help-28

[tame lark]

sentences

[neon geyser]

so a unit circle is a circle with radius 1

have u used a unit circle before?

[tame lark]

not for a question like this

but ive used it for like radians and such

[neon geyser]

so you do know that a unit circle has radius 1?

hence the name unit

[tame lark]

yes

[neon geyser]

and you notice that the triangle then has to have a hypotenuse of 1 right?

[tame lark]

its asking how i find tan b, but in that area isnt it cos and sin

[neon geyser]

since the hypotenuse is technically the radius

[neon geyser]

don’t jump the gun

[tame lark]

uh alr

so radius is 30

?

[neon geyser]

no i just mentioned that a unit circle has radius 1 by default

but your triangle has hypotenuse 30

[tame lark]

i thought you were talking about the triangle mb

[neon geyser]

how could you make a similar triangle

such that the hypotenuse is now 1

[tame lark]

divide 15 by 30

and 15sqrt 3 by 30

[neon geyser]

yes

that’s what i meant by scaling down

every side by 30

[tame lark]

ohh alr

[neon geyser]

now that it fits in your unit circle

you first find sin(theta)

then you find cos(theta)

[tame lark]

whats theta

[neon geyser]

angle B

[tame lark]

B?

ok

[neon geyser]

yes there were angles A,B, and C in your triangle

[tame lark]

sin b according to the new triangle?

[neon geyser]

yes

[tame lark]

with like the new values or

ok

[neon geyser]

and cos(b)

[neon geyser]

yes ofc

[tame lark]

well sin b is 1/2

wait i mean cos

[neon geyser]

and then use the fact that tan(b) = sin(b)/cos(b)

u just have to explain briefly ig

it’d worth the least mark but we spent the most effort on it

thus far

[tame lark]

sin b is a long decimal tho

[neon geyser]

use exact values

ofc it’s a long decimal

it has sqrt(3) which is irrational

so “long decimal”

[tame lark]

sqrt 3/2

[neon geyser]

yes

tan(b) = sqrt(3)/2

so b = 60 degrees

also no actually

[tame lark]

howd you get that

[neon geyser]

using the unit circle

you know that side b is

sqrt(3)/2 right?

[tame lark]

isnt it sqrt 3/2 over 1/2

[neon geyser]

which corresponds to sin(B)

[neon geyser]

yes

cos(B) = 1/2

you should already at least know that cos(60 degrees) is 1/2

u should’ve also known other basic angles like cos(30 degrees)

[tame lark]

yeah

[neon geyser]

and cos(45 degrees)

so yeah

[tame lark]

but what happened to the 1/2

[neon geyser]

an angle B such that sin(B) = sqrt(3)/2 and cos(B) = 1/2 is B = 60 degrees

[neon geyser]

[tame lark]

ok i think i understand

[neon geyser]

,,\frac{\sqrt 3}{2} \div \frac 12 \implies \frac{\sqrt 3}{2} \cdot \frac 21

[glossy valve]

nyxie9151

[tame lark]

and then that is

[neon geyser]

i hope you know what that is lol

it’s just multiplying fractions

cancel the 2s out

[tame lark]

its sqrt 3

[neon geyser]

yes

tan(B) = sqrt(3)

B = 60 degrees

,w tan(60 degrees)

[glossy valve]

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[tame lark]

okok

now its just part c

but i think thats pretty easy

[neon geyser]

a thing to note is that scaling down works because you just made a similar triangle

the angle is still the same

only the side lengths changed

even then the ratio of the side lengths are also preserved

[neon geyser]

i mean yeah just use 1/2 b*h

[tame lark]

i dont think my teacher would let me do that lmao

[neon geyser]

lmao

that’s literally what you’re supposed to do

[tame lark]

i think im prob supposed to use a law or smt

[neon geyser]

nah

u got 5 marks

for finding an angle which was easy to find in a right triangle

i think 3 marks is fair for the area

[tame lark]

prob smt similar to this

and i got that right

and i used sine law i think

[neon geyser]

yeah except you have stuffs here that allows you to use 1/2 b * h

making stuffs harder doesn’t mean more score

[tame lark]

i guess

[neon geyser]

in that case you should go back

and use law of cosines on question 1

[tame lark]

no ty

[neon geyser]

yeah so just use 1/2 b * h

there’s nothing really

lol

[tame lark]

just to check its 225 right

[neon geyser]

if u really want u can also use your other formula

[neon geyser]

this one

if for some reason u like making stuffs hard

but i feel like your teacher would find it weird that u used this formula when u identified it’s a right triangle in part 1)

[tame lark]

maybe

[neon geyser]

,w 0.5 * 15 * 15 * sqrt(3)

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[tame lark]

the base isnt the hyp?

