help-39

[midnight haven]

what is it then and how did u get it

[wanton latch]

0.25

1/4

Let me work out number one for you so it is easier for me to explain

[midnight haven]

are you using law of exponents?

[wanton latch]

Yes

[midnight haven]

which law brother

[wanton latch]

A negative exponent such as -1 means the number is turned into its reciprocal

2^-1= 1/2

[midnight haven]

so 256^4?

[wanton latch]

no no no

[midnight haven]

brother just do my work

ur edcuating a dog rn

[wanton latch]

The laws you need to work with in number one is both negative and fractional

I’ll work out number one and show you

[midnight haven]

ok

[wanton latch]

For the purposes of explaining I separated the -1 and 1/4 but you just need to recognize the negative

[grim fractal]

lmao

Here's a simpler exercise what is 4^(1/2)

[wanton latch]

This is the proof the step by step if this helps

[pearl pond]

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[sleek junco]

W is going to right, -w is going to left

Up + left =

Yes

Usually would draw D tho

B and D gives the same resultant force direction ,but D gives the correct order of v-w

I'm not sure

V+(-W) actually

[sterile crystal]

Both yield the same resultant vector

[sleek junco]

Juz go for D ig, B looks weird to me too but resultant force direction is same

[sterile crystal]

(-w)+v=v+(-w)

It does

The negative just reverse the direction for w

Which effectively makes it vector addition, and addition is a commutative operation

[sleek junco]

Oo

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[brisk trellis]

How to start this problem?

I have no clue

[acoustic path]

hint: ||am-gm inequality||

[brisk trellis]

I see

C?

[pearl pond]

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[kind palm]

Real question how do I find HP

[pearl pond]

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[steep condor]

Can someone help im stuck

[cyan bronze]

!da2a

[pearl pond]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[steep condor]

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[rough abyss]

confused on the last part

[unkempt yacht]

you dont know how to integrate or?

[pearl pond]

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[midnight haven]

Hey just making sure I did this right
Find the value of x which makes the shape a parallelogram. Given AE=EC and DE= 4x+ 3 and EB= 10x round answer to nearest tenth.

[AE = EC] [3x - 4 = x + 12]|
solved for x [2x = 16] [x = 8]

In a parallelogram the diagonals bisect each other,
Which means AE = EC and DE = EB
You Have EB and DE given, use those and find x

[midnight haven]

I thought it was 8 but people said it wasnt

[mint flax]

DE = EB, enter the values of DE and EB

[midnight haven]

0.5 ?

[mint flax]

Yeah

[midnight haven]

I feel dumb now

.close

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[potent elbow]

How does one go about answering the second part of this question

[worthy ginkgo]

find a zero of the polynomial ig

[potent elbow]

Ik i neef to find the factors of the cubic then sub them into the function just I havnt really learned how to factorise cubics yet

[worthy lance]

But u did part a

[worthy ginkgo]

if you can prove that its a factor

and know factor theorem

then i guess you can find the zero

since they're related

[worthy lance]

Ok let start by factoring obly the cubic and squared x

[potent elbow]

Yeah part a i just sub -0.5 in and if it comes out as 0 its a factor just ive got to the stage where its The cubic divided by the factor

[worthy lance]

Like this 100x^2(2x+1)

Now factor the swcond part

Factor now the -18x-9

[potent elbow]

9(2x - 1)?

[worthy lance]

No

[potent elbow]

-9(2x + 1 sorry

[worthy lance]

Yes

Now factor both together

[potent elbow]

Is it (10x -3)(10x +3)(2x + 1)?

[worthy lance]

No

Hou have this

100x^2(2x+1) - 9(2x+1)

Factor this

Wait

U did sorry

I didnt expect u to do the final result

Yes it is correct

[potent elbow]

Aight ty

So split the cubic into 2 then take (2x + 1) out as a factor then factorise the outside of the bracket parts is the method?

[worthy lance]

I do not believe in methods

You use your knowledge to answer the question

This is a way to solve it

Now i have a product equal to 0

So either one of those 3 multipliers are 0

[potent elbow]

Could i also use long algebraic division?

[worthy lance]

Knowing that 2x+1 is a factor yes

[potent elbow]

So -0.5, 0.3 and -0.3 are the values of x?

[worthy lance]

As u can see u can chose your path

Yes

But

Use fractions better

-3/10, 3/10, -1/2

[potent elbow]

Ok, one last thing. In algebraic long division does it only work with factors?

