help-13

[frigid frigate]

i dont know how the second last line of working out turned into the last line

is that right?

i saw another solution where the 3rd last line skipped to the last line of working out

how?

[wispy marten]

The 2nd last line is in my opinion useless

[cerulean sail]

Yep not exactly sure where that line came from, looks like they pulled it out of nowhere (might be a miswrite?)

[wispy marten]

You can get distribute a and b from the (a+b) to the sum term

[frigid frigate]

naw tahts the soln

from the 3rd last line to last line

[frigid frigate]

i can see why this is the case

[frigid frigate]

how would that work?

[wispy marten]

If K is any number, how do you expand K(a+b)?

[frigid frigate]

OH

OH

i c waht u mean

ok

[wispy marten]

o u c, gr8

[frigid frigate]

from the 3rd last line to the last line u distribute the (a+b) intro the sum

[wispy marten]

Yep

[frigid frigate]

but how does it then produce

the last line?

[cerulean sail]

A useful identity if you wanna go down that road is $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$ btw

[wraith dagger]

chartbit

[wispy marten]

This is not yet 😄 ^

[frigid frigate]

sorry hold 1min

pls

[wispy marten]

Well look at the powers of a and b in the two terms it gave, you should notice quickly

[frigid frigate]

,rotate

[frigid frigate]

thats what i tohught it would become?

how did 2 diff sums come into play?

[wispy marten]

You basically distributed a and b to the same term, it's like if you did:

(a+b)K = abK

Which is wrong of course

[frigid frigate]

OHHHHH

IM SILLY

OH OK

dumbing it down helps alot

ok ty

tysm

that helps so much

i have been stuck on this q for like a solid 2hrs

💀

.close

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[main oak]

How is this even possible?

[midnight lagoon]

It is possible

[main oak]

no

[midnight lagoon]

Yes!

[main oak]

like when you work it out you get (4,6)

[obsidian coral]

Post your work

[main oak]

ok

[obsidian coral]

What did you do to get that

[main oak]

So like

[midnight lagoon]

I guess you took HG/FG = 2/3

[main oak]

ok so

X2-X1 so 9-3=6 and then 2/3 multiplied by 6= 4 so x=4
Y2-Y1 so 13-4=9 and then 2/3 multiplied by 9= y=6
So (4,6)

you get it?

[obsidian coral]

Put spaces or \ before it when you're using * to show multiplication

Discord with default anything to italics if you surround text in *

[red pumice]

you forgot to consider the starting point

[main oak]

It is supposed to be (7,10)

[red pumice]

exactly

[red pumice]

as shown here as well

[main oak]

what?

[red pumice]

its (7, 10)

[main oak]

i don't get it

[red pumice]

you forgot to consider the starting point

[main oak]

i used the formula my teacher taught me

[red pumice]

dont blindly apply a formula without thinking

[main oak]

i mean

the question is direct

so I just applied it

in fact, all my classmates were so confused

[main oak]

can you show me the working?

[red pumice]

if the line started in (0,0) it would work

but it doesnt

you forgot the starting point

which is (3,4)

so if you add the points together

(3+4, 4 + 6) then you get (7, 10) which is the correct answer

[main oak]

got it thanks

😄

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[bright sorrel]

quick question

[bright sorrel]

why does (sqrt(2))^8=4

and not (sqrt(2))^4=4

[crimson delta]

who says it does?

[earnest socket]

,w calc sqrt(2)^8

[wraith dagger]

Results provided by WolframAlpha

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Requested by tushar#0942

[south tundra]

,w sqrt(2)^4

[wraith dagger]

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Requested by A Lonely Bean#7998

[earnest socket]

[bright sorrel]

I'm solving this

and it equals 8

it's sqrt(2)^x=4

[tropic oxide]

,w log(4)/log(sqrt(2))

[wraith dagger]

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Requested by Ann#0413

[tropic oxide]

says who?

[bright sorrel]

my textbook 😭

[red pumice]

its good you caught the mistake then haha

[tropic oxide]

can you show the problem and its answer key exactly as they appear in the textbook?

[bright sorrel]

give me a moment as taking a picture would require a lot of work

give me a sec

[tropic oxide]

okay, go ahead and take your pictures

[bright sorrel]

@tropic oxide @red pumice

you know the rest

should equal 4, right?

[tropic oxide]

you havent done as instructed...

[bright sorrel]

but the rest shows:

[tropic oxide]

do you have the textbook you pulled this exercise from on hand? Y/N

[bright sorrel]

it is exactly just this problem

this is the rest

[tropic oxide]

i asked you a yes-no question.

[bright sorrel]

Yes I have

[tropic oxide]

okay

is the textbook physical or electronic?

[bright sorrel]

physical

[tropic oxide]

okay

can you take a picture of the problem statement as it is written in the textbook, and send the picture here?

[bright sorrel]

no

[tropic oxide]

why not?

[bright sorrel]

because I already wrote 1:1 the solution as written in the textbook

and the problem itself

the context for the problem is

[tropic oxide]

so it's refusal and not inability?

[bright sorrel]

Ann

look

I promise you

nothing will change

even if I took a picture

because it's literally 1:1

the problem is

[earnest socket]

how did you get from here to log base sqrt(2) of 4?

[tropic oxide]

even if nothing DOES change.

i still want that picture.

if the textbook is screwed up then so be it.