[neon geyser]

well if it’s called the hypotenuse

why would u also call it the base

[tame lark]

;-;

adj is the base then

[neon geyser]

there’s no difference between what u consider as base or what u consider as height in a right triangle

multiplication is commutative so

3 * 4 = 4 * 3

[tame lark]

because i did 30 times 15

[neon geyser]

well i never said there’s no difference between the hypotenuse and the base/height

i’m talking about the base and height

hypotenuse is not interchangeable

[tame lark]

ok

thanks

[neon geyser]

sure

np

[tame lark]

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[past marten]

I need help solving question 3. I’ve tried using the Euclidean algorithm and I believe I ended up with minimal n is 1 and minimal m is 4 but I am unsure

[thin flint]

n = 1 is correct

m = 4 is too

[full forum]

@past marten Has your question been resolved?

[past marten]

Thank you for answering! To solve for the pair in each part
Do I just plug n into the equation to solve for the m in that situation and vice versa in b?

[thin flint]

for a it's easy because it's 100

and for b it's 236 iirc

but yes just plug in m = 4 in

234

1932/8

,w 1932/8

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[thin flint]

i made a mistake damn

1832

not 1932

,w 1832/8

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[thin flint]

now i've confused myself :D

[past marten]

Oh no

[thin flint]

it's 229

i confused myself :D

it's 1832/8

that's 1998 - 90 - 19 * 4

i have no idea where i got the 236 and 234 from

[past marten]

Thank you! So I did the work right then, just needed to plug it in to find the pairs?

[thin flint]

yes

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[edgy shale]

Use Pytagora's trigonometric identity

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@candid estuary Has your question been resolved?

[torn jolt]

hi

using

sin^2(x) + cos^2(x) = 1

divide through by cos^2(x)

tan^2(x) + 1 = sec^2(x)

using that you can rewrite 2sec^2(x)

so

2(tan^2(x) + 1) + tan^2(x) -3 = 0

and then just re-arrange so

3tan^2(x) = 1

tan^2(x) = 1/3

tan(x) = sqrt(1/3)

and then so on

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@candid estuary Has your question been resolved?

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@candid estuary Has your question been resolved?

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[nova crater]

I tried doing the multiplication rule: (5)(6)(8)(7) = 1680

but the answer key says 1736

[sacred sparrow]

ok let's view the bottlenecks first:

The last one has 5 possible values

The first one has 7 possible values

Second and third have 10 possible values

so we'll regard the last one first, as it's the most critical bottleneck

so we can definitely multiply our option count for the first three letters by multiplying by 5 for the last letter

now we'll consider the second bottleneck, the first letter

since for the last letter we chose from among 1, 3, 5, 7, 9

since 1 < 3

and 3, 5, 7, 9 >= 3

we have different cases depending on what we chose for the last

if we chose 1 for the last, then we have 7 options for the first

if we chose 3, 5, 7, 9 for the last, then we have 6 options for the first

so we can start splitting:

5 * [options first three letters] = 1 * 7 * [options for 2nd & 3rd] + 4 * 6 * [options for 2nd & 3rd]

the options for 2nd and 3rd aren't influenced by our choices for first or last, so for them we just have 8 possible values for one of them and 7 for the other:

= 1 * 7 * 8 * 7 + 4 * 6 * 8 * 7

@nova crater

oh oops I did even for the last

ok corrected

[nova crater]

wow, thank you for the deep explanation.

i think i get it now

[sacred sparrow]

in short: our choice for the last letter influences the choices for the first :) (that's why we need case splitting)

[nova crater]

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[placid patrol]

I need help with part b

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Show your work, and if possible, explain where you are stuck.

[placid patrol]

Im stuck on B

<@&286206848099549185>

Im just seeking for help for part b

thanks!

Anyone?

part b

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[ember siren]

hi could somebody help me I don't understand this at all

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[ember siren]

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[late flame]

Yo can I get help

[torn jolt]

yo

[silent quest]

show us the question bruh

[late flame]

Wag

Kk

What this mean

I don’t understand word problems

[manic jacinth]

square fence

[silent quest]

lol

[manic jacinth]

each side is the same length

[torn jolt]

10

[late flame]

Wait wait

Pause

Slow down

I meant formula

[torn jolt]

bro its class 5 problems

[silent quest]

x^2

[manic jacinth]

ah

[late flame]

What

[manic jacinth]

sqrt(100) ig

[silent quest]

but formula...

umm

[torn jolt]

area of sq = l^2

[late flame]

Ye

So it area thingy

[silent quest]

O

l = sqrt(A)

[late flame]

What is this

[torn jolt]

100 = l^2
l= 10

[late flame]

😭

I got to get one of the formulas from here

[torn jolt]

which grade r u in?

[royal temple]

how in gods name are you in pre uni class 😭

[silent quest]

BRUH

why u didn't show us this at the first sec

[atomic blade]

No dw I have seen college students who do worse

[late flame]

Bc i didn think it this

[atomic blade]

Just ask anyone else in my microeconomics class

The idea of triangles is foreign to fellow business majors

[late flame]

That made me feel better

Thank you kind sir

[silent quest]

A = s^2

ye

[late flame]

Yessir bro

[atomic blade]

That's also not an excuse to kinda overlook this kinda stuff though lol. Make sure you understand the formulas

[late flame]

Ima do that…Saturday

Ok nextvproblem

What does this mean

[manic jacinth]

hm

[torn jolt]

area of sphere

[manic jacinth]

surface area

of a sphere

[late flame]

Thanks bro

But how yall got it

[torn jolt]

4 pi r^2

[late flame]

Wise words

[torn jolt]

just understand math

[atomic blade]

You have a formula sheet

But think about it this way

[late flame]

Ye

[atomic blade]

If you were to wrap a ball in wrapping paper

You're gonna need to cover the sphere

Aka the outside of the ball

[late flame]

Mhm

[atomic blade]

So just the skin of the ball in a sense, which is legit just the surface

So you'll need the surface area of the ball

[torn jolt]

u got some nice teachers

[late flame]

Ohhhhhhh

Big brain

So is this surface area too

Wait no

Its volume right???

