help-10

[fading quest]

because its not 50 mm^2 to cm^2, its 50 (something) PER mm^2 to (something) PER cm^2 that you need to convert

that "per" makes a bit difference

just like we have 50 kilometers PER hour, which makes the hour be in the denominator

or 3 meters PER second, making the second in the denominator

etc.

[fading quest]

yep thats correct

you can just put this into a calculator, but you really should be able to do this calculation in your head

dividing by 1/100th is the same as multiplying by 100

thats just how that works

and it is something that you NEED to know

and its the reason why this works

[true lily]

but I don't understand what is my next step after doing this?

[true lily]

to get to this

[fading quest]

divide 50 by 0.01

or write 0.01 as a fraction and then use the rule

[true lily]

5000

[fading quest]

$$50\frac{B}{0.01 cm^2}=50\frac{B}{\frac{1}{100}cm^2}=50\frac{B}{1}\div\frac{1}{100}cm^2=5000\frac{B}{cm^2}$$

[warm shale]

Duh Hello

[fading quest]

after that its just asking, what is higher, 5000 B/cm^2 or 4500 B/cm^2?

[true lily]

ah so the convertion is only in the denominator

okay thanks for the help I should have figured it out sooner

.close

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[nimble prawn]

I am unable to find the values of a,b and c in question 5. Kindly help

[royal basin]

$x^3 + ax^2 + bx + c = (x-1)(x-2)(x-3)$

[warm shale]

Ann

[nimble prawn]

Oh! Got it

Thank you

I’ll close the channel

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[timid silo]

is this write if I wanna find the principal argument of Z

[timid silo]

I don't know the right ans to the question

Sorry

I got the ans

Is this right ?

[obtuse pebble]

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[undone ember]

why is my gradient negative?

[obtuse pebble]

Please don't occupy multiple help channels.

[tranquil arch]

the right side should be -6x^2

[obtuse pebble]

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[timid silo]

How do I do this

[spice citrus]

use integrals

the average value on the interval is the area under the curve divided by the length of the interval

[timid silo]

The bounds are a to 0

That confuses me

[spice citrus]

because you have to find the area under to curve on the interval 0 to a, so the bounds of the integral are 0 and a

[timid silo]

Yeah but when you sub in a what happens

Do you like isolate for it

[spice citrus]

you solve the integral, just like you normally would

$\frac{1}{a}\int_{0}^{a}\left(3x-x^{2}\right)dx=0$

[warm shale]

Jelle

[spice citrus]

you need to find a such that this is true

[timid silo]

It becomes

a^2-a^2/3?

[spice citrus]

what is the antiderivative of 3x?

[timid silo]

It’s just x^3

[spice citrus]

3x, not 3x^2

[timid silo]

Wait it’s

3/2x^2

[spice citrus]

yes, and the antiderivative of -x^2?

[timid silo]

-x^3/3

[spice citrus]

yes

[timid silo]

Then plug in a

[spice citrus]

yes and 0, but that cancels

[timid silo]

You get (3a^2/2 -a^3/3)

[spice citrus]

yes, so $\frac{1}{a}\left(\frac{3a^{2}}{2}-\frac{a^{3}}{3}\right)=0$

[warm shale]

Jelle

[spice citrus]

notice that a can't be 0, because then you divide by a

[timid silo]

Right

[spice citrus]

now just solve the equation

[timid silo]

Like isolate for what

[spice citrus]

you can first multiply through by 1/a

[timid silo]

3a/2 - a^2/3?

[spice citrus]

yes

you can even divide by a, because we already know a = 0 should not be a solution

[timid silo]

Wdym

[spice citrus]

divide the entire equation by a

$\frac{3a}{2}-\frac{a^{2}}{3}=0$

[timid silo]

Factor ?

[warm shale]

Jelle

[timid silo]

Oh yea

I did that

[spice citrus]

factoring also works

[spice citrus]

you can divide by a once more

$\frac{3}{2}-\frac{a}{3}=0$

[warm shale]

Jelle

Jelle

[timid silo]

It’s 9/2

[spice citrus]

yes

that's the answer

[timid silo]

Thanks can I ask another q

[spice citrus]

yes

[timid silo]

How do I do this

[spice citrus]

averages are linear, so you can just subtract the averages

nvm

[timid silo]

Like 14-10…

[spice citrus]

okay so:

[warm shale]

Jelle

Jelle

[timid silo]

Ohh

[spice citrus]

what we want is:

[warm shale]

Jelle

[timid silo]

Right

[spice citrus]

you can multiply the first 2 integrals by 5 and 2 respectively and then you can subtract them

[timid silo]

Why

[warm shale]

Jelle

[timid silo]

I’m confused about why you multiply

[spice citrus]

to be able to subtract the integrals they can have different coefficients in front

$\int_{0}^{5}f\left(x\right)dx=50$

[warm shale]

Jelle

[timid silo]

TYSM

.close

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[rich aurora]

Can someone show the solution part pls

[white thunder]

Need help in number 9 guys

[rich aurora]

bro i did the same mistake XD, #❓how-to-get-help just enter there

[rich aurora]

is A=4?

and how to solve trapezoidal rule ,o, (how to know exact h, we just need to find y at x=0)?