[bright sorrel]

in bulgarian?

can you translate it

?

The problem

is

[tropic oxide]

i speak enough bulgarian to understand the problem.

[livid hound]

and the math will still be math

[bright sorrel]

"The number [picture given below] belongs in the interval:"

Intervals being

(2; 5) (4;6) (7;9) (9;11)

and I just re-wrote the solution 1:1

in the picture above

ok asshats

I will take a picture

please tell me what changed

[tropic oxide]

,rccw

[tropic oxide]

what changed is that now we know 100% that it's the textbook's fault and not yours.

[bright sorrel]

no shit

[tropic oxide]

$4 \neq (\sqrt{2})^8$.

[wraith dagger]

Ann

[bright sorrel]

thanks for nothing I will email the publishing company

.clos

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[cosmic quartz]

i need help with these questions

[cedar kiln]

@cosmic quartz Has your question been resolved?

[cosmic quartz]

no

<@&286206848099549185>

[crimson bobcat]

Can u translate it pls

[cosmic quartz]

yeah sure

14. In a right triangle with sides a, b, c (and opposite angles A, B, C respectively) a is the longest side. Then applies:

(A) a=b.cosB + c.cosC (B) a= b.cosB + c.sinC
(C) a = b.sinC + c.sinB (D) a = b.sinB + c.sinC
(E) a = b.sinB + c.cosC

and

1. Which polynomial is not a divisor of (x-1)²(x³+x) ?

(A) x³ -x² +x -1
(B) x² -2x +1
(C) x² - x
(D) x³ -x
(E) x4 -2x²(x-1) -2x +1

<@&286206848099549185>

[cosmic quartz]

<@&286206848099549185>

[half forge]

do you need help with 14 or 1 or both

[cosmic quartz]

both

@half forge

[half forge]

is this question asking for the right one?

[cosmic quartz]

yeah

with explanation

[half forge]

did you find which of the angle is the right angle?

also what grade is this?

[cosmic quartz]

no clue

questions from 1997

[half forge]

you don't know what grade it is?

what grade are you in

[cosmic quartz]

we dont work with grades here

[half forge]

huh okay

[cosmic quartz]

i’m 16

what grade is that

[half forge]

i think it would be grade 10

[cosmic quartz]

grade 10 then

🙂

[half forge]

a is the longest side here, so the opposite angle would be A which make A the right angle

[cosmic quartz]

yeah

[half forge]

what does b.cosB mean tho

does it mean b multiplid by cosB

[cosmic quartz]

i have no idea

[half forge]

then I can't help you with that question

[cosmic quartz]

i dont understand what i need to do

[half forge]

I'll help you with 1. first

[cosmic quartz]

its just *

like 1 * 2

[half forge]

oh

so b.cosB means b multiplied by cosB iguess

[cosmic quartz]

ye

lets do other question

arent there some really smart people on here

[half forge]

k then lets start with the second one

1. Which polynomial is not a divisor of (x-1)²(x³+x) ?

(A) x³ -x² +x -1
(B) x² -2x +1
(C) x² - x
(D) x³ -x
(E) x4 -2x²(x-1) -2x +1

do you know how to factor (A)?

[cosmic quartz]

no

[half forge]

oof

(A)= x³ -x² +x -1 = x²(x-1)+x-1

can you factor it now?

[cosmic quartz]

its a divisor

you dont even need to factor it

[half forge]

aight then

[cosmic quartz]

you can tell its a divisor for x-1

[half forge]

can you tell if (B) is a divisor for (x-1)²(x³+x)

[cosmic quartz]

yeah it is

1 - 2 + 1 = 0

[half forge]

bro

[cosmic quartz]

what

[half forge]

do you know what a divisor is?

[cosmic quartz]

yeah its a trick

translate this

het is deelbaar door x-1 als de som van de coëfficiënten gelijk is aan 0

[half forge]

It's not asking if (A)(B)...(E) are divisible by x-1

[cosmic quartz]

oh

[half forge]

It's asking if (x-1)²(x³+x) is divisible by (A)(B)..(E)

[cosmic quartz]

ah fuck

[half forge]

can you do it now then?

[cosmic quartz]

A is (x-1)(x^2+1) ?

[half forge]

yeah

[cosmic quartz]

and what now

[half forge]

$(x-1)²(x³+x) = x(x-1)^2(x^2+1)$

[wraith dagger]

jay.

[half forge]

you need to factor (A), (B), (C), (D), and (E)

if you factor A, you get (x-1)(x^2+1)

you can check that (x-1) and (x^2+1) are both factors of x(x-1)^2(x^2+1), so you can tell that (A) is a divisor

[cosmic quartz]

okay give me a second

what is E

[half forge]

do you know synthetic division

[cosmic quartz]

no

[half forge]

oh alr

if you expand (E)

you get

x^4-2x^3+2x^2-2x+1

[cosmic quartz]

yeah

[half forge]

$x^4-2x^3+2x^2-2x+1=(x^4-2x^3+x^2)+(x^2-2x+1)=x^2(x^2+2x+1)+(x^2+2x+1)$

now u'll be able to do it

oh wait

$x^4-2x^3+2x^2-2x+1=(x^4-2x^3+x^2)+(x^2-2x+1)=x^2(x^2-2x+1)+(x^2-2x+1)$

[wraith dagger]

jay.

[half forge]

@cosmic quartz

[cosmic quartz]

yeah i dont get E

[half forge]

can you factor this?