[atomic blade]

Well

How would you measure a plot of land

[late flame]

With a those thingys

The measuring ape

[atomic blade]

Well I mean like what are you measuring

[late flame]

Tape

[torn jolt]

how much acres is meter?

[late flame]

1.32

So it’s volume???

Yessir

Boom shakalaka bang bang

No wait

[torn jolt]

1 acre = 4046 m^2

[late flame]

How you get that

[torn jolt]

1.32 acre = 5341 m^2

area of fotball ground = l*b=5341

[late flame]

Idk how you did that but I’m impressed

[torn jolt]

breadth given is 160 ft = 49m

[late flame]

So it’s squared….?

[torn jolt]

l *49=5341

[late flame]

SA???

[torn jolt]

l= 109 m = 358 ft

[late flame]

R u solving the problem

[torn jolt]

the ground is 109m long or 358 ft long

check if its correct

probably its correct

[late flame]

Wait how I check

[torn jolt]

ask ur friend or sth

[late flame]

Bromie…I’m sorry for making you do kinda ish pointless work but I just needed the formula

But the work

Was real impressed tho

A++

⭐️👍👍👍

Tadowwww

[torn jolt]

formula l*b=A

could u tell me which grade r u in

[late flame]

Ye

8

[torn jolt]

any more questions?

[late flame]

Ye

Just 2 left

1 2

[late flame]

Is this one V=lwh

@torn jolt you here…?

Yo??

Ima call for another helper sir

Sorry

But thanks for the help bro

<@&286206848099549185>

Yo

[steep dragon]

which question

[late flame]

Uhh

This one

Also is the formula for this one V=lwh?

[steep dragon]

perimeter of a rectangle is the length of all the sides added up

[late flame]

Ohhh

So 8

??

[steep dragon]

yea

[late flame]

Yessir

[steep dragon]

what do you think 18 is asking you for?

[late flame]

V=lwh??

[steep dragon]

yea

[late flame]

Yessir

That’s it

Fist bump

👊

Thanks for your help bro

Peaceeee

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[hushed arch]

hello! Can anyone help me with some math expressions i was given pls

[normal vector]

Send it

[hushed arch]

really

[placid patrol]

kana send your question so we can assist u

[hushed arch]

ok

[normal vector]

What is the question?

[hushed arch]

i was given some questions to work out and im a bit stuck where you have to simplify this expression

[agile sinew]

-9b -15

[manic jacinth]

distribute

[agile sinew]

-3 * 3b = -9b

-3 * 5 = -15

[normal vector]

(-3 * 3b) + (-3 * 5)

[agile sinew]

Yes

[hushed arch]

wait i still don't understand it

[normal vector]

What year are you? Like what level of math knowledge do you have?

[hushed arch]

im in highschool in yr 8

[hushed arch]

thank you for helping me

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[charred ivy]

how do I solve this

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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

[charred ivy]

well

I was thinking of plugging all the input values

and creating a system of eqn

but it seems too lengthy and I don’t think that’s the correct way to solve it

[fluid jacinth]

oh

[open igloo]

On the right side try getting the same denominator as that on the left side

The terms A,B,C and D are all independent of x, so you can substitute suitable values of x in the numerator terms to get the values of those 4

If you've done partial fractions before this would be familiar to you

[charred ivy]

So I can plug in any values of x?? @open igloo

[open igloo]

Yup

Also after getting the same denominator on the right, the new numerator is still p(x) and some of the values for p(x) have been given

[charred ivy]

so in order to find d, would I still need to find a,b and c

[open igloo]

Idts

Try substituting values of x for which all the other terms become 0 in the numerator

[charred ivy]

Ok so for now

[open igloo]

Except the one you want to find

[charred ivy]

I will sub p(1)=0

I willl show my work here

One sex

[open igloo]

p(1) isn't 0 tho, x-1=0 on substituting x=1, I think that's what you meant?

[charred ivy]

Is this correct?

[open igloo]

Yes

[charred ivy]

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[slender pond]

why did the use k = -2 and 2 when k can only be -1 0 1

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[prisma bear]

How do I solve this?

[prisma bear]

I know the answer but I'm not sure how to get to it

[slate violet]

Can you write out the third term? You will need an n choose 2

[brave blaze]

should be n choose 2

[slate violet]

Yeah ahhh that makes it so much easier

[prisma bear]

Im confused what do you mean choose?