[obtuse pebble]

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[rich aurora]

help

pls

<@&286206848099549185>

,o,

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[vital oracle]

What is the variance of the sample mean of 10 draws from the uniform distribution on [0,2]?

[obtuse pebble]

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[abstract ermine]

how do i add and subtract rational algebraic expressions with unlike denominators?

[mental plaza]

make the denom like each other

[abstract ermine]

i know that but how do i do it?

im beyond confused

[mental plaza]

depends on the denom u were given

got a picture?

[abstract ermine]

hold on

an example

[mental plaza]

okay

look at the first fraction

denominator has x

it needs to have a y in it too

so we multiply numerator and denominator of the first fraction by y

[abstract ermine]

so x+y^2 over x+y?

[obtuse pebble]

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[timid silo]

could someone help me solve this? https://imgur.com/a/UF7ljyS

[timid silo]

ans should be 2

[viral blade]

idk why there's an integral

[timid silo]

yea thats the thing

[viral blade]

it's just a sum

[timid silo]

its just so that x can be a variable

[viral blade]

double sum

Just sum from x=0 to infinity though

but yeah so

convert to a double sum

[timid silo]

got it

[viral blade]

then, say

call the term inside both sums f(k)

Find the number of times f(0) appears in the sum, f(1) appears, f(2) appears, etc

So how many times does 2^-7/(7+1) show up

[timid silo]

but the pattern of f(0) , f(1) , f(2) ... is pretty random like what does it tell me

[viral blade]

you can turn the double sum into one sum

maybe try making a grid of the terms to visualize it or something?

[timid silo]

yea ill try cause im not exactly following

wait but all of the terms after f(1) are going to evaluate to 0 in the floor function

[viral blade]

wym

[timid silo]

ok, how am i supposed to construct the one sum

i think i understood it wrong

[viral blade]

ok do you see how the integral is basically just a discrete sum

[timid silo]

yes

[viral blade]

ok

what happens when x=0

[timid silo]

k=floor(0)

[viral blade]

That's only where the sum starts though, right

[obtuse pebble]

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[timid silo]

(b) and (c)

[timid silo]

i already proved (a)

im not sure how to use (a) for (b)

remove the largest element of A to find the second largest element. Then remove that to find the third largest. Rinse and repeat via induction

what would my base cases be?

if im going backwards

induct on n

so base case is n=1

oh so i go forwards

the hypothesis would be "A can be labelled..."

[timid silo]

yeah, I was describing the inductive step

[timid silo]

does this proof make sense?

[obtuse pebble]

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[native flicker]

How do you calculate the speed of these algorithm?
f(n) = 3n^2 + n - 4
g(n) = n^2 + 2n + 2

The calculation my teacher gave us was
speed = number of comparisons + number of swaps / instructions per second

But I'm not too sure what this means

Any help is appreciated, I'm pretty new to this.
Thanks

[solemn osprey]

Can u plz clarify? What does the algorithm do?

[twin sapphire]

sorting algorithm most likely

[solemn osprey]

Thank Benjamin. Yeah idk this one

[twin sapphire]

but i think they gave the time of execution instead of the actualk algorithms

[obtuse pebble]

@native flicker Has your question been resolved?

[native flicker]

This is the exercise from the book. I'm not sure about determining the speed.

How would you answer this question?

This is what I have written so far.

How do you rotate the images again?

[twin sapphire]

,rccw

[twin sapphire]

,w plot 3x²+x-4 ,x>=0

[warm shale]

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[twin sapphire]

,w plot 3x²+x-4 ,x=0 to x=1000

[warm shale]

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[twin sapphire]

,w plot x²+x+2 ,x=0 to x=1000

[warm shale]

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[twin sapphire]

https://www.wolframalpha.com/widgets/view.jsp?id=1bd413de70f32142f4a33a94134c5690 use this to plot

multiple functions

and see with your eyes wichi one is faster

i gotta go sorry

[native flicker]

Ah okay I think I understand

Can you tell me if I'm thinking correctly here?

for this function
f(n) = O(n^2)

n = 1     n^2 = 1
n = 2     n^2 = 4
n = 3     n^2 = 9

and so on...

The value of n^2 is what is being plotted on the Y axis correct?