[cosmic quartz]

let me see

isnt it already factored

[half forge]

no

[jaunty pumice]

one more step required

[half forge]

factoring means you express the equation using only multiplication

[cosmic quartz]

what is it then

[half forge]

can you factor this?

[cedar kiln]

@cosmic quartz Has your question been resolved?

[cedar kiln]

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[mossy gull]

How I can factorize it?

[red pumice]

substitute u = t^3

then its just a quadratic

[vagrant elbow]

This is actually a triquadratic

You got bamboozled

[mossy gull]

? Wdym

[vagrant elbow]

What Gijs said

[red pumice]

u = t^3

u^2 = t^6

[mossy gull]

Hmm wait

[red pumice]

so you get 2u^2 - u - 15

[mossy gull]

(x+x1) (x+x2) with this formula

A special tripolynomial

[mossy gull]

Have I to make an example?

Like

a^4-7a^2-30 = (a^2-10)(a^2+3)

3 is the factor and - 10 is the sum

By the way now I go hope that someone knows it. Goodbye

[cedar kiln]

@mossy gull Has your question been resolved?

[cedar kiln]

@mossy gull Has your question been resolved?

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[random meteor]

hello can i please get some help with the following question

[languid nexus]

did u find d-theta?

[random meteor]

i got that, i'm struggling with what to do after actually subbing in 4sin as x

[languid nexus]

when u sub it in you get sqrt(16-16sin^2theta)

then you can take a factor of 16 out

and it becomes 16(1-sin^2theta)

do u know how to simplify the above?

[random meteor]

oh yes, i understand now

thank you so much

:))

[languid nexus]

np

[random meteor]

.close

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[minor rose]

Cant understand the question

[minor rose]

<@&286206848099549185> Pls help

@vague moth help

[crimson sedge]

hi

[minor rose]

Hello

[crimson sedge]

can you provide a better picture please?

[minor rose]

Ok

[crimson sedge]

and what grade are you in?

[minor rose]

7th

[crimson sedge]

alr

[minor rose]

alr?

[crimson sedge]

in this question

you need to get the edge that meets AN

and DE

[minor rose]

But the problem is A and N are on the same edge

How will they meet?

[crimson sedge]

NM is perpendicular

on AN

[minor rose]

Yes

Now?

[crimson sedge]

so

a edge

means angle

[minor rose]

Ohk

Gotchya!

Thx

[crimson sedge]

so AN and NM meet

[minor rose]

Thanks Sir!

[crimson sedge]

no problem

[minor rose]

.close

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[obsidian vapor]

Can anyone help me simplify the one on the top to get the one on the bottom?

[obsidian vapor]

I simplified it until I got
(n+1)cos(x+(nπ)/2)-xsin(x+(nπ)/2)
But idk what to do after this

.close

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[near marsh]

Hi I am going to need help with this math equation is (4/9)-1/2 and my work of this equation I did is 1/9 1/2 = 1/2 square root 9 = 3 I am confused how to get the exact answer I reference the text but unfortanley I am still lost and I am confused on other math questions also.

[cosmic steppe]

Huh?

I need the original equation

Or whatever you're tryna solve

Either put it in mathematically or through latex

[south tundra]

The equations you are referring to are unclear, could you show the problem stated?

[near marsh]

the original equation is (4/9)^-1/2

[south tundra]

That's not an equation but alright

Recall that generally (a/b)^-n is the same as (b/a)^n

So in this case we have (4/9)^-1/2 = (9/4)^1/2, right?

[near marsh]

sorry I couldn't put the fraction inside the parenthesis would a picture help noticed the math problem much better?

[cerulean sail]

a picture would help better please

[cosmic steppe]

$\parens{\frac49}^{-\frac12}$

[wraith dagger]

Umbraleviathan

[near marsh]

here is the picture.

[cerulean sail]

Then this here

[south tundra]

$$(\frac{4}{9})^{-1/2} = (\frac{9}{4})^{1/2} = \frac{9^{1/2}}{4^{1/2}}$$
Also remember that raising to the power of 1/2 is the same as taking square root of the number, meaning that this is the same as
$$\frac{\sqrt{9}}{\sqrt{4}}$$
Which is just
$$\frac{3}{2}$$

[wraith dagger]

A Lonely Bean

[south tundra]

How do I center text in LaTeX?

[runic garnet]

Amazing

[mossy mango]

bean doesnt know latex

[south tundra]

No, I just know math

[mossy mango]

[south tundra]

The question is still there btw

[near marsh]

.close

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[dark orbit]

could someone please help me otu with 2b

[long arrow]

have you tried solving an equation?

[cerulean sail]

Looks like they gave you part a for a reason

[dark orbit]

yeah im using part a

i equated 3x^2-10x-4=x^2-3x-4

and got 2x^2-7x=0

[long arrow]

but

[cerulean sail]

Why did you equate these?

[dark orbit]

in part a i used the synthetic division to asnwer it

and the quotient was x^2-3x-4

[cerulean sail]

Yeeaaaa that's not what I meant by the fact that part a would be helpful

[cerulean sail]

You mean you get that as the quotient? The remainder should be zero?

[dark orbit]

yeah thats what i meant

sorry forgot the names

[cerulean sail]

What happens when these two intersect?