[brave blaze]

$n \choose 2$

[glossy valve]

Herels

[prisma bear]

Maybe this is some other method of solving
I haven't seen this, shouldn't it be like this?

[brave blaze]

C_n^k is the same as n choose k

[prisma bear]

oh

Okay but how does that help me solve n?
I know I got n choose 2 and 1

[slate violet]

Yeah so you need the ratio of these two

[brave blaze]

$${n \choose 2} (\sqrt{2})^{n-2} (2\sqrt{3})^{2}$$

wtf

[slate violet]

You should write out the third term (n = 2) and the second term n = 1

And then simplify the ratio down using algebra

Also note that n choose 2 = n (n - 1)/2

[glossy valve]

Herels

[slate violet]

It follows from the definition, $\frac{n!}{2! (n - 2)!}$

[glossy valve]

south

[slate violet]

Expand n! and terms cancel

[prisma bear]

Yeah but apparently I'm supposed to obtain this somehow

[slate violet]

Step by step: you won't understand anything if you keep rushing and looking at the solution when you didn't write anything down yet

[slate violet]

This is the expression for the 3rd term, n = 2

Can you write out the expression for the 2nd term, n = 1?

[brave blaze]

$${n \choose 1} (\sqrt{2})^{n-1} (2\sqrt{3})$$

[glossy valve]

Herels

[slate violet]

Dude

!nosols

[full forum]

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

[brave blaze]

he just asked me the second term, not the answer

[slate violet]

Still you're not helping them by doing key steps for them

Instead of letting the person try things out themselves

[brave blaze]

there's nothing about steps if i'm just applying the same formula he sent himself (?)

[brave blaze]

.

[slate violet]

They need to be able to apply the formula

Not just memorise it

Applying formulas might seem easy to you but they seem to be lacking in that skill

So they need practice

[prisma bear]

I've done this for both the second and the third term

I don't know what to do next

[brave blaze]

the ratio of the third term to the second term is 5sqrt6

[prisma bear]

Yeah I get that
But what do I do next 😭

[brave blaze]

well use this information to solve for n ?

:x

[prisma bear]

how

[brave blaze]

third term / 2nd term = 5sqrt6

if i still know about ratio

first :
$$\frac{{n \choose 2}}{{n \choose 1}} = \frac{\frac{n!}{2(n-2)!}}{\frac{n!}{(n-1)!}} = \frac{(n-1)!}{2(n-2)!} = \frac{n-1}{2}$$

[glossy valve]

Herels

[brave blaze]

second :
$$\frac{12\sqrt{2^{n-2}}}{2\sqrt{3} \sqrt{2^{n-1}}} = \frac{6}{2\sqrt{3}} \times \sqrt{2^{n-2-n+1}} = \sqrt{3} \times \frac{1}{\sqrt{2}}$$

[glossy valve]

Herels

[brave blaze]

at the end we have :
$$\frac{\sqrt{3} (n-1)}{2\sqrt{2}} =5 \sqrt{6}$$

[glossy valve]

Herels

[brave blaze]

@prisma bear welp you didn't know how to do it

[prisma bear]

Well this gives n=21 but the answer is n=11

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[tired steppe]

is there a way to solve this without the quadratic formula?

[cerulean cargo]

complete the square

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[bleak vigil]

Hello! I’m having an issue regarding hookes law in my calculus 2 class, here is the question for reference, when I solve this I get to 1000(0.18^2 x 0.08^2) getting 26J but the answer was 50? Am I doing this wrong or is the answer wrong?

[glacial pasture]

that seems about right to me

[bleak vigil]

Is this answer incorrect then? This was my professors solution I’m assuming he just rushed it and got it wrong?

[glacial pasture]

the solution is ignoring that the natural length is 12

the extensions arent .3 and .2

rather 0.18 and 0.08

fix that and it would be fine

x is extension, not just the spring length, thats their error

[bleak vigil]

Yeah that’s what I thought, thank you! Got a final test tommorrow so this feels good that I got it right

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[leaden pollen]

Hi! In my Calculus 2 class improper integrals are solved in this way. In this picture, this integral is defined in (-1, 1) and my teacher chooses an arbitrary x_0 (in this case, 0) and calculates the two limits approaching the bounds. How should i choose the arbitrary x_0? In this example its easy, but what should be my reasoning in regards to chosing the x_0? Thanks!

[ivory cairn]

the choice is arbitrary since the function is continuous between -1 and 1, so the integration will work.
So the it comes down to choosing the easiest number to work with.

which is usually 0

[leaden pollen]

So any number in the interval will do?

[ivory cairn]

yep.

[glacial pasture]

could use 0.2

[leaden pollen]

Oh

[glacial pasture]

just an odd choice

[leaden pollen]

Well that explains it. Thanks guys!