[obtuse pebble]

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[noble berry]

says its wrong

[civic zealot]

you messed up with the negatives

[noble berry]

where?

[obtuse pebble]

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.close

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[solar trellis]

Show that if x is in S then x is in f^(-1)(f(S))

[timid silo]

im not really sure how to dissect f-1(f(s)) to demonstrate so

[merry geode]

f-1(f(Y)) = {x | f(x) is in f(Y)}

[solar trellis]

x is in that iff there exists y in f(S) such that f(y) = x

[obtuse pebble]

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[dawn mist]

hello. have you tried something ?

[solemn osprey]

Rewrite 2 using a logarithm base 10

[sonic crown]

what?

[dawn mist]

can you show what you have done ? then i or someone else might guide you

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[tepid urchin]

can someone walk me through this?

[errant basin]

find the derivative of f

and then find the critical numbers

then plug in for #s between critical numbers to see + or -

for f'

and then plug in for endpoints as well just in case

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[timid silo]

is the anti derivative of sin(x) ..
-cos(x) + C

[low pecan]

yes

[timid silo]

yay

.close

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[rough egret]

Hi, need help with this quadratic inequalities problem, specifically from part G onwards.
got to use the numbers 1 - 9 at the bottom to answer the questions, can be used more than once.
Answers (using the numbers below)
A = 4, B = 3, C = 1, D = 1, E = 1 , F = 7. G = 6, H = 4, I = 1, J = 4, K = 1, L = 7, M = 3
I have inputted my current workings, do let me know how I can better do the earlier parts too if what I am doing is inefficient! Haven't done mathematics for quite awhile but am currently preparing for an exam so do need some help! Thank you so much

[obtuse pebble]

@rough egret Has your question been resolved?

[obtuse pebble]

@rough egret Has your question been resolved?

[obtuse pebble]

@rough egret Has your question been resolved?

[obtuse pebble]

@rough egret Has your question been resolved?

[obtuse pebble]

@rough egret Has your question been resolved?

[rough egret]

For this question, I am unable to get the part of E, F and thus H (could get G as it was just asking for minimum value, upon further inspection I realised i made a careless mistake and the minimum should be =a^2=4a-3)
Unsure of how to proceed to get part E and F.
E = 3, F= 5. H-5 (the number corresponds to the answer number using 0-9)

For this, clearly clueless on how to proceed from part 3 onwards. I do not understand where teeter is suppose to be (which I think is the limiting factor to me being able to answer the question)
D = 7, E = 9, F = 4, G = 2, H = 9, IJ= 80, K = 2, L = 3

Not sure how to manipulate the equation further, or if I even manipulated it right to begin with. Cannot tell where does m come into play for this equation, and hence unable to continue to answer this question.
O, P and Q = 2. R and S = 4. T = 1. U = 3, V = 2. W=1, X = 2, Y = 3, Z =4.

Have quite a weak foundation in trigonometry... Just stuck for the entirety of this question... unable to answer them at all.

[rough egret]

answer for this

So sorry for the spam of questions, am preparing for an exam this Sunday. Thank you so much for helping!

[rough egret]

<@&286206848099549185>

[obtuse pebble]

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[onyx berry]

.close

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[sacred oar]

Hello

[obtuse pebble]

Please don't occupy multiple help channels.

[sacred oar]

I need help with the 3 and 4

[obtuse pebble]

@sacred oar Has your question been resolved?

[sacred oar]

<@&286206848099549185>

[obtuse pebble]

@sacred oar Has your question been resolved?

[timid silo]

@obtuse pebble

[fading quest]

what do you think should be in 3.?

[fading quest]

#❓how-to-get-help go here and read, dont post question in other peoples channels

[timid silo]

ok

[obtuse pebble]

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[timid silo]

lost on this

[solemn osprey]

Another way to write problem 55 is this: given the function A(x) = f(x) + 2 * g(x), what is A’(3)?

U understand that A is a function right? Just starting at the most basics just to make sure

@timid silo

[timid silo]

yeah i know its a function

[solemn osprey]

So

Before we find the derivative for x= 3, let’s find the derivative. What’s the derivative of A(x)?

[timid silo]

Is it f'(x) + 2g'(x)?

[solemn osprey]

Yup!

Ok now let’s plug in x=3

[solemn osprey]

For x=3, what does the derivative evaluate to?

[timid silo]

is it 2

[solemn osprey]

Look to the table of values for help

[timid silo]

I got B

[solemn osprey]

Yup!

Ur correct!

[timid silo]

this helped me a lot ty

I visualized it as A(x) = f(x) + 2 * g(x)

[solemn osprey]

Wait May I ask, how did it help u?

