help-10

[robust sleet]

Do you have a question, or you opened this channel just to say that?

[mellow wraith]

What channel was your doubt before in 💀

[robust sleet]

some simple equations where u just had to use (x+a)²

to equate both sides

[obtuse pebble]

@hexed stirrup Has your question been resolved?

[obtuse pebble]

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[hollow sail]

Hi can someone point out where I got wrong for 28e

[tawny fog]

Don't see any mistakes there

,w d/dx(x³ln(cos²(x)))

[warm shale]

Results provided by WolframAlpha

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Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[tawny fog]

That also simplify into this

[obtuse pebble]

@hollow sail Has your question been resolved?

[frosty pendant]

Your teacher simply pulled out the ² from the ln so it becomes ln(cos²(x)) = 2ln|cos(x)|

You're correct

[obtuse pebble]

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[mellow wraith]

is G(1) minima

Ofc it is

[shell prairie]

so k=ln2

[obtuse pebble]

@shell prairie Has your question been resolved?

[obtuse pebble]

@shell prairie Has your question been resolved?

[obtuse pebble]

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[leaden jetty]

Integers (I think)
2 +(1/(a +1/(b + c/d))) = 61/23
a,b,c,d are positive and integers
what is the smallest possible a+b+c+d?

[leaden jetty]

text is from another language but I translated it

[tired skiff]

you can turn 2 into 46/23

[leaden jetty]

btw answer is 17

but I found 10

[tired skiff]

what were your a b c d values

[leaden jetty]

I found
a=4
b =1
c=2
d=3

[tired skiff]

that's incorrect unfortunately

[leaden jetty]

yeah

I'm told that the answer is 17 but I dont know why

2 = 46/23
I need 15/23 but there is 1 not 15 in the top so 23/15 must be the bottom

then a =1

and the top should be 8 but it is 1

so the bottom should be 8/15

[visual plover]

try manipulate the denominator using reciprocals

[leaden jetty]

wow english please

lol

oh okay I got what you mean I think

[visual plover]

1/(1/a) = a

that's just an example

careful with your denominator

[leaden jetty]

I was trying to do the same thing

I didnt know the word "denominator" so I used bottom

[visual plover]

you know what let me try it out now

[leaden jetty]

okay

[nocturne sun]

so after writing 2 as 46/23 as g4j said, you can move it to the other side and cross multiply

15a + 15 / (b + c/d) = 23

this is what you get

[leaden jetty]

okay

[nocturne sun]

since a, b, c, d are positive integers, the fraction can't be negative and a must be 1

that leaves

15d / (bd + c) = 8

since 15 has no factor of 2, we want d = 8 and want the denom. to cancel the 15

so bd + c = 15

d was 8 so 8b + c = 15

(1, 7) is the only positive integer solution to this

so a = 1, b = 1, c = 7, d = 8

[leaden jetty]

You write 61/23 as 2 + 15/23, and the 2s in the equation go from there. You have 15/23 left on the right, but 1/ on the left, so you write 15/23 as 1 over 23/15. So the expression a + 1 over b + c/d is 23/15. You write this expression as 1 full 1 divided by 15/8 and it comes out as 1 because a is the whole part. b + c/d became 15/8. You write it as 1 full 7/8. From here, we get b:1, c:7 and d:8. The answer is 17 from the sum of 1+1+7+8

I think this is the answer

oh

okay thank you

[visual plover]

I got

oh wait

you guys solved it

lol

nice question

[leaden jetty]

lol

thanks for helping

I just realised something about maths

every country has their own type of math questions

so some really easy math questions is different for other countries

and other way around

anyways,

thank you both for helpinh

bye

!close

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[limber eagle]

i don't understand the question

<@&286206848099549185>

.close

[obtuse pebble]

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[spark shore]

I was told to determine $\alpha,\beta,\gamma,\x ( \xi \in]0,1])$ so that the quadrature formula $\int_{-1}^{1} f(x),dx \approx \alpha f(-\xi)+\beta f(0)+\gamma f(\xi)$ is exact for all polynomials with the greatest possible degree n

[warm shale]

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

planeboi

[spark shore]

but I'm thinking since we're only working with 3 points and two sub-intervals n won't be that big anyway

[tardy epoch]

Sounds hard

[spark shore]

oh yeah! but we have the interval and we have h so that should make things easier

creating a system for the error function for each degree would work too no?

[tardy epoch]

Did you already figure it out for f= x^n

[spark shore]

I did it for n=2k and n=2k+1 and it formed a non linear system I wasn't able to solve

[obtuse pebble]

@spark shore Has your question been resolved?

[trail musk]

Willing to bet that n is relatively small

No more than 2

In fact I'd wager on 1

[spark shore]

I figured it out!

just need to use lagrange base

and calculate weights using this formula: \ $A_i ^{n} = A_{n-i} ^ {n} = \frac{(-1)^{n-i}h}{i! (n-i)!} \int_{0}^{n} \Pi_{j=0_{j\neq i}}^n (u-j)du$

[warm shale]

planeboi

[obtuse pebble]

@spark shore Has your question been resolved?

[obtuse pebble]

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[prime kraken]

find the domain of the function with equation y = 4x/ln(-4x-4)

[prime kraken]

i dont have any work

i just dont get what to do and how

[tardy epoch]

https://youtu.be/djT6-YamHaA

[prime kraken]

Thanks

.close

[obtuse pebble]

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[tardy epoch]

https://youtu.be/WfgYJy035k0

[obtuse pebble]

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[spice chasm]

guys

[spice chasm]

what i do wrong

[spice chasm]

i put 3 as the base to cancel the log

[dark stirrup]

oh wait, I missed that you removed the logs

oh still ain't right

[spice chasm]

yea idk what i did wrong

the answer is x=5

[dark stirrup]

a*log(b)=log(b^a)

[spice chasm]

Why can’t i remove the log by making 3 the base

[warm shale]

Mehdi_Moulati

[spice chasm]

oooh

but don’t you use the exponent properties tho

[dark stirrup]

You use this exponent property

[spice chasm]

oops i used the wrong property

[dark stirrup]

The one you just deleted is wrong

[spice chasm]

yea xd

ok ty

.close

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[nimble harbor]

why does the inequality turn into an equality?