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[fast epoch]

Hi! So I have an exam on limits tomorrow, and I'm going pretty well with it. However, there is one type of them that is making me struggle a bit. It's this one:

[fast epoch]

I figured out I had to equal it to e, but to do so I'm resting 1 to that 2 and adding 1 at the end inside the parenthesis so the result does not change

However I can't think of a way ti make the division positive

[granite torrent]

what u do is say x/(x + 1) = (x + 1 - 1)/(x + 1) = 1 - 1/(x + 1)

that’s where the positive comes from

[fast epoch]

how?

what I mean is I'm doing this

[granite torrent]

do u agree that x is equal to x + 1 - 1

[fast epoch]

Yes

[granite torrent]

is there a part of this line of reasoning that doesn’t make sense

[fast epoch]

no but where I'm doing that is outside the division

like rest 1 to everything and then adding it again

The division remains the same, but its still negative and idk how to make it positive

[granite torrent]

well u do basically the same thing

u rewrite 1 as (x + 1)/(x + 1)

[fast epoch]

oh I think I see the mistake

so after that I should have [1 - (x+x+1)/(x+1)]^x

This is how I made the whole exercise

[granite torrent]

no you should have a configuration where the x terms in the numerator cancel

[fast epoch]

I can't find how then

I need the 2 to become a 1 but if I rest 1 then I have to add one and that's what I get everytime I try

[granite torrent]

2 - x/(x + 1) becomes 2 - 1 + 1 - x/(x + 1)
(you wrote something down similar to this so i assume this step makes sense)

now, what you do is rewrite the + 1 as (x + 1)/(x + 1)
then, the expression is 2 - 1 + (x + 1)/(x + 1) - x/(x + 1)
which is 2 - 1 + (x + 1 - x)/(x + 1)
or 1 + 1/(x + 1)

[fast epoch]

ah

yep that makes sense

seems like I was forgetting the - when introducing the +1 into the division

thank you

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[vivid gyro]

im not sure why my answer is wrong

[stiff musk]

did you actually perform an integration?

[vivid gyro]

yes

@stiff musk

[stiff musk]

show your work

[vivid gyro]

k hold on

[stiff musk]

that is not the taylor series for ln(1+x)

[vivid gyro]

i havent learned taylor series yet

ive only done geometric series

like power series

[stiff musk]

but you got this from somewhere:

[vivid gyro]

can

ln(1+x) be rewritten as 1/(1+x)

thats the derivative

[vivid gyro]

and then it becomes this because of geometric series?

[stiff musk]

1/(1+x) and ln(1+x) are not the same thing

1/(1+x) is the derivative of ln(1+x)

[vivid gyro]

yes

does the integral cancel that out after?

or is taht not how it works

[stiff musk]

you have an x^2 complicating things

so not directly no

[vivid gyro]

how would i solve it then?

[stiff musk]

use the correct series for ln(1+x)

,w taylor series for ln(1+x)

[glossy valve]

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Requested by bungo.the.original

[stiff musk]

you can write more compactly as $$\ln(1+x) = \sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}$$

[glossy valve]

Bungo

[vivid gyro]

so would i do the same process i did but with that series instead?

[stiff musk]

yes

[vivid gyro]

okay

lemme work that out

so

after integrating does n become = to 1 for the series or does it remain 0? @stiff musk

i was told when taking the derivative it changes to 1 but im not sure about integration

[stiff musk]

it becomes 1 when taking derivative because the n=0 term of the original series is constant, so the derivative of the n=0 term is zero

for integrating you don't lose an n

[vivid gyro]

oh okay

i got

C+ the series from n=1 to infinity of (-1)^n * (x^(n+2))/(n-1)(n+2)

is that right @stiff musk

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[wanton sun]

how do i find the answers for the first question

[wanton sun]

Am i doing this right

[full forum]

@wanton sun Has your question been resolved?

[wanton sun]

No

Now what

[noble sedge]

diagonals of a rhombus are perpendicular bisectors i think

so m<ptq is 90

@wanton sun

[wanton sun]

What about the rest of them

[noble sedge]

you can figure them out with basic algebra

all the side legnths are equal because it's a rhombus

[wanton sun]

So m<trs is 28

?

[noble sedge]

no

it is 62

angles in a triangle sum to 180

so 90 + 28 + x = 180

118 + x = 180

x = 180 - 118

x = 62

[wanton sun]

Ohhh

Ok thanks

How do i find the sides

[noble sedge]

all the sides are 18 because the definition of a rhombus is a quadrilateral with 4 equal sides

[wanton sun]

Is m<srq gonna be 62 also

[noble sedge]

m<srq is 124

its double 62

[wanton sun]

So is m<tqp 62 since its just a half?

[noble sedge]

tqp is 28

diagonals of a rhombus bisect angles

[wanton sun]

my bad

Did i do this right

[noble sedge]

yes

[wanton sun]

For number 2 is m<cra 90

[noble sedge]

yes

[wanton sun]

Is m<cat 90

[noble sedge]

yes

[wanton sun]

I think i am get the hang of@it now

Thank you a lot

[noble sedge]

np

its the only thing i can help with ngl

im too dumb for any of this other stuff

have a good one bro

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[brazen tulip]

When does a limit not exist?

For a does it not exist since as the limit approaches -1 the graph approaches the whole at 1 and the point at -2?

It's approaching the whole and the point on the graph at the same time for question a so does that limit not exist?

