

[keen harbor]

oh, I see where you're comming from, now. In order to result in the same value in the resulting vector, if the basis is different, then the coefficients are/may need to be different. Is this a correct interpretation of what you're saying?

[humble torrent]

Inherently these don’t have coordinates

It’s once we decide on a basis we can relate them to tuples

[keen harbor]

If my interpretation of what cloud said is correct, then that perfectly answers the question of why the coordinates are primed.

I understand now.

Thank you, both, for the education and the hearty defense of your points. I appreciate it.

.close

[midnight plank]

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[drifting root]

Im doing test rv (this is a past test not cheating)

[drifting root]

this is all they gave me when going through ans

im so very stuck

[midnight plank]

@drifting root Has your question been resolved?

[midnight plank]

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[calm quest]

im a little confused on where to go next to prove that 2^n > x^n

[surreal moon]

what is x?

[calm quest]

oh im not sure i think its supposed to be n?

this is the question, it may be a typo

[surreal moon]

Yeah. I assume it meant n^2, not x^2

Okay. Throw x in the trash

do n instead

[calm quest]

yep thats what i did

i dont think i even noticed it said x lmao

[tribal temple]

(for the modified question, 2k^2 is k^2 + k^2, of course, and if you expanded out (k + 1)^2, you get k^2 + 2k + 1, so we at least have one k^2 that we want )

[calm quest]

ohhh

[tribal temple]

it might also help to notice that k^2 is k * k

[calm quest]

here where im at

i dont really know where else to go

[tribal temple]

Well remember how one of your assumptions is that n, and so k, is at least 5?

[calm quest]

yes

ohhh

ok let me have a look at that

[tribal temple]

Yep give it a go, and let us know how it goes

[calm quest]

i tried this and it worked but im not sure if i can do it this way

uh oh

[tribal temple]

It isn't entirely clear how you got here

[calm quest]

ohh ok so should i replace that with (k + 1)^2

[tribal temple]

It isn't that it's wrong or needs to be replaced (you can do that later), but it isn't justified enough imo to me why is k^2 + k^2 > k^2 + 2k + 1?

One of the suggestions why I mentioned that k^2 was k * k, is that if k >= 5, you could state from there that k^2 >= 5k

These don't imply the line we need

[calm quest]

i seee

[calm quest]

do you mind explaining what u mean here i am a little confused

[tribal temple]

Do you at least agree with the statement I said:

"[As] k^2 [is] k * k, is that if k >= 5, you could state from there that k^2 >= 5k"

[calm quest]

ohhhhhh wait do you mean like this:

[tribal temple]

Well, more that you can get to 2k^2 = k^2 + k^2 >= k^2 + 5k

You still need to keep one of the k^2, but that 5k, we can do a bit of work to it to get where we need, it's closer to the extra 2k + 1 we need

[calm quest]

ohhhhHHH

also does the inequality sign change from greater than to greater than or equal to?

[tribal temple]

Well, for now it's greater than or equal to, because we had k >= 5 as like our "starting point" (where that one is a "...or equal to")

We do have strict inequalities at other points though, so in the end the overall inequality is strict at least

[calm quest]

sorry for all the questions! but im still feeling a little stuck 😭

tried factorising the k but i dont think it gets me anywhere

[tribal temple]

(don't be sorry at all, after all, you're here to learn )

And, well, do you agree that 5 is bigger than 3?

[calm quest]

yep!

[burnt idol]

break 5k into 2k + 3k

[tribal temple]

Well, so you know 5k > 3k, right?

[burnt idol]

easier to visualise

[calm quest]

yes

[tribal temple]

Cool, now remember we needed a 2k, and a 1 as our end goal, do you have any ideas yet as to what you can do?

[tribal temple]

(disagree that it makes a significant difference imo, but anyways-)

[calm quest]

ohh so we can split up the 5k?

into 2k and 3k?

ohh and then we prove that 3k is greater than 1?

[tribal temple]

That works

[calm quest]

omgmg ok ill have a go

[tribal temple]

(this should be super easy to do! )

[calm quest]

is this better?