[trim locust]

I believe it is because x=11/4 and -0.25 are the critical value, it’s the value from which the function changes signs, it’s easy once you sketch the graph to find the inequality from these 2 points

[obtuse pebble]

@nimble harbor Has your question been resolved?

[trail cloak]

Also

After finding these values, you still have 2 inequalities to write

To show the values of x that keep y negative

[obtuse pebble]

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[sly sleet]

At the Willowbrook Country Fair, Camille will pay an entrance fee in addition to any tickets she buys for rides. The total amount Camille will pay depends on how many ride tickets she buys.
This situation can be modeled as a linear relationship.
What does the y-intercept of the line tell you about the situation?
A. if camille buys 5 ride tickets, she will pay $25 in all
B. the entrance fee is $15
C. each ride ticket costs $2
D. camille will buy 5 ride tickets

[obtuse pebble]

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[cursive arrow]

Q: x-y=12 and xy=6. Solve for 'x' and 'y'

[cursive arrow]

I really can't solve this equation pls help

[atomic bobcat]

or put x=12+y in xy=6

and solve quadratic equation

😂

[cursive arrow]

okay so it's pretty simple then

I'm just dumb lol

thanks!

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[dark stirrup]

what have you tried

[pulsar breach]

i integrated to find velocity

and then made the velocity = 0 and subbed in t=1 to find C

[dark stirrup]

cool

then what

[pulsar breach]

i integrated again to find displacement

but idk what to do with it

[dark stirrup]

wdym

[pulsar breach]

how do i use the t=3 to find out x

[amber meteor]

Hint: stationary means v=0, at the origin means x=0

[dark stirrup]

You need to solve for your integration constant first

[pulsar breach]

i solved to find C and got -1/2

[pulsar breach]

isnt this same thing

[dark stirrup]

I'm assuming this is your integration constant for your first integral

[pulsar breach]

yes

[dark stirrup]

You have one more integral to go

[pulsar breach]

i integrated again and i have another constant?

[dark stirrup]

Yes

Every time you integrate

[amber meteor]

No, stationary means not moving, origin means the starting point

[dark stirrup]

You get a new constant

[pulsar breach]

do i make the displacement equal to 0 and sub in t=1 to find the new constant

oh

yes

i got the answer

ty

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[lavish crow]

What's the deal with this?

[obtuse pebble]

Please don't occupy multiple help channels.

[lavish crow]

i don't get what to use to solve this

probably a natural logarithm, but how

just to multiply each side by ln?

[lost tree]

what do u want to solve for

[lavish crow]

this G, i know S

but i don't know how to solve this e to this power

especially not how to solve it on a calculator even

[lost tree]

if alpha=0.6 and mu=0.5, then alpha*mu=0.6*0.5=0.3

so e^(alpha*mu) = e^0.3

?

[lavish crow]

but how do i calculate that on a calculator, can i put 2.71 to the 0.3

or do i need to use e

it's a really dumb question

but had to ask someone

[lost tree]

put e to 0.3, if ur calculator has a button for e

ig

[lavish crow]

okay thank you

i'll try 😂

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[flat lodge]

Determine whether the points A(5; -1; -2), B(-1; -2; -5) and C(-2; -8; -2) are in an equilateral triangle
ABC vertices.

[flat lodge]

I just don't know where to start to complete this task

[royal basin]

do you know what an equilateral triangle is?

[gleaming ridge]

Use the property of equilateral triangle

[flat lodge]

i know how it looks

[royal basin]

that doesn't answer my question

do you know the definition of the word "equilateral triangle"?

[flat lodge]

sorry no

[royal basin]

an equilateral triangle is, by definition, a triangle in which all sides have the same length.

[flat lodge]

ohhh

[royal basin]

in some languages other than english, the word for "equilateral" even translates to "equal-sided" or similar.

that's kind of also the case in english, except it's obscured behind the fact that it's a word made of Latin roots.

anyway, do you understand how to proceed?

[flat lodge]

thanks for explanation what is a equilateral triangle but i still don't know how to do my task

[royal basin]

well

again

an equilateral triangle is a triangle in which all sides have equal length.

you have the coordinates of points A, B and C.

[atomic bobcat]

do you know distance formula?

[royal basin]

do you know how to find the side lengths of your triangle?

for example, if i told you to find the length of AB, would you be able to do it? @flat lodge

[flat lodge]

to find the lenght of ab i need to use
Pythagorean theorem?

[royal basin]

in some sense, yes.

do you know in general how to find the distance between two points?

(and do you understand that the length of a line segment and the distance between its endpoints are one and the same?)

[flat lodge]

yes

[royal basin]

ok,

then you should have no trouble calculating all three side lengths for ABC and telling if they are equal or not

[flat lodge]

so wait my first step is to find lenght of AB

[royal basin]

sure if you want

it does not matter which order you do the sides in

if you want to find BC first then you can do that

[flat lodge]

and i will know all lenghts of tirangle because the lenght of a line segment and the distance between its endpoints are one and the same

[royal basin]

...why don't you just do it now?

[warm shale]

Artutu

[outer maple]

(x, y, z) is a vector and not a point

[flat lodge]

oh so i use this formula on AB

[royal basin]

@outer maple careful not to cause any more confusion here.