[dark orbit]

i though they equaled eachother

idk

[cerulean sail]

Yes, equate those [via their y values being the same], what happens?

[dark orbit]

i then cleaned it up

to 2x^2-7x=0

[cerulean sail]

How did you clean that up?

[dark orbit]

like took the left away frm both sides yk

[cerulean sail]

Alright, to be clear, you started with $x^{3} - 3x^{2} - 5x + 8 = 3x^{2} - 10x - 4$ from here, yes?

[wraith dagger]

chartbit

[dark orbit]

no i used the quotient part of the first part

which was x^2-3x-4

[cerulean sail]

Where did you use that?

[dark orbit]

i equated to 3x^2-10x-4

[cerulean sail]

Can't do that, that's very naughty

And that's not how you use part a

[dark orbit]

k

[cerulean sail]

You first start off with this here

When you rearrange it so that you have zero on one side, you'll see the polynomial you get is the same as in part a, yes?

[dark orbit]

yes

[cerulean sail]

Now from there, use part (a)

[dark orbit]

kk ty

[cerulean sail]

That'll find you your x-coordinates

[dark orbit]

i got x=-1, x=4

[cerulean sail]

Think there should be another one too?

[cedar kiln]

@dark orbit Has your question been resolved?

[cedar kiln]

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[near marsh]

I have a question how will you solve this equation to get the answer unfortanley I am still lost?

[radiant fjord]

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

[near marsh]

All I was able to get is to know the prime factor of 16 is 2 x 8 = 16 and the prime factor of 81 is 3 x 27 =81 for the rest of there I am totally lost

[radiant fjord]

okay

[crimson sedge]

Okay continue doing that

Prime factorise 16 and 81 completely

What do you get

[radiant fjord]

do what alex says, and then it's important to remember the power of a quotient rule

[long swan]

Exponents distribute over multiplication

[radiant fjord]

it's also important to remember what fractional exponents mean

[near marsh]

I am honestly lost and I am more confused on those examples

[cedar kiln]

@near marsh Has your question been resolved?

[near marsh]

.close

[cedar kiln]

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[dreamy vapor]

Solution A has a pH of 5.6 and solution B has a pH of 10. If there are 2.5 litres of solution A, how many litres of solution B would you need to add to neutralize the solution (pH 7).

[dreamy vapor]

the equation for pH is pH=-log(h3o)

[cedar kiln]

@dreamy vapor Has your question been resolved?

[dreamy vapor]

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[cedar kiln]

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[near marsh]

I was wondering if you can please help with my issue I having above in the text channel because I am still lost?

[near marsh]

.close

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[near marsh]

How would you solve this equation because I wasn't able to find a textbook example to solve this similar math equation?

[modest token]

what does it want?

[near marsh]

the title of beginning of the page say rational exponents evaluate each expression

[stark quest]

do you know the exponent rule for multiplying exponents with the same base?

[modest token]

thats just adding them up together

[stark quest]

i was asking OP but yes when two exponents with the same base are multiplied, you add the exponents together and keep the base

[near marsh]

I do not know the exponent rule of multiply it together because I learning it as brand new to these math topics which I have never heard about

[stark quest]

okay so youve never seen $x^a\cdot x^b=x^{(a+b)}$?

[wraith dagger]

light

[near marsh]

would the answer be 5^3/6? This answer can be wrong also

[stark quest]

no, thats not how you add the fractions

[modest token]

no it would be 5^(1/3+2/3)

[near marsh]

I have a question how will you add the fractions would it be vertically to each other?

[modest token]

let me use white board for a sec

$a^3 \times a^2 = a^5$

[wraith dagger]

Duck Vibin

[near marsh]

would the answer just 5 by itself and what will happen to exponent of 5?

[modest token]

yes its just 5

[near marsh]

okay thank you for the help

.close

[cedar kiln]

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[tame wave]

im so confused as to how to solve this question, could someone just guide me as to the steps that i should do please

[dire geode]

Do you know when two vectors are parallel?

[cedar kiln]

@tame wave Has your question been resolved?

[tame wave]

when their cross product is zero right

but i dont think i can use that here

[lethal jackal]

honestly that actually probably works

much more painful than it has to be, but maybe you should give it a try and see what you get

[dire geode]

Not the best solution here

[dire geode]

Vectors are parallel if they're a scalar multiple of each other

Say v1 and v2 are your two parallel vectors. Mathematically that means there's a constant C such that v1 = C * v2

[tame wave]

hey would this work;

so first i took the parametrics which are

x= 1 + 2s
y= 2 +3s

so then I made it so 1+2s = 12 and it ended up with s being 5.5

but when i plug in that 5.5 for s when trying to solve for the y (or k) coordinate, I get 18.5

[dire geode]

The point 1,2 is irrelevant to your problem

Your first vector is (2s, 3s) and you need to figure out the second vector from 12x+ky=0

[tame wave]

ooooh, so would s be 6? since 2(6) is 12. therefore my answer wouldbe 18?

[dire geode]

No

You skipped calculating the direction vector

[cedar kiln]

@tame wave Has your question been resolved?