So i can do it with (2x + 1) but not (3x + 5) for example

[worthy lance]

If u divide bu 3x+5 u can do the long division but u wont have a exact division

Its like dividing 4/2 u have 2

In natural nimbersy

But divide 4/3 is 1 + R

[potent elbow]

So what do i do where it isnt an exact factor?

[worthy lance]

U dont do long division if u want to find the roots

[potent elbow]

Oh

[serene jacinth]

Today is 18th march of 2024 and the time right now is 9:20PM,,, What will be the time if we add 12 days 3h 10mins to it (Please say the answer in 24 hours format)

Pls solve it i dont have brain cells or neurons anymore

[potent elbow]

The way i was truing to do it was do algebraic long divison to get a quadratic then factorise that cause its easier to visualise the factors

[potent elbow]

The time would he 0:30AMin the morning and either 31st of march of 1st of april idk if march has 31 days in it

[serene jacinth]

U son of albert eingstien?

[potent elbow]

What?

.close

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[sharp smelt]

Try finding z

use eular's form of a complex number to do so

[brisk trellis]

I find only

One

√2 e^i(π/4)

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[strong furnace]

How would you show that all groups with an order of 6 is either an isomorphism to the group Z/6Z or the Symmetric group of {1,2,3}

[pearl pond]

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[kindred geyser]

Hello

[safe lance]

https://tenor.com/view/0001-gif-17282391190974969363

[kindred geyser]

For the 7 g). Anyone can show how to simplifly ?

[sterile tusk]

since it looks like an arcsine situation, what would we substitute for sine theta

[kindred geyser]

🤔

[sterile tusk]

actually it isn’t arcsine

[kindred geyser]

In the corrected version.

They add e^x

but idk how

[frosty knot]

use substitution u^2=e^2x -1

[kindred geyser]

to look like arcsexcx

[sterile tusk]

yeah kinda

actually you can prob try that

if you multiply e^x to the top and the bottom

that’s be the x in the denominator

id say try subbing $sec(\theta)=e^x$

[jolly parrot]

y0shi

[sterile tusk]

trig subbing is usually the solution to these types of problems

[kindred geyser]

can i do that in exam ?

[sterile tusk]

certainly

since you’re basically multiplying by 1

and you can definitely just u sub if you do that

and get it into an arcsec situation

[kindred geyser]

in that case we multiply e^x right ?

k thx

[sterile tusk]

to the numerator and the denominator yes

that way we don’t change the value of the fraction

[kindred geyser]

ty again

.close

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[ornate granite]

working thru my calc 2 study guide and i ended up with the wrong answer here

[ornate granite]

im not very confident on what we've been learning this semester, but the answer the study guide provides is that the function converges to 448/3

[pearl pond]

@ornate granite Has your question been resolved?

[ornate granite]

<@&286206848099549185> ._....

[ornate granite]

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[errant flax]

yo can someone help me with a question plz?

[errant flax]

.close

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[rocky axle]

Need help

[rocky axle]

I entered the equation into a calculator but it said it was wrong still

[digital cargo]

do you know how to compute hydrostatic density in general?

[maiden seal]

it asks for $\rho \times g \times \int dx$ did you multiply by these values?

[jolly parrot]

sadkid

[pearl pond]

@rocky axle Has your question been resolved?

[rocky axle]

how do I?

[rocky axle]

no

[pearl pond]

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[flint musk]

.close

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[fallen tartan]

i have AB = 20

and A is one of the sides of the hexagon

but i could try finding the area of the two white triangles and do some intuition from that

wait is it js 30?

[grizzled dust]

what do you know about the inner angles on the inside of the hexagon?

[fallen tartan]

the interiror angles?

[grizzled dust]

as in the angles on the inside of the figure

the 6 angles

what's the sum of angles in a triangle?

[fallen tartan]

180

[grizzled dust]

and in a rectangle?

[fallen tartan]

369

360 lol

[grizzled dust]

and a pentagon?

[fallen tartan]

540

[grizzled dust]

and hexagon?

[fallen tartan]

and hexagon is 720

[grizzled dust]

yes

and the key here is that it's a regular hexagon, what does that mean?

[fallen tartan]

all sides congruent

[grizzled dust]

yep, and all angles

[fallen tartan]

and all angles

yea lmao

we type at the same time lol

[grizzled dust]

so what would be the measure of one of the inner angles of the hexagon?

[fallen tartan]

120?

or 60

[grizzled dust]

indeed

720 / 6 = 120

[fallen tartan]

wait can you construct special right traignels on the polygon

[grizzled dust]

aha good thought

so what triangle do you have in mind?

[fallen tartan]

30 60 90 right

[grizzled dust]

what segment do you want to draw in though?