[royal basin]

again. just do it.

[flat lodge]

ok

i got √58

close.

close

[trail cloak]

Hmm

Are you sure?

[flat lodge]

i don't know i think my answer is wrong

nvm i will ask again later i think i know how to do my task

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[obtuse pebble]

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[tiny flicker]

No.3

[tiny flicker]

I just can't understand the question

Here's an example

[timber fox]

do you know that $\sqrt{x}=x^{\frac{1}{2}}$

[warm shale]

Clarkie

[tiny flicker]

Yes

[gleaming ridge]

If $x = (-125)^{\frac{1}{3}}$ then this means $x^3 = (-125)$

[warm shale]

numbpy

[timber fox]

soo if you have x^(1/3) its not asking for the square root but theeee

[gleaming ridge]

So, you want to find a number where cube is -125

as clarkie said, it is basically the cube root of -125

[tiny flicker]

what

my brain is all knocked out

[fierce lagoon]

$x^\frac{1}{n} = \sqrt[n]{x}$

[warm shale]

Umbraleviathan

[fierce lagoon]

There

[timber fox]

^^^

[gleaming ridge]

Do you know how to solve $125^{\frac{1}{3}}$

[warm shale]

numbpy

[tiny flicker]

I don't think so

[raven knot]

So $125=555=5^3$

[warm shale]

SmokingPuddle58

[gleaming ridge]

okay, are there no examples like this in your book? There should be some lesson on radicals and surds

[tiny flicker]

3(weird shape) 125?

[tiny flicker]

at this point I don't even know what ur saying

Ignore the other language, I'm forced to learn it at school

[fierce lagoon]

Well you gotta know your perfect cubes

[tiny flicker]

Sir what cubes.

[fierce lagoon]

8, 27, 64, 125, 216

[gleaming ridge]

Do you know $\sqrt[3] {125}$?

[fierce lagoon]

Those are your perfect cubes

[fierce lagoon]

Forgot brackets

{}

[warm shale]

numbpy

[tiny flicker]

125 1/3

[fierce lagoon]

Perfect cubes are when you take the cube root of them, they output an integer.

1, 8, 27, 64, 125, 316 are perfect cubes. If you take the cube root of each of those numbers, you get:

1, 2, 3, 4, 5, 6

Memorize these because they'll appear a lot more often

[rotund canyon]

1^3, 2^3, 3^3, 4^3, 5^3
anything to the power of the three is a cube

[tiny flicker]

Oh yes that

I'm sorry guys I'm trying my best to understand its 1am here😭

[tiny flicker]

Yes I remember that

I'm studying index rn

[fierce lagoon]

You know what the 3 means here?

[tiny flicker]

1/3

[gleaming ridge]

yeah it's the cube root

[fierce lagoon]

Well ... it's basically indicating cube root

And so

[gleaming ridge]

alright, so if you do $3^3 = 3 * 3 * 3 = 27$ then it's reverse is $\sqrt[3] {27} = 3$

[warm shale]

numbpy

[fierce lagoon]

$$\sqrt{x} := \text{square root of x}$$
$$\sqrt[3]{x} := \text{cube root of x}$$
$$\sqrt[4]{x} := \text{quartic root of x}$$
$$\sqrt[n]{x} := \text{nth root of x}$$

[warm shale]

Umbraleviathan

[gleaming ridge]

Do you understand a bit @tiny flicker ?

[tiny flicker]

Got it

[tiny flicker]

Ig so

.close

[obtuse pebble]

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[vital rain]

is it mathematically acceptable to use this method to solve asymptotes? https://www.youtube.com/watch?v=qJrrZQgSkO8&t=181s&ab_channel=NancyPi

[vital rain]

because my teacher uses limits, and the method I found on youtube is much less tedious than doing limits?

[fierce lagoon]

Well I'm not gonna watch the video but I'm assuming they're comparing the numerator and denominator's leading term's powers, correct?

They're basically doing a limit ... through analysis instead of actually showing the limit

It's what I do anyways, and it can be proven using limits

[fierce lagoon]

@vital rain basically this

[vital rain]

basically this

but it should be mathematicaly acceptable in exams to do what she just did

[green epoch]

Yeah that IS the method

It's just the final result divided into 3 cases

[novel knoll]

No you shouldn’t do smth you haven’t done in class

[fierce lagoon]

I mean ... they kinda generalize a "prediction", for lack of a better term and my brain being deep fried

But generally stick to what your teacher shows you, and then verify with whatever tricks you'd like to use on your own

[novel knoll]

And why exactly is calculating limit so tedious? $$\lim_{x\to \infty}\frac{x+1}{x^2-1}=\lim_{x \to \infty} \frac{(x+1)/x}{(x^2-1)/x}=\lim_{x \to \infty} \frac{1+1/x}{x-1/x}=0$$

[fierce lagoon]

If you learned about properties of infinite limits, you know you can literally erase low order terms from the numerator and denominator

[warm shale]

ScapeProf

[fierce lagoon]

Perhaps your teacher showed you this?

[vital rain]

hm true, how would you do for vertical, she was saying that you should try do factor everything out, and then set denominator to = 0.

[vital rain]

this is what my teacher did

[novel knoll]

Limits like this

Divide through with leading term in num (or denom)

(Or anything in between those)

[novel knoll]

Like I did above I divided through with x

[vital rain]

that s how you calculated horizontal? how would you do for vertical?

[novel knoll]

Like you said above? What was wrong with that

[vital rain]

I don't know how would they do vertical asymptote on this using limits, without setting x^2-4 = 0

is it because vertical asymptote is based on the definition of the function?