[tame wave]

i dont understand

[dire geode]

12x+ky=0 is not a direction vector

It's the equation of a line through the origin

Write the line in parametric form

[dire geode]

Similar to r

[tame wave]

sorry, im not really sure how to do that

[dire geode]

Let x = t and find a vector v(t)=(t,...) that equals your line

The second component will depend on k

[tame wave]

so youre saying that the first component would be 12 then? im not really following

[dire geode]

Do the reverse of what she's doing
https://youtu.be/QE72-1X1jTc

You have a line in slope intercept form. You need to convert it to parametric

So you can compare it to (2,3)s

[dire geode]

I mean your answer's close, you're just off by a minus sign I think. But your approach is flawed from the start

[tame wave]

which one is in slope intercept form? i thought 12x+ky=0 wasnt intercept form.

[dire geode]

It isn't. That's why your answer's wrong

[tame wave]

so do i have to convert it to intercept form and then parametric?

[dire geode]

Yes

[dire geode]

Oh yea I was wrong here with my first sentence. My bad

[tame wave]

ok i think i finally got it,

i just turned it into intercept form so it was -12/k x = y

then i used the ratio 3/2 from the first equation which is (2,3)

and then i made 3/2 = -12/k , and ended up getting -8. is this also a good way to solve this question or is this just a coincidence?

[dire geode]

Perfect

[tame wave]

awesome. was that what u were trying to tell me tho? or did i just like figure a different way

because i didn't really use parametric

[dire geode]

It's slightly different. Your way is shorter so it's better

[tame wave]

okok thank u so much

.close

[cedar kiln]

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[cedar kiln]

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[junior timber]

a grocer packed 2/3 of his oranges in box 1 and the remaining oranges in box 2 and 3 in ratio 1:2, if box 1 holds 56 more ornages than box 3, how many oranges are there in total

[junior timber]

so i got box 1 = x
box 2 = y and
box 3 = z

so x = 2/3

y+z= 1/3

x - 56 = z

2y = z

someone help pleasee

[nimble veldt]

let a be the total number of oranges. then x = 2/3 a, y+z = 1/3 a, x-56 = z, 2y = z

[junior timber]

okk

and then what

oh nvm got it

[cedar kiln]

@junior timber Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[rancid pollen]

i honestly got no clue how to do it

i was thinking of changing base to 1944 and 486root2 but im not sure what to do after that

[lethal jackal]

nah just use some base like log base e

$\frac{\log 1944}{\log 2n} = \frac{\log 486 \sqrt{2}}{\log n}$

[wraith dagger]

Saccharine

[lethal jackal]

and I'm sure you can figure out the rest

well also $\log 2n = \log n + \log 2$

[wraith dagger]

Saccharine

[rancid pollen]

o i see

okay thanks

.close

[cedar kiln]

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[cedar kiln]

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[crude python]

Need help, The equation is a parbola so how would one calculate the average slope for that?

[crude python]

for reference*

The equation I created for the parbola is
y=0.1x^2 - 2.8x + 9.6

[shell cloud]

average slope is (final point-starting point)/(time)

[crude python]

Yes I do know that

but doesn't the slope for parbola change?

[shell cloud]

it does but we still keep with the same formula because they want the average

[crude python]

oohhh so the way to calculate slope in this case is litteraly the same?

[shell cloud]

average slope yes

[crude python]

so in this case difference from 9.6 to whatever the y axis is on 14 yup?

[shell cloud]

yup

[crude python]

kk ty!

and last question but the average rate of increase and decrease would be litteraly just negative yup?

decrease would bbe negative while increase would be positive

[shell cloud]

yes but they would be different numbers

[crude python]

?

[shell cloud]

so -a vs +b

not -a vs +a

[crude python]

wait huh

[shell cloud]

bc of how the function ends at a different y

[crude python]

ohhhhh

yup just got that

so one of the x intercepts is the highest point of the parabola

i was wondering how to put the values of the increase temperature

Thank you so much!

.close

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[prisma gull]

I found this in a pre-calc textbook, but unable to recollect which part of calc was this in

is it some sort of trig sub ?

The book contains expression used in calculus,

As in here, I was able to figure out that this was applying first principle to x^3+7x +C

similarly, where the other expression shows up in calculus ?

[dire geode]

You're not trying to do any problems?

[prisma gull]

I'm actually just doing the algebra from the textbook

like the exercise only care out factoring

but it's mentioned they shows up in calc, i've already taken calc but couldn't recollect

[dire geode]

76 is a difference quotient for derivative of a function at x

[rugged eagle]

this one shows up when you're doing derivatives using the limit definition

[prisma gull]

yes, that i understood

[violet flume]

you can draw a triangle for this one, unless im just dumb\

[rugged eagle]

the first one?

[dire geode]

,w int sqrt(1+x^2/(1-x^2))

[rugged eagle]

usually an internal

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by riemann#2529

[rugged eagle]

integral*

a yucky one

[prisma gull]

This will boil down to 1/sqrt(1-x^2) which is actually the differentiation of sin^-1 x

[dire geode]

You can probably do almost anything with functions in precalc

Differentiate, integrate

[dire geode]

It's kind of a vague and pointless sentence

[prisma gull]

no my question is, is this some specific setup like first principle ?

[dire geode]

Probably arclength

[prisma gull]

as in this was a specific setup for first principle

[prisma gull]

yeah that makes sense

arclength of the function integral (x/root(1-x^2))

because arclength L= integral (1+(dy/dx)^2)

Arc length setup for $-\sqrt{1-x^2}$

[wraith dagger]

.doc

[prisma gull]

Thank you @dire geode , Have a great day.

[cedar kiln]

@prisma gull Has your question been resolved?