[pallid coral]

a limit does not exist when it approaches different values from the left and from the right

[brazen tulip]

Ohhh interesting

Have not heard that before

But they do exist when you specify what direction the limit is approaching from, correct?

[pallid coral]

yes

[brazen tulip]

Okay

[pallid coral]

as long as there is actually something on that side

[brazen tulip]

So the the limit for a. would not exist?

[pallid coral]

yep

[brazen tulip]

Same with b?

B is approaching a hole and a point at the same time

But it's the same for each direction of the limit

[pallid coral]

it doesn’t matter where f(1) actually is, just look at the function when it is close to 1

so the limit would exist because you approach the same value from the left and from the right

[brazen tulip]

What value are we approaching though?

Wouldn't it be approaching 1 and -1 simultaneously? @pallid coral

[pallid coral]

what do you mean by approaching

[brazen tulip]

Like as x approaches 1

The graph approaches 1 and -1

[brazen tulip]

So why does the point at (1, -1) not matter?

Like why are we ignoring that point

[pallid coral]

the limit does not care about the f(1), but rather f(x) when x is really close to 1

[brazen tulip]

OHHHH

Interestinggg

So interesting okay

That helps a lot thank you

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[torn jolt]

(12th grade, Prob. And Stats. Geometric distributions. Just wanting some explanation here. Why exactly does it come out with one? I'd expect zero? is the lesson saying that it cannot use that value because a probability of 1 is incorrect? or some other reason.

[brazen tulip]

Why would you expect 0?

I mean algebraically it comes out to 1.

[torn jolt]

yeah

[torn jolt]

but, why does it come out as one instead of zero? I know that zero behaves weirdly. (obviously. It doesn't exist!)

but if this formula calculates the geometric probability, why can't it handle zero?

I mean, you could edit the formula so that it excludes zero, i think, but i guess that's just uneccessary at this level of math? (would be more useful in a coding application.)

[brazen tulip]

What does that variable respresent?

If it were 1, 2, 3, etc. What is that variable representing?

[torn jolt]

number of trials before expecting a success.

hold on i was gonna post the reference formula.

[brazen tulip]

Not sure haven't taken this class yet 😂

Just saw the algebra and how it comes out to 0

[brazen tulip]

Oh

[brazen tulip]

So why would you select 0 students?

You can't select 0 students

[torn jolt]

Okay, but why would that be different than a binomial distribution?

Because the binomial distribution can do it just fine.

if you do zero trials, you get an output of zero.

and not one

[brazen tulip]

Yeah I'm not the person to ask mb

Sorry about that

[torn jolt]

No problem! thanks for trying.

[brazen tulip]

Good luck though

[torn jolt]

I needed to post this info anyways lol

[torn jolt]

double checking this btw

[torn jolt]

k is number of trials, p is probability

binomial distribution

okay so i think i figured it out. Because a factorial of zero (required in binomial distribution.) results in one (weird af...), and the p(k) (k=0) is still 1, but you're subtracting 1. it all comes out to 0? So eli might have had the right answer, it doesn't produce a valid answer with Geometric because i'm drawing a sample of zero, a binomial distribution is actually in a way worse because i can't have a sample of zero, but i CAN choose 0 of a sample

Is this correct?

i think this confirms it. (Geometric)

could i get some confirmation on that?

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@torn jolt Has your question been resolved?

[torn jolt]

<@&286206848099549185> been 30 mins, just need a quick confirmation to make sure I understand the topic.

ah. it's not incorrect at all with binomial. It's spot on. the function DOES work with 0. That's why I needed confirmation. Because when you have nCk, the chances of you not getting a success from a population of 0 is 100%, or 1. However, in geometric, it doesn't actually work

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[wicked cliff]

help

[wicked cliff]

how i can find e and f

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[wicked cliff]

Note that if you divide the zeros of the function by 2 it will give you the maximum and minimum function.

It is always so?

x = -pi/3 , 0 , pi/3 are zeros

but if you divide to 2

you get maximum and minimum

[pallid coral]

since -1<= sin(3x+pi) <=1, you can get -5 and 5 as the minimum and maximum

[wicked cliff]

is -pi/6 and pi/6

[pallid coral]

yeah those are possible points

[wicked cliff]

how

[pallid coral]

wdym how

[wicked cliff]

ohh i see

but how i can calculate possible points

[pallid coral]

hmm

so normally sin has a period of 2pi right

but since we have sin(3x+pi) the period would be 2pi/3

the possible x coordinates for the minimum would be -pi/6 + 2kpi/3 for k integer, and the maximum would be pi/6 + 2kpi/3 for k integer

[wicked cliff]

oh okay

you can determinate period?

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[late storm]

I'm having some trouble making up a function that fits my data. I am writing code where the variable T starts at 10, and decreases to 0.1. Close to 0, it decreases rapidly, and close to 0.1 it decreases slowly. So in a sample size of 10 it should look something like:
10, 6, 4, 3, 1.4, 0.7, 0.4, 0.2, 0.12

[late storm]

I have been thinking for quite a while but I'm stuck

this is my best answer but because it is defined recursively, the only time both sides of the equation are satisfied is at exactly 1 point (so T doesn't change as it should)

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@late storm Has your question been resolved?