[civic zealot]

sort of, just because the denominator is 0 at certain x values does not mean that there is a vertical asymptote.

for example $\frac{x^2-9}{x-3}$, the denominator is 0 at x=3, but there is no vertical asymptote

[warm shale]

Zybikron

[vital rain]

but you would need to factor out the function and then you set denom to 0

[civic zealot]

yes

[vital rain]

so we can't do denom = 0 because it is 1

but how can we use the limits to find out the vertical

[civic zealot]

use limits when you can't factor, or when you don't have a polynomial to reduce.

mostly just to verify that they are actually asymptotes

[vital rain]

but how would I use the limits to find the vertical asymptotes for this one? my teacher says x= 2 when f(x) --> + infinity and x = -2 when f(x) ---> - infinity

[elfin burrow]

the rule in the youtube video is an observation about the results that the limits give for various powers of the numerator and denominator

[civic zealot]

look at the left and right limit around x=2 and x=-2

because what you said isn't necessarily true.

[vital rain]

I said wrong

f(x) -->> goes to infinity when x=-2 or x = 2

but that is because there is due to the definition of the function

[civic zealot]

right

but you're saying f(x) ---> infinity means you're using a limit. You're saying lim_(x -> 2) f(x) = +/- infinity

[vital rain]

yes true

but that means if we have (sqrt(2-x)/(x-4)) , this is our vertical asymptotes, 2 and 4 right?

[civic zealot]

basically you're saying 'by the definition of the funciton' because you learned that in algebra. Now in calculus you're learning the reason that it's a rule is because of the limits

[civic zealot]

no, there aren't any vertical asymptotes in that function

[vital rain]

there are for sure vertical asymptotes?

[civic zealot]

ok, what happens around 4?

[vital rain]

it is 0, but that's the thing when x -- > 4 the function = infinity

[civic zealot]

that would require 4 to be in the domain of the function

sqrt(2-x) isn't defined (over the reals) for x>2

[vital rain]

true

but if we had

(sqrt(2-x)/(x-1)) would our vertical asymptotes 2 and 1?

[civic zealot]

just 1

[vital rain]

why not from sqrt(2-x)

[civic zealot]

what happens when the numerator is 0?

[vital rain]

0

[civic zealot]

ok, doesn't sound like an asymptote to me

[vital rain]

but I thought of vertical asymptotes is for when x is undefined

when x = 2 it is undefined

f(x) -->> goes to infinity when x=2

[obtuse pebble]

@vital rain Has your question been resolved?

[obtuse pebble]

@vital rain Has your question been resolved?

[nova stump]

sqrt(2-2) = 0

f(x) doesnt go to infinity when x=2, nor does it approach infinity as x approaches 2

[obtuse pebble]

@vital rain Has your question been resolved?

[golden cave]

Could someone help me with a question on combinations?

[obtuse pebble]

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[silk elk]

Two identical circular sheets must be glued together at their edges and blown up to form an ellipsoid balloon, without stretching. the ellipsoid must be 132.2 units in diameter across, and 80.6 units tall when fully expanded.

Find the neccesary radius of each sheet to fulfill this criteria.

[silk elk]

not sure how to go about solving this problem, my professor posed it to us as a challenge for our research topic over the break. any ideas?

[dark stirrup]

What are you trying to find

Oh

I see it

[silk elk]

the radius of each sheet before they're blown up

yeah

[dark stirrup]

Sorry I'm blind

[silk elk]

nw

lol

any clue how?

[dark stirrup]

Hold on

So when it say circular sheet, does that mean like a hollow cylinder? Like this?

Or is it just a flat circle?

I'm assuming the latter

[obtuse pebble]

@silk elk Has your question been resolved?

[silk elk]

the latter

[wooden cipher]

make an ellipse, then use calculus to find surface of revolution

[dark stirrup]

You can't really find the analytic surface area of an ellipse though

Also, how does it get stretched?

[wooden cipher]

it doesnt

oh wait

[dark stirrup]

Does the joined seem stay a circle, or does it deform to an ellipse?

[wooden cipher]

yeah i think so

[dark stirrup]

Approximations may be allowed since the values are just given to 1 decimal

[wooden cipher]

oh if youre just trying to find the radius, you can find arclength

[dark stirrup]

surface area of an ellipse can be approximated to ~1%

[wooden cipher]

calculators are allowed right?

[dark stirrup]

They have to be

I would never multiply by pi without a calculator

I'm assuming the seem stays circular

And I'm gonna guess the ellipsoid vertically since it mentioned how tall it must be when expanded

So that means the sheets will be horizontal

yeah what class is this for @silk elk, what are you allowed to know?

Cuz this is gonna be a lot of approximations

[silk elk]

approximate anything

it isn't for a class, it's a puzzle related to a research project my group was working on

[obtuse pebble]

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[dark stirrup]

.reopen

[obtuse pebble]

✅

[dark stirrup]

Well then best advice I can offer is this:

The surface area of the ellipsoid must be equal to the area of the two circular sheets

So find the surface area of the ellipsioid

You're given the length of each of its axes

[obtuse pebble]

@silk elk Has your question been resolved?

[obtuse pebble]

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[bold siren]

I need to figure out how to find the highest common factor and lowest common multiple of 2970, 4950, 6930

I have the answers but i dont know how to get to them

[tranquil sonnet]

Prime factorize, and compare factors

[loud tangle]

pain to do but yeah

[bold siren]

Yes i dont know how to do that

[nocturne minnow]

Can you prime factorize something like 40?

[tranquil sonnet]

It’s like how 6=3*2

[bold siren]

I mean yea, but i dont see how to do that to a number as big as 2970

[tranquil sonnet]

Do you know divisibility rules?