[timid silo]

I feel like the original question was written weird

"f + 2g"

[solemn osprey]

Yeah the AP calc BC exams won’t allow that way of writing functions either lol

[timid silo]

ty for the help

[solemn osprey]

Np

[timid silo]

.close

[obtuse pebble]

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[burnt thunder]

whats 8.6 csc 41 degrees

[burnt thunder]

how do i do that

[timid silo]

bruh

trig in degrees

miss me with that

[burnt thunder]

ha

isnt it 1/sin 41?

[obtuse pebble]

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[zinc cargo]

can anyone help?

[zinc cargo]

<@&286206848099549185>

.close

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[trim crane]

Idk how to find db

<@&286206848099549185>

pls

<@&286206848099549185>

Pls

[obtuse pebble]

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[trim crane]

<@&286206848099549185>

[obtuse pebble]

@trim crane Has your question been resolved?

[noble berry]

I think I have an idea on how you could solve it?

Ec:ae = 2:1

Draw line from point D to C

[obtuse pebble]

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[wicked moat]

The number of ways to select a committee of eight that contains A and not B is... we have 15 to choose from

[trail musk]

What

[wicked moat]

we have a student council wiht 15 people. we need to pick 8 for a sub council and that council must have person A and not person B

finding the number of ways to select a committee of eight that contains A and not B

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[undone obsidian]

can i have help please? I found the big circle's radius using pythagorean theorem (4/9) but can't find the others. i drew radii but don't know how to proceed

[obtuse pebble]

@undone obsidian Has your question been resolved?

[undone obsidian]

<@&286206848099549185>

[timid silo]

Wtf

[final sky]

this shit is a fractal of circles

mandlebrot set type shit

[timid silo]

What's the area of the big semi circle? @undone obsidian

[undone obsidian]

pi/2?

[timid silo]

Bruh

[final sky]

i mean

yeah pi/2 makes sense

[timid silo]

How can it be 180

If that circle has 1 unit

Which is already smaller then the big 1

[final sky]

im out 💀

[timid silo]

L

[undone obsidian]

damn

[obtuse pebble]

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[undone obsidian]

thanks guys

[obtuse pebble]

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[spice plank]

how to do this question? i tried forming the two eqns for s2=15 and sum of infinity=27

[spice plank]

not sure how to proceed

[floral canopy]

get a new channel this one is occupied

[floral canopy]

well, the first equation will be:
a1*q=15.

[obtuse pebble]

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[timid silo]

can someone help me Inverse Laplace this equation?

[obtuse pebble]

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.close

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[obtuse pebble]

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[cloud nimbus]

!claim

[stable rain]

lol

u shld enter ur q first

not

!claim

u want

P(A|B) = P(A)

and

P(B) = P(B|A)

also note

P(A|B) = P(A and B) / P(B)

[obtuse pebble]

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[latent knoll]

Help with q2 b iii

[obtuse pebble]

@latent knoll Has your question been resolved?

[latent knoll]

<@&286206848099549185>

[tranquil arch]

just find $n$ let $\frac{n \pi}{3} \in [0,2 \pi]$

[warm shale]

秋水

[latent knoll]

?

[tranquil arch]

you have done ii

[latent knoll]

Yeah

[tranquil arch]

it's the result you got

[latent knoll]

I’m confused how to do iii

It says state all the solutions

But idk how with 2 variables

X and n

[tranquil arch]

let n=0,1,2,...

[latent knoll]

Oh

[tranquil arch]

when n=0, x=0 is a solution

[latent knoll]

Wait so n can’t be like 1.1 or anything?

Has to be whole number right

[tranquil arch]

n is an integer

[latent knoll]

Oh yeah I forgot

Lol tysm

!close

.close

[obtuse pebble]

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[regal swallow]

Question: Prove that every odd number may be rewritten as the difference of two squared numbers.

I am aware there exists an acceptable proof that makes use of true examples that seem to prove the question:
\begin{align*}
2n-1 &= \
2^2 - 1^2 &= 3,\
3^2 - 2^2 &= 5, \
4^2 - 3^2 &= 7, \
5^2 - 4^2 &= 9, \
6^2 - 5^2 &= 11...
\end{align*}
However, I was wondering if there were a more rigorous proof? I tried solving using induction and honestly I'm not convinced.
For context, this problem was given to a friend who is still in high school. Been a while since last I dabbled in Math, but can anyone help out?

[warm shale]

Marin

[timid silo]

2n-1 = -((n-1)-n)((n-1)+n)

[regal swallow]

The numbers are different

Another way to write it would be something like this

Prove that: $\forall 2n+1, n\in\mathbb{Z}, \exists(a,b)\in\mathbb{Z}\mid a^2-b^2=2n+1$

[warm shale]

Marin

[tranquil arch]

$$2n+1=(n+1)^2-n^2$$

[warm shale]

秋水

[tranquil quiver]

exactly what toby said ^^

[regal swallow]

How do we know that the numbers a and b are based on n?