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[cedar kiln]

Channel closed

Closed due to the original message being deleted

[latent bloom]

Okay?

Your channel will close soon tho

Sure

Wait till this closes

Or actually you can ask now

[atomic vapor]

this may help

happy to help

[cedar kiln]

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[solar tendon]

How to prove this

[solar tendon]

proving fundamental identities

here is what i tried btw

[cedar kiln]

@solar tendon Has your question been resolved?

[solar tendon]

<@&286206848099549185>

[crimson sedge]

so basically you didn't expand (1+sin(a))^2 and (1-sin(a))^2 correctly

@solar tendon you here?

can you try re-expanding $(1+sin(a))^2-(1-sin(a))^2$ for the numerator?

[wraith dagger]

starlight

[crimson sedge]

hmm i guess not

[solar tendon]

I'll try

[crimson sedge]

[solar tendon]

i dont know if its right

[crimson sedge]

(1+sin(a))^2 is not 1+sin^2(a)

and (1-sin(a))^2 is not 1-sin^2(a)

remember the expansion (a+b)^2 = a^2+2ab+b^2

and (a-b)^2 = a^2-2ab+b^2 (basically same thing as first expansion)

[solar tendon]

??

[halcyon oyster]

you need to start from here

[crimson sedge]

wait wait lets start from the beginning alr

from here

[solar tendon]

ok

[crimson sedge]

you basically multiplied the first fraction by 1+sin(a) on both numerator and denominator

and multiplied the second fraction by 1-sin(a) on both numerator and denominator right

to make the denominator common

right

that would make the fraction $\frac{((1+sin(a))^2-(1-sin(a))^2)}{(1-sin(a))(1+sin(a))}$

damnit

can't use texit sorry

[solar tendon]

what i did to the first part is kinda like this

and that wrong?

[sonic oyster]

This is wrong

It should be 1/6

[wraith dagger]

starlight

[crimson sedge]

YES

I GOT IT TO WORK

[crimson sedge]

so do you understand why the fraction would turn into this?

[solar tendon]

You multiply the denominator and crossmultiply?

[crimson sedge]

yep

so do you understand that part?

so assuming that you do, (1-sin(a))(1+sin(a)) = cos^2(a) like you wrote yourself right

[solar tendon]

basically what i did here except i added the denominator

[crimson sedge]

bc it becomes 1-sin^2(a)

$\frac{(1+sin(a))^2-(1-sin(a))^2}{cos^2(a)}$

[wraith dagger]

starlight

[solar tendon]

That is equal to this?

[crimson sedge]

no

you would have to expand the numerator

(a+b)^2 = a^2+2ab+b^2

(a-b)^2 = a^2-2ab+b^2

if we look at (1+sin(a))^2 first

if a = 1, b= sin(a)

can you expand it ?

[solar tendon]

Im gonna try

[crimson sedge]

very close

if you look at the second parenthesis'

bc it's (1-sin(a))^2

it would be -2sin(a) inside the brackets

instead of 2sin(a)

$\frac{(1+2sin(a)+sin^2(a)) - (1-2sin(a)+sin^2(a))}{cos^2(a)}$

[wraith dagger]

starlight

[solar tendon]

oof messed up on the sign

[crimson sedge]

but anyways do you understand this?

bc if you do, if you expand the negative sign

you would be able to quite quickly get the desired expression

[solar tendon]

?

[crimson sedge]

yessss

now you know sin(a)/cos(a) = tan(a) right

[normal pawn]

dang

[solar tendon]

The denominator has a ^2 on it

can it still be applied?

[normal pawn]

it's just 4sin(a)/cos(a)*cos(a)

[crimson sedge]

yep, cos^2(a) = cos(a) x cos(a)

but cos(a) will remain

[crimson sedge]

?

[normal pawn]

then it's easy

[normal pawn]

nthing

[crimson sedge]

yep

but basically this is same thing as 4 tansec(a)

[solar tendon]

Thanks

really appreciated

now i need to do 9 more of these

goodbye

@crimson sedge

[crimson sedge]

[solar tendon]

yes

.close

[cedar kiln]

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[crimson sedge]

Need full explanation: Task 1
Emma makes a straight prism with an equilateral triangle as its base from a net labeled as shown in the adjacent figure. She uses the prism as a 'thrower, where the letter at the bottom is considered to be rolled.
Show: If the probability of rolling M A T H E is greater than the probability of rolling EM M A, then the probability of rolling T E E is greater than 1/64

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

.close

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[modest lichen]

The positive integer n n is such that n^2+20 is exactly divisible by n+2. What are the possible values of n?

[cedar kiln]

Channel closed

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[modest lichen]

The positive integer n is such that n^2+20 is exactly divisible by n+2. What are the possible values of n?

How do i do this

[south tundra]

Do you know modular arithmetic?

[modest lichen]

nope

i got to (n-2) +24/(n+2)

[south tundra]

Yup, so n + 2 must be divisible by 24

[modest lichen]

oh

wait is that it

[south tundra]

So n + 2 can be either 3, 4, 6, 8, 12 or 24

Yeah

Note that n + 2 can't be 1 or 2 because it'd make n equal to -1 or 0

[modest lichen]

ahh

thank you

.close

.close()

im so lost

[south tundra]

It's already closed

[modest lichen]

kk ty

[cedar kiln]

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[pastel ridge]

in the table, for the first row, should the cost of 0 business cards be 0, or 25 because the design costs 25??