[loud tangle]

if you really dont want to could also do euler's alg for lcd and use the fact that lcd*hcm = number productes

[nocturne minnow]

A trick, just keep dividing by 2

[tranquil sonnet]

You could use them to see that it’s divisible by 9 and 2

[loud tangle]

or lcm * hcf

[nocturne minnow]

Then if it's not even, divide by some other prime, like 3 or 5, etc

[loud tangle]

you can take 10 out of all of them at the very start then divide by 3

since it works

by 9 actually

all of them

[tranquil sonnet]

Do you mean Euclidean alg?

[loud tangle]

euclidian sry

the one that gives lcm

i didnt scroll over computer science in a while and i forgot

[bold siren]

Never heard of it

[loud tangle]

its an algorithm that gives the lcm of any numbers

not lcm hcf

ugh i hate this stuff i always mix the terms uppp

[bold siren]

Not sure, the name doesn't ring a bell

[loud tangle]

here we say highest common divisor and lowest common multiple not lcm and hcf and jesus christ it gets confusing with abbreviations

[tranquil sonnet]

Just keep dividing by 2 and once it becomes odd, divide by 3 or 5 like dldh said

[loud tangle]

ik for sure you can divide by 2 and 5 once and 3 twice on all of them

[bold siren]

OH

[loud tangle]

so hcf has to be a multiple of 90

[bold siren]

Ok i can kind of see that in the numbers now

Uh

Ok that works until i hit 55

[nocturne minnow]

Divide by the next prime that goes into it

[bold siren]

So 5

[nocturne minnow]

Yes

[bold siren]

I see

Omg 3 hours searching the web and 5 min on this server answers everything

[loud tangle]

do say when ur finished

[bold siren]

Im just checking things to be sure

It says a mark is awarded for showing no common factors

But that seems wrong?

[loud tangle]

of 3 5 and 7 yeah

theyre coprime

[bold siren]

Well thats a "duh" why would they award a mark for stating something like that...

[nocturne minnow]

To make sure you can identify that

[bold siren]

I feel like im missing something...

[loud tangle]

it does say show

how woule you prove that

3 5 and 7 are prime in general meaning their only divisors are 1 and themselves

so their highest common factor/divusor would be 1

which stands for every number

[nocturne minnow]

Also, it's pointing out those 3 values, since 3, 5, and 7 are alone in each column, hence no common factors, if you were talking about the HCF part

[bold siren]

Ok im not quite sure why those numbers are alone in each column?

[nocturne minnow]

It just organized the factors from here

So 2970 = 2 * 3 * 3 * 3 * 5 * 11

And it just lined up all the factors

[bold siren]

So becouse 2970 is the only one that has 3x3x3 4950 has 5x5 and 6930 has 7x

You can tell no common factor?

[nocturne minnow]

Do you see how, here, it lists out the factors of each number?

[bold siren]

Yes

[nocturne minnow]

When you line each factor up, you will notice that 3, 5, and 7 are alone, like this

[bold siren]

Ok?

[nocturne minnow]

Since 3, 5, and 7, aren't in all three columns, then those aren't common factors to all three numbers

[bold siren]

Even tho all 3 numbers are all rationally divisible by 3?

[nocturne minnow]

2970 = 2 * 3 * 3 * 3 * 5 * 11
4950 = 2 * 3 * 3 * 5 * 5 * 11
6930 = 2 * 3 * 3 * 5 * 7 * 11

Do you agree?

[bold siren]

Yep

[nocturne minnow]

If you line up each factor, for each number, so the column is filled, it would look like
2970 = 2 * 3 * 3 * 3 * 5 * 11
4950 = 2 * 3 * 3 * 5 * 5 * 11
6930 = 2 * 3 * 3 * 5 * 7 * 11

Do you agree with that?

[bold siren]

Yes

[nocturne minnow]

Do you notice how, there is a 3, 5, and 7, in each column that only has one in each?

[bold siren]

Yes

[nocturne minnow]

Like there is a column with only one 3, one 5, and one 7

Because there is only one in each column, those three factors are only unique to each number

Because of the column that only has the one factor in each, that's why the key says 3, 5, and 7 aren't common factors to all three numbers

[bold siren]

Ok i think i understand that part

Can you explain what this part means?

[nocturne minnow]

The factors 3, 5, 7, just multiply two of the together in all the possible combinations

Like 3 * 5, 3 * 7, and 5 * 7

[bold siren]

?

[dark stirrup]

Late to the party, but did anyone mention the gcd algorithm?

If a>b, then gcd(a, b)=gcd(b, a-b)

You can also do remainders if you're good at division

[bold siren]

->'

Division rules work better thx

[nocturne minnow]

Tbh can't really explain that part sorry

[bold siren]

Ok, whats special about 35,21 and 15 that this is pointing them out

[dark stirrup]

no clue

[nocturne minnow]

Well, what I'm assuming it's their method to take 3, 5, and 7 and multiply two of those digits in all the possible combinations, then stating that those 3, aren't all factors 2970, 4950, and 6930

[bold siren]

Ok? Weird...

[dark stirrup]

Yeah that's nasty

[nocturne minnow]

Probably just another extra sanity method that is pointless, the first one was good enough, imo

[bold siren]

Well if it gets me a mark who am i to complain

[dark stirrup]

I'm curious

What method did you use?

Just get the prime factor of each number?

[bold siren]

Thats what this wants so ima do it this way

[dark stirrup]

for sure if that's what the assignment wants

[nocturne minnow]

I believe just dividing by 2, then the next divisible prime

[dark stirrup]

Not great but I can't change that

[bold siren]

What about this question

A2

[dark stirrup]

Looks like you need to know how exponents multiply and divide together

[bold siren]

Exponents?

As in the "^5" and "^15"?

[dark stirrup]

Yeah

Do you know how to multiply two exponents with the same base?