Sorry I'm just confused. If the case is built upon 'n' then that's easy

But we have no info about the squared numbers

We just know they're squared

Or are we basing the supposition on the proof I mentioned?

[tranquil quiver]

the squared numbers dont matter

and im guessing you mean squared integers

but it can be any squared integers

[regal swallow]

Yeah, we'll suppose they're integers

Yeah exactly

So how do we know 'n' is a common variable?

[tranquil quiver]

so this form is showing you can rewrite it like that

[tranquil quiver]

wdym?

all you need to know is

n + 1 and n are integers

and n is by definition

[regal swallow]

Sure

[tranquil quiver]

n + 1 is of course

[regal swallow]

And how do we know that these numbers are one apart?

[tranquil quiver]

spoiler n + 1 - n = 1

it doesnt matter what the n is

(n + 1) - n will always be 1

[regal swallow]

Again, you're saying that $2n+1 = (n+1)^2 - n^2$

How do you know that the righthand side is true?

[warm shale]

Marin

[tranquil quiver]

$(n+1)^2 - n^2 = n^2 +2n + 1 - n^2 = 2n + 1$

[warm shale]

dog.

[regal swallow]

Oh you rewrote it

[tranquil quiver]

yes its that simple

[regal swallow]

Ugh sorry lol

[tranquil quiver]

they are equal algebraically

so its true for any n

[regal swallow]

Alright thanks lol

Have a good day

.close

[obtuse pebble]

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[hot phoenix]

i love maths

[timid silo]

.close

[obtuse pebble]

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[timid silo]

Let F(x) = [f(x)]^2 + [f'(x)]^2 ; F(0)=6 ; f(x) is thrice differentiable such that |f(x)| =<1 for x in (-1 , 1) ; then prove that there is no point of local maxima of F(x) in (-1,1) ; also prove that for some c in (-1,1) , F''(c)=<0 and F(c) >= 6

this question was done in class by our math teacher and i couldnt understand it whatsoever. i straight up dont know what happened so i have no intial thought process to give [ please explain thru Lagrange's Mean Value Theorem and Rolle's Theorem]

[obtuse pebble]

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@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[clear solstice]

i should find the derivative

this is what ive done so far

[spring flax]

use product rule and x^1/6 i wrong

[clear solstice]

sry i got confused but 1 times X^ -1/3 its still x^-1/3

?

[spring flax]

yes

sorry mb use power rule not product rule

like work out the product

then use power rule

[clear solstice]

wym

ive simplified it so far

[spring flax]

(x^1/2 +1)x^-1/3

then x^-1/6 +x-1/3

the use the power rule

[clear solstice]

bro im adding the exponents

1/2 + ( -1/3)

its 1/6

not -1/6

[spring flax]

misclick

shit mb

[clear solstice]

ye

and also 1 × x^-1/3

its x^-1/2

[spring flax]

whut

[obtuse pebble]

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[restive plinth]

a canoe travels at a speed 3km/h in a direction due west across a river flowing due north at 4km/h. find the actual speed and direction of the canoe.

[restive plinth]

can someone help me with this?

[unique spear]

This involves vectors I believe

[final thunder]

Just Pythagorean theorem

Then a bit of trig to find the direction

Draw a diagram first

[unique spear]

Why Pythagorean theorem?

is it because it doesn’t ask for resultant velocity?

[final thunder]

It is asking for the resultant velocity

[unique spear]

so shouldn’t it be the sum of both velocity?

[final thunder]

Indeed

And we find the magnitude by using Pythagorean theorem

And direction with a bit of trig

[unique spear]

my school doesn’t teach like that?

my school taught us to use it when it’s on a Cartesian plane

[final thunder]

It’s just vector addition

[unique spear]

Then wouldn’t it be 7km/h

or 5km/h?

[final thunder]

5

[unique spear]

but you just said it’s addition

addition and Pythagorean aren’t the same thing

[restive plinth]

why is it 5 and not 7?

[final thunder]

vector addition

[unique spear]

so you guys use Pythagorean theorem for vector addition?

[restive plinth]

i do

idk how to draw this, im new to this

[final thunder]

Vector additions when both vectors are perpendicular, yes we use Pythagorean theorem

[unique spear]

what if they are non-parallel

do we just add them?

[final thunder]

Sure

[obtuse pebble]

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[formal thunder]

💀 please someone help with this, ive fried my brains doing these questions, and i live in a metrics system

[formal thunder]

help is very appreciated, thank you in advance

[pallid saffron]

Hello guys, I have this problem:

[earnest elk]

#❓how-to-get-help

[formal thunder]

rip brain fly high

[pallid saffron]

ohh it was free sorry

[earnest elk]

what exactly are you struggling with in this?

i believe you are given values for conversions

[formal thunder]

so as i said i live in a metrics system, so still not familiar with the us standards, and ive been doing a lot of other questions like this

the last few

solved everything except for 30

someone can help with that right 🥺

[obtuse pebble]

@formal thunder Has your question been resolved?