[uneven ingot]

so the difference

of the that sequence is what

how does it decrease

any reference

@late storm

[late storm]

I am basically running a loop for 1 million iterations

and T needs to go from 10 to 0.1

I dont have any other reference

I need to make up a function that fits this

@uneven ingot

[uneven ingot]

okay

1 million iteration

you can use for I in range

and if the number gets to a point

u can minus it by smaller number

that's one logic u could use

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[worn creek]

Okay ik it's very ez but help .

Which of the following is greater
√13 - √12 or √14 - √13

PS ik that the first one is greater but why and pls do some steps so as to why it is the ans....

Pls help this Lil niga

[uneven ingot]

take smaller case

hol on

[thick hedge]

check the nature of $\sqrt{x+1}-\sqrt{x}$

[glossy valve]

ƒ(Why am. I here)=I don't know

[uneven ingot]

😂

[worn creek]

Wtf is goin on

[uneven ingot]

if id be a

guy solving

[uneven ingot]

this is the type of thing I'd do

idk if u can use calculator

but try small values

[worn creek]

No calculator

[thick hedge]

I meant using calculus

[uneven ingot]

like 1 2 3 4 5

oh

[uneven ingot]

he got cool idea idk how he'd do it, learn him

[worn creek]

That's how I knew that first one is greater since √2 - √1 > √3 - √2

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[torn jolt]

trying to find the cricial points of this

[torn jolt]

I found the derivative to be 2sin(x) - 1

the points I found are 0, pi/6, and 2pi

There is a fourth point but I can't figure out how I am supposed to find it

[stiff musk]

solve 2sin(x) - 1 = 0

[torn jolt]

@stiff musk

why you keep deleting messages

[stiff musk]

i deleted one wrong message

[torn jolt]

oh that was someone else who did it a second ago

nvm

anyways, i solved it and i got pi/6

[stiff musk]

that's one solution, there's another

sin(x) = 1/2

[torn jolt]

how is there two solutions?

[stiff musk]

there's a solution in the 2nd quadrant

[torn jolt]

oh

yeah s

yeah so

0 1 2 3 4 3 2 1

0 30 45 60 90 135 150 180

so its

180 degrees

no

[stiff musk]

no

sin(180 deg) = sin(pi) = 0

[torn jolt]

150?

i gotta write this out

[stiff musk]

yea

5pi/6

[torn jolt]

see how i did this

sin goes in pattern

0 1 2 3 4 3 2 1 0 -1 -2 -3 -4 -3 -2 -1 0 1 2 3 4 3 2 1 0 etc

square root of that divided by 2

eh its stupid though

is there a better way?

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[turbid stone]

Does this proof seem correct?

[inner bobcat]

I think so. Since the bisector of AC guarantees that the angles are equal. Since the angles are equal and the lengths BA and AD are equal, respectively, the distance from C to B and D are also equal. So we get that CD=CB, <1=<2, <3=<4 => triangle ABC is equal to triangle ADC.

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[marsh ridge]

Looked everywhere online and couldn't find anything on
$\sum_{n=1}^{N} \left( 1 + \frac{1}{n} \right)^n$

[glossy valve]

7aman

[marsh ridge]

Anything would be greatly appreciated

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[hard kestrel]

How do you show that the integral goes to infinity?

[fathom saddle]

cos(0)/0 = 1/0 so you know this is an improper integral, at least

[ivory cairn]

probably l'hopital and FToC

[hard kestrel]

yeaah, but like, I need to have a infinity/infinity-expression before I use L'hopital

[ivory cairn]

change x to dividing by 1/x

[hard kestrel]

Yes

But the integral

[hard kestrel]

So just because the integrand is that, it must be improper?

[fathom saddle]

ye

But I get what you're saying - integral might still be finite. How do we prove it isn't?

[ivory cairn]

FToC

[fathom saddle]

There's no need to take the derivative at all if the integral is finite

[hard kestrel]

yeah exactly

[hard kestrel]

Yup, but how can I take the derivative (from l'hopital) without first proving that we have an 0/0 or inf/inf expression?

Ah wait

[ivory cairn]

standard improper integral methods don't work? (honest question, i haven't thought too much about it)

[hard kestrel]

$\frac{\cos x}{x^2} \leq \frac{1}{x^2}$, and since $\int_0^1 \frac 1{x^2}$ diverges, the one with cos must also diverge. Isn't that enough to argue that the integral approaches infinity?

[glossy valve]

Michael

[hard kestrel]

Not sure myself

[ivory cairn]

you need the integrand to be smaller than cos(x)/x^2, which you can still find. it'd just be a constant multiple of 1/x^2

[hard kestrel]

Ah yeah

Which constant multiple can that be?