[bold siren]

Uh... No

[dark stirrup]

Okay well that's what this question is asking

So you will learn

Think of 2^3

That is 2*2*2

And think of 2^4

That is 2*2*2*2

So what is (2^3)*(2^4)?

Why, it's (2*2*2)*(2*2*2*2)

Oops I wrote this wrong. Deleting it

I meant this

[bold siren]

Well yea, i know what powers are, but i thought it would be a simple case of applying the powers then multiplying the result

[dark stirrup]

Nah too much work.

I don't think a calculator could reliably do 7^15 anyway

The trick here is to know how to group exponents together

[bold siren]

Ok

[dark stirrup]

Yeah I'm sure you know what powers are, I'm using that knowledge to help you understand how to multiply to powers together

Look at the 2^3*2^4 example

You multiply 2 three times, then again 4 times

So overall, you multiply 2 seven times

[bold siren]

So its just 2^3 +2^4

To give you 2^7

[dark stirrup]

I think you made typo here

[bold siren]

Ah yes

Ment x

I just mean your simply adding the powers together

[dark stirrup]

Yes

In general

[bold siren]

I see

So for the question

7^20\7^16

Would be 7^4

Ey

What about the 2nd part where none of the bases are the same

[dark stirrup]

You need to break each exponent base (15, 3, 2, 10) into its prime factors

[bold siren]

n>`

[dark stirrup]

As an example, consider 6³.

6³=(2*3)³
  =(2*3)*(2*3)*(2*3)
  =(2*2*2)*(3*3*3)
  =2³*3³

[bold siren]

Oh that's weird

Makes sense tho

[dark stirrup]

yup

In general

[bold siren]

So what are the actual steps then, givin 15^4x3^6 / 2^10 x 10^4

I can see 15^4 = 3^4 * 5^4

But that still leaves us with a 5 on top

[dark stirrup]

That's fine. Is there something saying you need to remove the 5?

[bold siren]

Well i can see the answer

I just have no idea how to get to it

"=2^(-14)x3^(10)

[dark stirrup]

you've done 15, but what about the other numbers in the fraction?

[bold siren]

Ok so

Why does it become 2^(-14) * 3^(10) tho

Is it just how you write it in the format?

[dark stirrup]

It comes from how how define a^0

Remember that we found (a^b)*(a^c)=a^(b+c)

So assume b=0

Then we should have (a^0)*(a^c)=a^(0+c)=a^c

So you get a^0=1

[bold siren]

Uh..

[dark stirrup]

It's okay, ask questions

[bold siren]

I just dont get why we are making b=0 and why a^0=1

[dark stirrup]

Let me format this, hold on

[cobalt acorn]

cuz a^0 means a^(1-1) right?

which also means a^1/a^1

and ofc its 1

for all a belongs to Real numbers without 0

[cobalt acorn]

cuz u cant divide by zero

[cobalt acorn]

@bold siren

[cobalt acorn]

website name sir? O_O

[dark stirrup]

Microsoft Word

[cobalt acorn]

tyy

[cobalt acorn]

OH

i thought its a website in google XD

[bold siren]

Wait i think there's been a misunderstanding

My question was why we are turning 3^(10)/2^(14) into 2^(-14) * 3^(10)

[dark stirrup]

yes

I was making my way there

[bold siren]

Is it that the latter is simply considered simpler?

[dark stirrup]

I just wanted to show you how a^0=1

That'll help get you there

[bold siren]

I feel it would be clearer if instead of using a and c you used numbers

[dark stirrup]

Sure let's try it with real numbers

Let's say we want to know what 2^0 is

We know 2^1=2

And we know that (2^0)*(2^1)=2^(0+1)=2^1

So it must be that 2^0=1

[bold siren]

Ok, kinda makes sense

[dark stirrup]

In short, you just need to know that a^0=1 and a^-b=1/(a^b)

That's it

I was just trying to give you the motivation behind it

[bold siren]

OH

OK WRITTEN DOWN IT MAKES SENSE

kinda

Not sure where the 3^10 from the question comes into this but it makes sense on why
/2^(14) = * 2^(-14)

Still not sure why you need to write it like that but whatever

[dark stirrup]

You're not sure where 3^10 came from?

[bold siren]

No more where it fits into the formula you just posted

[dark stirrup]

Since you already have the answer, I'll just show you

[bold siren]

Ok just wondering

Why invert the power

Is it litterly just that its considered more simple that way

[dark stirrup]

Either way is fine

I suppose it's a requirement of your problem. "form as powers of primes numbers"

[bold siren]

Ah

Ok

Well that's that then

Ty for your help c:

.close

[obtuse pebble]

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[valid dew]

so when x is 1, y will be 0

and when y is 2, x will be zero

okay

lets solve it

I will do it for you

oh right

wait

mhm

[nocturne minnow]

Don't solve the problem for people

[valid dew]

And why is that ?

you can derivate it for an easy way out

[nocturne minnow]

Because if you read #❓how-to-get-help

• When asking for help, do not insist on getting just the answer; we are here to help you learn, not do the work for you. Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer.

[valid dew]

but I guess you are a highschooler and you are expected to solve it without derivation ?

[nocturne minnow]

Meaning don't do the work for people

[valid dew]

Oh I get it dldh06, thanks!

Okay so you have a problem with it ?

So this solution looks right to me

oh okay.

[nocturne minnow]

Why not?

[valid dew]

Okay I have to figure out how to help you out without solving

[nocturne minnow]

Then sleep

And try again later

Here's the thing, lack of sleep is not good, especially before a test. You need the rest so you aren't fatigued

Better than 0, 1, or 2 hours

[valid dew]

1 hours of sleep is 1 hours of sleep

[nocturne minnow]

Yeah... that's not really healthy

Anyways

Find the triangle enclosed by graph of y=-x²-2x+5 ,touching line on x=1 y=2 and coordinate system
Makes no sense, so can you post the actual question? Like is it from a book or review sheet?