[obtuse pebble]

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[polar idol]

For this would I just find the xyr values and use them to solve the equation?

[desert sinew]

You can draw 3-4-5 triangle based on what quadrant those angles are in, and then find the corresponding trig value.

[polar idol]

Okok

for this

C = +6

A = pi/3 -6 which = 17pi/3 (I think)

no K value since period is 2pi

How would I find the x value that shifts the max to the right

feel like the point the curve passes through might help but I wouldn’t know how

current equation is 17pi/3cos(x-x) +6

<@&286206848099549185>

Nvm I’m an idiot

I sub it in right?

Could someone help confirm my values tho?

yo no one’s gonna help?

<@&286206848099549185>

[obtuse pebble]

@polar idol Has your question been resolved?

[obtuse pebble]

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[lucid meteor]

.reopen

<@&286206848099549185>

[obtuse pebble]

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[royal flint]

...physics?

[knotty crow]

method?

[lucid meteor]

i think its just 1/250 + 1/100

and then

1/300 + 1/225

idk if i should use any theorems

[knotty crow]

but what are doing, calculating equivalent resistance?

[lucid meteor]

calculating the total resistance

[knotty crow]

yes

this is what I was thinking about

[lucid meteor]

🫡

[royal flint]

Aren't R4 and R5 in series

[knotty crow]

but wtf

R(45) = 100 + 225 = 325

R(23) = 125 + 175 = 300

etc.

[lucid meteor]

would it not be parallel?

[knotty crow]

now R(23) and R(45) are in parallel

so u do 1/R(2345) = 1/R(23) + 1/R(45)

[lucid meteor]

is r1 not included?

TwT

[royal flint]

R2345 in series with R1

To get final ans

R2345+R1

[lucid meteor]

oh i see i see

and r1 also have

1/250

[knotty crow]

??

Firstly find R(2345) and tell me your result

[lucid meteor]

1/250 + 1/300 + 1/325

[knotty crow]

nope, it's false

[lucid meteor]

1/156?

[knotty crow]

yes

so R(2345) = 156

[lucid meteor]

ohhh

[knotty crow]

what is R(12345) now?

[lucid meteor]

and 156 + 250

sorry

[knotty crow]

yep

[lucid meteor]

406

is total resistance?

[knotty crow]

yes

[lucid meteor]

ty!

i'll try solving the rest

[knotty crow]

np, just Ohm's law

and KCL

[lucid meteor]

ty ty!

[lucid meteor]

what law would i have to use for total current?

current divider law?

I1=R2/R1+R2

[knotty crow]

you have total R

and V

by Ohm's law I = V/R

[lucid meteor]

oh so basically just I1= 15/250

[knotty crow]

15/406 *

[lucid meteor]

OH

cause its the total current

[lucid meteor]

wouldn't this be for current through each resistor

[knotty crow]

how each resistor?

current is different for different branches (and resistors)

[lucid meteor]

oh cuz

15/250 15/125 15/175 ?

[knotty crow]

absolutely not

[lucid meteor]

oh

what law would be good

TwT

[knotty crow]

you have total I

and R1

find voltage drop on R1

[lucid meteor]

hmmm

ok

[knotty crow]

obv, you've already calculated currect through R1

and it's total current

[lucid meteor]

i got

15/406 * 250

which is 9.2

seems v big

drop

[knotty crow]

it's correct, 9,236

V

[lucid meteor]

so I1=9.236

[knotty crow]

nope

[lucid meteor]

oops

[knotty crow]

it's voltage drop, not current

[lucid meteor]

oh yeah

oops

[knotty crow]

current through R1 is equal to total current

and it's 15/406

,calc 15/406

[warm shale]

Result:

0.036945812807882

[knotty crow]

approx 36,94 mA

[lucid meteor]

would the current be the same for everything @knotty crow

[knotty crow]

nope

since we have different branches

[lucid meteor]

so

36.94-9.236?

[knotty crow]

aka sum of current flowing into a node is equal to the sum of current flowing out that node

you'll use that

[knotty crow]

nope, wtf?

what is it

[lucid meteor]

😭

i thought the voltage drop effected it

[knotty crow]

Current through R1: I(1) = I = 36.94 mA

Voltage drop on R1: V(1) = 9.236 V

now lemme draw a circuit containing R(2345) and R1

[lucid meteor]

tysm again :)

i wish my lecturers could have taught me as well as this 😢

[knotty crow]

so we have

can you calculate voltage drop on R(2345) ?