[pastel ridge]

alright, for the rest of the rows I should just multiply the x by 0.05 and then add 25 to fill out the table?

[nimble veldt]

yes

[pastel ridge]

alright

is this correct?

[nimble veldt]

looks good for me

[pastel ridge]

alright

how do we find the equation of an undefined slope which passes through (-1,2)

it would be a vertical line where x doesnt change, right?

how would we make that an equation though

[nimble veldt]

is this a new question or should there be a connection to the business cards?

[pastel ridge]

its a new question

[nimble veldt]

what means "undefined slope"?

[nimble veldt]

ok, i would say you are right with a vertical line where x doesnt change. then the equation will be x = -1.

[pastel ridge]

wait why would it just be -1?

[nimble veldt]

i thought (-1/2) is given?

[pastel ridge]

yes it is

but how do we get -1 from it?

oh wait

nvm i understand

thanks for the help

.close

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[sand cradle]

When we have a function with more than one extremum, is it always the case that a maximum follows a minimum and the other way around?

[sand cradle]

It seems so, how would you prove it though?

f'' is negative when a maximum occurs, so the graph is concave down. It will be concave down and thus no other extremum occurs until it changes the concavity to concave up, then there will be a minimum

[tired perch]

i think u could also do it with a contradiction. Like suppose f(x) is a continous function on the interval [c,d] and has a maximum at x = c. Suppose that the next extremum point is another maximum that occurs at x = d. f'(c+a) where a is a small value will be negative for f(c) to be a maximum. We also know that f'(d-a) will be positive for f(d) to be a maximum. Then you can apply Intermediate Value Theorem which just says that there must be a point where f'(x) = 0 and since we know it is decreasing on the left and increasing on the right, that point must be a minimum. So we have a contradiction and there must be a minimum in that interval.

probably not the best proof tho

[rustic siren]

(between 2 max if the function doesnt go "down" and only keeps growing, then the first one wasnt a max)

[sand cradle]

Thanks

.close

[cedar kiln]

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[maiden valve]

My answer is on the right, why is it wrong, i used: log_a*a^x=x

[drifting marlin]

It's wrong because you aren't in a log_a (a^x) situation yet

[snow zodiac]

is 3 = 9?!?!

glitch in the matrix confirmed

[maiden valve]

oh lol, so can i go 3^2^(x-7)

and then =2(x-7) then 2x-14 oh ok

[drifting marlin]

That's the same thing that they did

They just used more steps

[maiden valve]

but they used log_ay^r=rlog_ay, its a different formula so using log_a(a^x)=x is probably valid as well?

[crimson sedge]

it's the same logic basically

knowing log_a (a) = 1

[drifting marlin]

log_a(a^x) = x follows because of their rule

(And log_a(a) = 1 as starlight mentioned)

[cedar kiln]

@maiden valve Has your question been resolved?

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[cedar kiln]

Please don't occupy multiple help channels.

[crimson sedge]

Anyone?

Is the answer 9 ?

[cedar kiln]

@crimson sedge Has your question been resolved?

[manic acorn]

I think 9

[wind maple]

yeah, sorry my battery died,i was talking about proportionality , k

y=kx

k=constant

[crimson sedge]

But the answer is given something else

[manic acorn]

I could be wrong i havent done much des

My idea was multiplying both sides by the thing to the power of 6

[crimson sedge]

Yes. That is what I thought.

[manic acorn]

Honestly degree questions are just a scam

Are they saying the answer is 3?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yes. They are saying the answer is 3. How did you get that?

[manic acorn]

They squared both sides

Which is a very weak logic

Imagine if someone cubes both sides

The answer becomes 2

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yeaa...

Hmmm

[crimson sedge]

Yes.

[crimson sedge]

No. The answer becomes 4.5

[manic acorn]

Yea mb, Well then whats stopping us from saying the degree rn is 1.5 💀

[crimson sedge]

It is not in the options given 😂

[cedar kiln]

@crimson sedge Has your question been resolved?

[cyan wren]

They cannot be fractions, and iirc they didn't square both sides

I remember learning about this, and it's a bunch of bs stuff you have to do to make into non fractions

But in this question, it looks like it can be solved simply by taking the 6th power on both sides, then subtracting that by the given equation by a constant

Finding the constant is your job 🙂

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[jagged charm]

How many triples $(x,y,z)\in\mathbb{N}^3_0$ (so $x$, $y$, and $z$ must be non-zero) exist such that $x+y+z=60$?

[wraith dagger]

Pierre de Fermat

[manic acorn]

ur asking x+y+z=60 for x,yz belong to natural numbers?

How many*

if its that then this is done using the "stars and bars" thing

[sacred grail]

what's the subscript 0 for if they must be nonzero

[jagged charm]

I don't know it

[manic acorn]

I have to go sorry someone else can explain

[jagged charm]

Okay.

[mossy mango]

Ur daughter

[mossy mango]

U choose like 3 out of 63

There are 63 twinkling stars and u choose 3 to be the bars

How romantic

[tropic oxide]

you write N_0 for naturals WITHOUT zero???