[valid dew]

was about to ask the same thing

so touching the line on x=1 y=2 meaning ?

what line are you talking about

[nocturne minnow]

Post an image of it, then translate it

[valid dew]

cause the equation is on second order

which is always parabolic

[nocturne minnow]

There's gotta be something that has the proper question

[valid dew]

agreed

[nocturne minnow]

Because x = 1 y = 2 doesn't make sense

[valid dew]

they can be two seperate functions, a point or like 1000 different things

[nocturne minnow]

And neither does "find the triangle enclosed" because there isn't a triangle

[valid dew]

okay what we are going to do is

draw both of the graphs

so you have a line

and a parabol, right ?

and then you can find the area remaining inside

[nocturne minnow]

Can you please just post the original question, so it can logically make sense?

[valid dew]

let me draw it for you without solving it

nah I got what he is trying to say

[nocturne minnow]

That makes it a bit more clearer because you sayin x = 1 y = 2 made zero sense and seeing that, you're referring to a coordinate point of (1, 2)

Because that's not what the parabola looks like

[valid dew]

listen dude

there is no triangle

Okay I am glad you figure it out man!

Cause I still have no clue about your question

find the area of the big triangle = A1

then the are of the small triangle=A2

and then A1-A2 will be the yelow area

yellow

okay

I misstyped circles

[nocturne minnow]

I'm assuming that paper is a review for your exam, was there an answer key for that?

[valid dew]

it will be triangles lol

dude, you wanna find the area

that you pointed out right ?

[nocturne minnow]

A previous year's test?

Or a test you are currently taking?

[valid dew]

alright, I am out of this, sorry.

[nocturne minnow]

You understand that what you are doing is academically dishonest, right?

[obtuse pebble]

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[obtuse pebble]

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[next reef]

Let $V={(a_1,a_2,a_3,\dots,a_n):a_i \in \mathbb{C} \text { for } i=1,2,3,\dots,n}$. Is V a vector space over the field of real numbers with the operations of coordinatewise addition and multiplication?

[kind hawk]

What have you tried

What does V need to satisfy

[next reef]

I see it is a vector space over the field of complex numbers

Does that imply it is a vector space over R as well?

[kind hawk]

Yes but why

[next reef]

It passes all that eight properties to be a vector space

[obtuse pebble]

@next reef Has your question been resolved?

[warm canopy]

Prove it

[obtuse pebble]

@next reef Has your question been resolved?

[obtuse pebble]

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[burnt briar]

I need to find median but I forgot the way for it

[valid crown]

4×4sqrt(3) = 8×m

[burnt briar]

Ty

.close

[tawny fog]

The Size of the half of the hypotenuse is equal to the median

[obtuse pebble]

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[stuck pilot]

The answer is -10, but I can only think of -8

[trim locust]

What have you done

[stuck pilot]

With sin being 0 and cos being -1

[trim locust]

Id turn it into harmonic form

[stuck pilot]

What is harmonic form?

[trim locust]

R sin(x+alpha)?

Are you familiar

[stuck pilot]

What is R?

[trim locust]

Constant

R Ane alpha are constants

[stuck pilot]

I think I'm not familiar then

[trim locust]

Um ok, are you familiar with the formula that lets you rewrite sin (x+a) where a is just some number?

Sometimes called the sin addition formula

[stuck pilot]

Umm can you give an example?

Do you mean
sinx.cosa + sına.cosx?

[trim locust]

Perfect

[stuck pilot]

So how can I use this?

[trim locust]

Now multiply the entire thing by a constant R (this is so that the coefficients of sin x and cos x aren’t stuck between 1 and 0) and we can represent any combination of sin x + cos x

[stuck pilot]

Makes sense

[trim locust]

Ok

So

We need to pick the correct r and a constants to make this woek

R sin(x+a) =R * Cos a * sin x + R* sin a * cos x

I took your expression you put above and multiplied by R

We want 6 sines and 8 cos

Xan you form some expressions for x and a

That would make that be the case

[stuck pilot]

And how do we do that?

[trim locust]

R * Cos a * sin x + R* sin a * cosx
6 sin x + 8cos x

I’m not sure if that helps you to see

You need to compare coefficients for sin and cos

[stuck pilot]

But for that, don't I need to know every single coefficient possible?

Like I think that R should be 2 in this case

And cosa be 3 and sina be 4

[trim locust]

Can cos a =3

[stuck pilot]

No

But what else can I do?

[trim locust]

Try to set up some equations

What does R sin a need to equal to

[stuck pilot]

I went like this:
Rx = 8
Ry = 6
R = G.C.C (8, 6) = 2

[stuck pilot]

6

[trim locust]

Not quite, R sin a = 8

Do you see how I did that

[stuck pilot]

Oh right

Yes

[trim locust]

Can you do r cos a

[stuck pilot]

It should be 6

[trim locust]

Yes nice

Now we have 2 eq and 2 unknowns

You can solve them

Any ideas how

[stuck pilot]

Lemme see

We have Rsina = 8
Rcosa = 6

Are these our equations?

[trim locust]

Yep

[stuck pilot]

No I can't solve it?

[stuck pilot]

I mean I did here, and seemingly went wrong

[trim locust]

Ok, do you know any formulas that relate sin x and cos x

[stuck pilot]

And don't we have 3 unknowns?

[stuck pilot]

I do, but it depends on where a is. If it's been 0-90, 90-180, 180-270 or 270-360

[trim locust]

No Becsuse don’t forget we want this to be in the form Rsin(x+a) where R and a are constants, forget x for a minute

Also

Your equations are in terms of a

a doesn’t change

Sin a is just a number

Cos a is just a number

Why don’t you rearrange one of them for R?