[lucid meteor]

is this the same as

battery

[knotty crow]

yes

[lucid meteor]

ty

[lucid meteor]

hmm kk!

[knotty crow]

so can you compute it?

[lucid meteor]

i definitely

think i did smth wrong

15 x 1/156

[knotty crow]

Ohm's law: V = RI

hence V(2345) = R(2345) * I1 = 156 * 0.03694 =

,calc 156 * 0.03694

[warm shale]

Result:

5.76264

[lucid meteor]

ahhh

[knotty crow]

now u have your original circuit

with following data

[lucid meteor]

hmm

why is there the current from r1 the multiplying 156?

[knotty crow]

from here u only need to calculate I2 and I3

use Ohm law's again

[knotty crow]

for elements connected in series current is same

e.g. for R2 and R3 current is same

for R4 and R5 current is same also

(but obv different)

than I2

[lucid meteor]

i got 8.98

[knotty crow]

as what

[lucid meteor]

as I2

[knotty crow]

total current is 0.03694, how any other current in the circuit can be higher

[lucid meteor]

oh

okey

[knotty crow]

I = V/R

[lucid meteor]

i did V= 5.7624

[knotty crow]

good

[lucid meteor]

OH

okay so

its

1.92

mA

[knotty crow]

you've divided by sum of resistance, right?

[lucid meteor]

cause its 5.7624/(125+175)

[knotty crow]

yes

,calc 5.7624/(125+175)

[warm shale]

Result:

0.019208

[knotty crow]

= 19.2 mA

not 1.92

[lucid meteor]

oh whoops

[knotty crow]

yeah, so I2 = 19.2 mA

it's current on both resistors R2 and R3

I3 can be calculated as the difference between I1 and I2 or same why as you've just did to compute I2

[lucid meteor]

so its just

36.94-19.2

:D

17.74

[knotty crow]

yeah and look at this

,calc 5.764/(100+225)

[warm shale]

Result:

0.017735384615385

[knotty crow]

approx 17.74 mA

same thing

[lucid meteor]

thank you!!
i think the question has been solved now

[knotty crow]

it's current on R4 and R5 (BOTH)

[lucid meteor]

wait

but also

why both R4 and R5?

is it because its in series?

[knotty crow]

yes

exactly

[lucid meteor]

oh okay! ty :)

.close

[obtuse pebble]

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[knotty crow]

,w tangent to x^2 * arctan(y) = x^3-8y+pi at (x,y) = (2,1)

[keen heart]

can anyone look over my work and tell me if I'm wrong?

[warm shale]

Results provided by WolframAlpha

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Requested by Modus#7591

[keen heart]

wow

how do I get to that

where did I go wrong

[obtuse pebble]

@keen heart Has your question been resolved?

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[misty kiln]

what does it mean by the definition of a taylor series?

is it like

f'(x)(x-c)^n / n!?

is it that?

.close

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[spice bay]

In Euler's theorem we have that if gcd(a, n) = 1 then $a^{\varphi(n)}\equiv 1 \pmod{n}$

[warm shale]

Blesedi

[spice bay]

But if gcd(a, n) is not one, then for what m does this work?

[kind hawk]

do you know bezouts identity?

[spice bay]

yes I do

[kind hawk]

what does it say

[spice bay]

It is about the solutions of ax + by = d

Not sure what it has to do with the problem

[kind hawk]

let b=n, rearrange

ax=d - ny

[spice bay]

Hmmm

So if I assume that a^m = kn + 1
is a solution,
can I choose x = a^(m-1)
and y = -k?

Then ax = d - ny becomes
a^m = d + kn

am I on the right track?

but d is not equal to 1, since gcd(a,n) is not one

[kind hawk]

what exactly is the statement of bezout?

[spice bay]

from wikipedia
Bézout's identity — Let a and b be integers with greatest common divisor d. Then there exist integers x and y such that ax + by = d. Moreover, the integers of the form az + bt are exactly the multiples of d.

Yeah, what I did is wrong...

Hold on

Am I on the right track?

[kind hawk]

well depends on where you wanna go

bezout says: ax=d-ny is solvable if and only if gcd(a,n) | d

for which d are you interested in a solution

[spice bay]

Oooooh

that's a pretty solution

I think I understand it now

ax=d-ny is solvable if and only if gcd(a,n) | d
We want solutions for d = 1, right?
gcd(a,n) | 1
which is impossible.
Now, is this correct?

since gcd(a,n) is not one

[kind hawk]

yes

[spice bay]

wonderful!

thank you so much!

.close

[obtuse pebble]

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[agile cobalt]

Hello,

[agile cobalt]

I saw this in a text book

I just wanted to clarify my understanding

^ would that be correct?