[sacred grail]

[jagged charm]

I am not familiar with this method. Can you advance?
It's in the chapter of the conbinatory

[mossy mango]

Look at to the stars in ur backyard for inspiration first

Then try to use smaller numbers and draw it out

And then if u don’t get, just@search on YouTube

I gotta shower

[jagged charm]

Yes it is private from 0

[mossy mango]

[tropic oxide]

"private from 0"...?

where are you from / what is your native language

[jagged charm]

France

[tropic oxide]

bc that's literally the reverse of the convention i've seen

[mossy mango]

Je parle francais

[jagged charm]

Parfait

[tropic oxide]

i've seen N for naturals starting from 1 and N_0 for naturals starting from 0...

[mossy mango]

Oui

[jagged charm]

Bah c'est quoi le trucs des barres je capte rien là.

[mossy mango]

Too complicate

[sacred grail]

bruh chris stop trolling

[mossy mango]

I can’t handle this level of French

[jagged charm]

Y'a pas un trucs plus simple ?

[jagged charm]

You're not from France ?

[mossy mango]

Oui, Je etudiete francais pars pars je in Canada

[tropic oxide]

alors on veut compter les trios d'entiers (x,y,z) strictement positifs qui satisfont à l'équation x+y+z=60

[mossy mango]

Je suis stupide

[jagged charm]

Oui

[tropic oxide]

ça doit être équivalent à compter les trios non strictement positifs dont la somme est 57

avec un changement de variables x' := x-1, y' := y-1, z' := z-1 ...

et ça peut se compter avec une méthode purement combinatoire

[jagged charm]

C'est comment en combinatoire ?

C'est ce que je cherche

[tropic oxide]

l'ensemble de tous ces trios est en bijection avec l'ensemble de tous les dispositions en une ligne de 57 pierres blanches et 2 pierres rouges

donc ça fait C(59,2)

[jagged charm]

C'est donc la combinaison de 59 parmis 2

[tropic oxide]

2 parmi 59, non ?

[jagged charm]

Ah oui

Bah non sinon c'est 0

Merci en tt cas

[tropic oxide]

$\binom{59}{2}$

[wraith dagger]

Ann

[jagged charm]

Yes

[tropic oxide]

j'avais pensé que ça s'appelle «2 parmi 59»

[jagged charm]

Non c'est le contraire.

[cedar kiln]

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[nova snow]

How can I rewrite this in a more “tidy form”?

[ashen geyser]

looks tidy enough to me, lol

[nova snow]

The question wants me to rewrite it in a tidier form though

I’m a little confused lol

[dull oxide]

Well you got a c² in the sqrt

That's all I can think of

[ashen geyser]

well you could rewrite the surd as $\frac{\sqrt{1-2GM}}{\sqrt Rc}$

[wraith dagger]

Anon581

[nova snow]

Should I be removing the c^2?

[ashen geyser]

but does this look tidier? 😄

[nova snow]

I’m not too sure haha, I’m looking at the sample answer and it removes the c^2

[dull oxide]

This is all I can think of

[nova snow]

I’ll write it here hold on

That’s what the sample answer shows

[dull oxide]

Impossible

[nova snow]

that’s what it shows

[dull oxide]

Then some relation is not being shown

[ashen geyser]

what's the question then lol

seems like you might have started with the wrong expression

I've noticed... a lot of people asking questions here often make the XY fallacy: https://xyproblem.info/

makes it harder to debug the work lol

[dire geode]

Yea, it's a pretty common problem

[nova snow]

Although the answer to question one is 2.8

in the answer for problem 2

It’s 2.8 replacing the GM

but the c^2 is gone?

[ashen geyser]

so what's the time dilation formula?

[nova snow]

This

And the answer for problem 2 is this

[ashen geyser]

Say you evaluated 2GM/c^2 = k for your problem

then c^2=2GM/k

Replace that in your equation and simplify

[nova snow]

So does what you say make this solution make sense then

[ashen geyser]

for your problem, it does.

since you evaluated 2GM/c, the only unknown in the resulting equation should be R, yes

[nova snow]

Ohh ok

[ashen geyser]

assuming that your calculations were correct, your final answer does look ok

[nova snow]

👍 thanks

.close

[cedar kiln]

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[torpid urchin]

could someone verify if my steps were right?

homework is not accepting my answer but i feel like its right>?

[violet night]

your answer is correct I think it can be simplified further

sin(arcsec(x)) has an identity

[torpid urchin]

oh ok

[violet night]

,w what can sin(arcsec(x)) be written as with pythagoras

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Clarkie#5208

[violet night]

meh

worth a try

[torpid urchin]

will just throw into symbola

and find the identity that way

thank you

[violet night]

symbolab will tell you the identity

[torpid urchin]

just wanted to make sure i was right in the first place lol

.close

[cedar kiln]

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[violet night]

@frosty ocean

[cedar kiln]

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[crimson sedge]

Find the measure of each angle formed by a tangent and a chord drawn to the point of tangency if the intercepted arc has measure 38 degrees.

[crimson sedge]

can someone help draw diagram of this question for me

i am not able to visualize or draw what it say properly

[fallen moat]

this? im not sure

[crimson sedge]

hm

oh wait

[fallen moat]

oh wait

my bad

lemme redraw

[crimson sedge]

ok

[fallen moat]

is this better?

[crimson sedge]

wait

i think teh first drawing was correct

[fallen moat]

still no?

[crimson sedge]

because its chord to point of tangency

or am i mistaken

[fallen moat]

but it says the arc has measure 38 degrees

[fallen moat]

what does that mean to you?