[stuck pilot]

sina = 8/R

cosa = 6/R

(64 + 36)/R^2 = 1 -> R = 10

[trim locust]

Good

That was probably the hardest way 😂

[stuck pilot]

😅

[trim locust]

I normally turn mine into tan but you got there

Tbh

You don’t need to find a

You now have 10 sin(x+a)

Is the same as your original expression

So what is the max value of 10 sin (x+a)

[stuck pilot]

And if I think sin(x+a) = -1

Then -10 would be the answer

Alr thanks I got it

[trim locust]

Yeh

[stuck pilot]

But lemme review once

To make sure I understood

[trim locust]

Ofc

[stuck pilot]

I'll go through the messages rn

[trim locust]

Ok

[stuck pilot]

Ah I'm so grateful

Thank you for helping me solve it, being patient and explaining everything fully

🙏🙏🙏

[trim locust]

Np, there is a quick way for harmonic form I can show you if you like

[stuck pilot]

Sure!

[trim locust]

It saves you having to do all the algebra, does r sin a and r cos a remind you of any shapes

[stuck pilot]

Circle with the radious R?

Okay wow

Ye

Makes sense

I kinda saw it when I solved for R (but didn't realize till now)

[trim locust]

I meant a triangle but I guess it is also a circle and a sector

That’s good

[stuck pilot]

So basically a triangle, and then Pythagorean theorem

[trim locust]

Then it’s just Pythagorean

[stuck pilot]

Right?

[trim locust]

It’s a lot quicker than rearranging

And you make less mistakes

[stuck pilot]

Yeah

[trim locust]

Then a is just arctan (opp/adj) just in case you ever need to do it again

[stuck pilot]

Haven't learned arctan yet and seems like I don't need it

I'm studying for an exam and I' solving tests for it

[trim locust]

Ah ok

Fair enough

[stuck pilot]

Happens that I come across questions that need me learn something new

If I saw an arc tan, I'll learn it

For now I have learned enough and have enough things to learn (learned Matrix, a lil bit of derivative, need to learn derivative better and learn integral)

Anyways, thank you very much for your help ❤️

[trim locust]

Good luck

No probs

[stuck pilot]

Take care

[trim locust]

You too

[stuck pilot]

.close

[obtuse pebble]

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[earnest dagger]

Hi

[lost tree]

what is your question

[earnest dagger]

I need help with a 3d trig question

[lost tree]

send it here

[earnest dagger]

Question 11 please

[lost tree]

,rotate

[lost tree]

what have u tried

[earnest dagger]

I’m not sure how do draw 3 stuff

3d*

[lost tree]

can u imagine it in ur head

[earnest dagger]

Yeah but I have to write working out by drawing

[lost tree]

u can draw it from three sides

draw the view that a person in the sky would see

and then a view that a person standing would see if they look towards the north/south

and the view a person standing would see if they look towards the east/west

[earnest dagger]

oh separately

I get it thank you

[lost tree]

yes

[obtuse pebble]

@earnest dagger Has your question been resolved?

Channel closed

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[silver plover]

why is this -n

they said differentiate further (n-1) times

[royal basin]

guess they actually meant to take the (n-1)th derivative of 1/(1-x)...?

[unreal musk]

Differentiate (1-x)^{-2} for (n-1) times

[royal basin]

each differentiation lowers the power by 1.

[unreal musk]

Or equivalently, differentiate (1-x)^{-1} by n times

[royal basin]

i think sa's point was that this would leave us with (1-x)^(-n-1)

with some coefficients

actually yeah what's going on here? we're supposed to accumulate a factor of (n-1)! there aren't we?

[silver plover]

tbh im just confused how hes got (1-x)^-n

[royal basin]

can you talk to the author of this document?

,w d^n/dx^n (1-x)^-1

[warm shale]

Results provided by WolframAlpha

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Requested by Ann#0413

[silver plover]

ok its my lecturer

[unreal musk]

Yea this is weird actually

[silver plover]

hmmm

wait a sec

imma msg my lecturer

[unreal musk]

Yea scratch what I said earlier, that further doesn’t help in the way they worded it

[silver plover]

on email

[unreal musk]

Guessing they meant more to differentiate the original by that amount but it still doesn’t hold

[silver plover]

its on a video

and he isnt even sure

if hes correct

he said 'lets come back to it'

💀

lemme see what the rest of the video says

[unreal musk]

Fair fair 😂 sometimes it happens I guess, even to lecturers

I guess if they make mistakes like that then you can’t feel too bad when you end up making them too(!)

[obtuse pebble]

@silver plover Has your question been resolved?

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[fluid meadow]

why are the accelerations different here

but in a system like this theyre the same

[trail cloak]

Ooo I see now

Okay notice how the string is connected fully between A and B

And A has a bigger mass than B

And M is pulling both A and B

[fluid meadow]

yea

but i thought since theyre connected

theyll move the same amount in the same time no?

[trail cloak]

Well A and B have different masses

The forces they have will be different

And M will have to resist a force from A and a force from B

And the result will be an uneven force reaction towards A and B

And so A and B will have different accelerations

@fluid meadow

[fluid meadow]

hmm

but

say the M goes down by 5 meters

A and B would surely also have to move 5 meters

since its all connected

no?

[trail cloak]

Since A is heavier than B, M will have to give more effort to push it as far as it will push B

[fluid meadow]

right

[trail cloak]

And since mass stays constant, the acceleration result will be different.

[fluid meadow]

yeah

i think i get it

[trail cloak]

Happy you do

[fluid meadow]

since tension is the same