[ivory cairn]

note that cos(x) is decreasing on the interval (0,1), so cos(1) < cos(x) for all x in (0,1)

[spiral vine]

Have you worked with non elementals integrals before? Like sinx/x

[hard kestrel]

Ah yes, right, and then since the integral of cos1/x^2 diverges, cosx/x^2 must also diverge

[hard kestrel]

Hmm no, don't think I have

[spiral vine]

Cause if you did you can integrate by parts

And then calculate the limits

[spiral vine]

Fair enough

[hard kestrel]

or what does it mean that it is "non elemental"?

Have never heard of that term at least

[spiral vine]

Improper

Might be in english

[hard kestrel]

Ah right, yes, I have worked with improper integrals

[spiral vine]

So IBP

u=cosx; v=1/x^2

You end up with -cosx/x - integral of sinx/x

With t variable sorry

[hard kestrel]

Oh alright

[spiral vine]

-cos(t)… etc

[hard kestrel]

yeah no worries

[spiral vine]

Now the second integral is Si(t)

[hard kestrel]

Oh um, well that I haven't worked with before, no...

[spiral vine]

So u have [-cos(t)-Si(t)] from 0 to 1

[hard kestrel]

Aren't improper integrals just that they approach infinity/aren't defined?

[spiral vine]

Ok my english is not the best

I will look for translation

[hard kestrel]

Yeah I'm not english either

but again, no worries

[spiral vine]

Non-elementary was the name

[hard kestrel]

Oh okay, well in that case. No, haven't worked with them (except when differentiating the integral containing those functions)

[spiral vine]

If you want you can study that and then calculate the limit with this

$\int_{a}^{b} f(x) , dx = F(b) - F(a) = \lim_{x \to b^-} F(x) - \lim_{x \to a^+} F(x)$

Wait

[hard kestrel]

Yep, I can take a look at it

[proven mango]

honestly we can just use lhopital’s rule

since the integral diverges to infinity

and if we change x to divided by 1/x

then we’ll have infinity over infinity

[hard kestrel]

Yes, I know

Just wanted to listen to Mercurial's method also

[proven mango]

alright

[glossy valve]

Mercurial

[hard kestrel]

But thanks still!

[spiral vine]

No problem, this one takes a bit for me to write here, but this was the main idea

[hard kestrel]

No worries 👍 But alright, I see!

I'll take a closer look at these non-elementary integrals, so I can probably try to do both methods

Was it anything more you wanted to include, or shall I now close the chat?

[spiral vine]

You can close, cause i have a lot to say about this and i dont think is the moment to do

[hard kestrel]

Okay, thanks again

.close

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[full forum]

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[torn jolt]

any ideas ?
ive found the slope of the rad
idk what to do next

[torn jolt]

i have a slope and a center
and need to find x and y values

[slow canopy]

Ma’am let me ask my sir @quaint prawn.

[torpid pawn]

equatuon of slope =【 √（x2-x1）²+（y2-y1）²】
so u just open the quadratic then you will get the equation of a circle

[torn jolt]

thats the distance formula

slope is
y2-y1/x2-x1

[torpid pawn]

opp sorry

but your circle radius must be same right？

so u just find the shortest distance between the tangent and the centre of circle

[torn jolt]

radius is unknown

[torpid pawn]

use formular（ ︳ax+by+c︳）÷（√a²+b²）

[torn jolt]

ohh

make sense

[torpid pawn]

it can help u to find the shortest distance between the circle and tangent and your ans is radius

then use the distance formula ，your ans is here

[torn jolt]

so from the prep distance and the center we find the eq ?

[torpid pawn]

yep

[torn jolt]

thank you

.close

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[simple brook]

Penelope is 36 years old. She noticed that the sum of her age and here father's age is 5 times the difference the difference in her age and her father's age. Find Penelope's Father Age.

[simple brook]

Fred placed 19 blue marks on a pole that divided 20 equal sections. Karen placed 16 red marks on the pole that divided the pole into 17 equal sections. The distance between adjacent blue marks is m perecent less than the distance between the adjacent red marks, where is M is a postive integer Find M.

[full forum]

@simple brook Has your question been resolved?

[simple brook]

.close

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[jade radish]

What is a partition of a group??

[jade radish]

this is all the info about this we were given

When I think partition, I think parts of something

so I mean, yes cosets of H are clearly parts of G, but is that it

?

[kindred grove]

well they are parts of G yes

partition is a bit more specialized

[jade radish]

like cosets of H are just elements of G composed with elements of H ( which are from G )

[kindred grove]

I'll start with a picture

[jade radish]

and so it'll still be within G, since G is closed under compsoition

[jade radish]

okay

[kindred grove]

let's say there's 3 cosets of H in G

[jade radish]

hm

[kindred grove]

saying that "the cosets of H partition G" means that

• the cosets have no elements in common
• the union of all the cosets covers all of G (or in other words, everything in G belongs to a coset)

here we're talking about cosets of a subgroup, but partitions are also a thing just w/ plain sets

[jade radish]

hmm

I see

[kindred grove]

https://en.wikipedia.org/wiki/Partition_of_a_set for reference

[jade radish]

This makes sense

Thanks

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[untold vessel]

I have no idea how to do this, and the hints don't help me much

[sacred sparrow]

Have you had tangential line integrals before?