[cedar lichen]

Yes

Or {2} ∈ {{1},{2}}

[agile cobalt]

Thanks!

.close

[obtuse pebble]

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[misty kiln]

did I make a mistake somewhere along the way? I'm really just not seeing any possible relationship here to form the series

idrk where to look

[civic zealot]

wdym? That's the series

[misty kiln]

yes, but I have to write out the formula

not just have the expanded form

[civic zealot]

why?

[misty kiln]

because my professor is evil

[civic zealot]

well, the directions at the top say 'find the taylor series'
Those 6 terms are the taylor series, there isn't an infinite sum because the terms after (x-2)^5 are 0

[misty kiln]

oh

well

maybe im being dumb then

ty

.close

[obtuse pebble]

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[proven karma]

Hello; I am not sure how to answer this question, I was absent for the lesson in which my teacher taught me how to do these expected values problems and i only know how to calculate E(X) from known probabilities, not the other way around

[obtuse pebble]

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[supple epoch]

Let $A \subseteq \mathbb{R}$ be a nonempty set bounded below and $a \in \mathbb{R}$ chosen arbitrarily. Show that $a = \text{inf}A$ if and only if
$$
a = \text{sup}{c \in \mathbb{R} \ | \ x > c \ \text{for all} \ x \in A }
$$
is satisfied.\
\
How can I approach this?

[warm shale]

Levens

[obtuse pebble]

@supple epoch Has your question been resolved?

[supple epoch]

.close

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[vague mural]

how do I stop making silly mistakes? mistakes such as forgetting to flip negitive/addition sign, forgetting that a variable was squared not cubed etc.
I find that this contributes to the majority of quesitons i answer incorrectly

[timid silo]

after you finish solving any question

read it again slowly or solve on another paper and compare it with yours at the end (if you have free time)

or just solve slowly and focus more

[vague mural]

every time? that seems like a lot of time spent checking if you made a mistake

[timid silo]

U get used to it

[vague mural]

is this problem so common that thats required?

[timid silo]

practice more

[mental plaza]

u can’t prevent it 100%

but what u can do is check ur work ig

[vague mural]

thats it?

[mental plaza]

yes

as u said

[vague mural]

so just check ur work and practice?

[mental plaza]

those are silly mistakes

[mental plaza]

yeye

[timid silo]

yeah what did you expect black magic

[mental plaza]

[vague mural]

tbh i thought yall were gonna say aderal but thats just me

[timid silo]

its nuetella for me

To sound deep for a second, "I fear not the man who has practiced 10,000 kicks once, but I fear the man who has practiced one kick 10,000 times" - Bruce lee

Actually kinda has some truth to it

[vague mural]

thought that was from Barack

[mental plaza]

don’t be a jack of all trades but a master of none

~my chem teacher

he got it from somewhere but idk where

[timid silo]

" If you practice more topology and logic, you'll make more mistakes"
-Sun Tzu. The Art of War.

[vague mural]

lol hes got the knowledge

solve problem get clout, thats what my friend says

LOL

[timid silo]

thats a quote i would live by

[vague mural]

i gotta dip lol

^

[mental plaza]

HAHAHAHA

[mental plaza]

peaceee

[vague mural]

alright well peace

.close

[obtuse pebble]

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[smoky zenith]

My answer is incorrect but i cant find my issue

[knotty crow]

have you used tan(2x) formula

[smoky zenith]

Yes sir

[knotty crow]

wait, u did right triangle with side lengths √2, 3 and √5?

[knotty crow]

then it's untrue obv since (√2)^2 + 3^2 = 2 + 9 = 11

not 5

[smoky zenith]

ah i see, so when ever im doing a right triangle i have to check if it all adds up correctly

Would i keep the rooted five or just square the 5?

[knotty crow]

you want to have sin(angle) = 3/5

so idk why u did √5 instead of 5

[smoky zenith]

do i only root the angle for cos angles? When do i root?

[knotty crow]

why would you root the angle? you just choose side lengths such that trig ratio is true

then you find 3rd side using Pythag

[smoky zenith]

I saw it in one of the steps towards a solution

so i guessed if im given the Radius i must square root

[knotty crow]

this is related to definition of trig functions I guess

but taking a root is still same as doing Pythagoras

[smoky zenith]

Should i just opt out of using the normal pythagorean theorem formula and just use this one?

[knotty crow]

notice that r is derived from Pythagorean Theorem, legs are x and y

so just use Pythagorean Theorem and everything will be fine

[smoky zenith]

k

[obtuse pebble]

@smoky zenith Has your question been resolved?

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[spice chasm]

guys did i do this right

[obtuse pebble]

Please don't occupy multiple help channels.

[spice chasm]

oop

i forgot to close the other one