[fallen tartan]

oh idk how to draw on this thing

dy have that feature?

sry

[grizzled dust]

so consider the diagonal of the rectangle, it would be perfectly splitting the hexagon in half right, so what is it doing to that 120 degree orange angle?

[fallen tartan]

it would split to 60 degrees

[grizzled dust]

yup exactly so these angles here are 60 degrees, see if you can now solve the problem with this

[fallen tartan]

oh so it would be a * a(sqrt3) = 20?

as a as one of the sides of the rectangle

[grizzled dust]

ya exactly and a is the side of the hexagon

[fallen tartan]

wait idk if that would work tho

i got somth wierd

[grizzled dust]

now i dont know the formula for the area of a regular hexagon based on its side (there probably is one), but with that you know all the sides of those triangles to either side of the hexagon so you can find out their areas and with that you can get up to the area of the hexagon

[grizzled dust]

?

[fallen tartan]

wait what would A equal

the side?

[grizzled dust]

well unfortunately whoever wrote this problem did a terrible job making the numbers come out nicely, yes, $a=\sqrt{\frac{20}{\sqrt{3}}$

[jolly parrot]

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

[grizzled dust]

i would just label the sides of the rectangle a and asqrt(3) then work out the areas of the triangles in terms of a then substitute that ugliness in for a at the end 😄

[fallen tartan]

oh

ok

theres 6 equalateral triangels

@grizzled dust

can you help me out here 😭

i don't think the answer is an integer

[grizzled dust]

find the area of this (hypotenuse is a, the side on the right along the recetangle is half of the rectangles length so a * sqrt(3)/2 and you can find out the other leg from that

[grizzled dust]

no i dont think so either, why were you expecting it to be? 🤔

[fallen tartan]

oh i'm using mathcounts trainer which is liek competion math

[grizzled dust]

actually i think it will randomly come out to an integer, i just looked up the formula of the area of a hexagon 🙂

[fallen tartan]

and all answers are integers

oh

wait is the anser 30?

[grizzled dust]

how did you get that

it is in fact 30 😄

[fallen tartan]

oh i actually didn't do that wierd square root thing

and like

there are triangle symetries ig

then i did some intuition

[grizzled dust]

yes, there are various ways to think about this

[fallen tartan]

got it right yay

thx

have a nice day bro

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[midnight haven]

Okay here is the problem, the topic is about the Binomial Theorem for context.

now, just to get things straight from the gate go, i am not struggling to get the asnwer, i already got it in fact here it is.

So, in what exactly do i need help if i already got my answer? (I checked with my teacher btw)

I need help to understand why that process that i did is good, why the term that i picked (the 15 (x^2/2)^2 (a/x)^4) is the one that gave me the right answer. Why couldnt i just did it with another term? why did it had to be that one? i tried with another and it didnt quite work. I believe there is something am missing because tomorrow i got a test and i cant rely on luck like i did here and expect to get the answer first try. (

[midnight haven]

excuse me for my poor handwriting

[marble sigil]

that's when the x's cancel to give a constant

[midnight haven]

wait

watchu mean

[marble sigil]

15 (x^2/2)^2 (a/x)^4
this one's good because it has x^2^2 and a 1/x^4

[midnight haven]

so the trick is to pick a term in which both thingys cancel each other to make the equation easier when = with 960?

[marble sigil]

idk if easier but that's what constant means, there's no x's in the term

like all the others will have x's, the full thing is like 7x^-2 + 960 + 5904x^2

[midnight haven]

like the one i picked? because in the one i picked x4 and x4 cancel each other making it alot easier

[marble sigil]

if it asked for the x^2 term you'd choose a different one

[midnight haven]

ohhhhhh sooooo i have to pick a term in which x vanishes?

like cancels itself out?

[marble sigil]

right

[midnight haven]

thats it?

[marble sigil]

yea that's all it is

[midnight haven]

AYEEEEE

[marble sigil]

constant is when the x's cancel

[midnight haven]

Thank u bro

[marble sigil]

np 👍

[midnight haven]

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[thorny agate]

im back, still no idea how to do 36-39

[cosmic charm]

,rccw

[rough stream]

You're bored, I see

[thorny agate]

Look

My dm just gave me a new character to play

I was excited alright

[cosmic charm]

do you know what cos * tan is

[thorny agate]

No

[cosmic charm]

do you know how to write tan in sin and cos?

[thorny agate]

also no

[cosmic charm]

actually wait nvm you dont need to do that

have you learnt the quadrant signs?

which function is positive negative in which quadrant

[thorny agate]

Ah yes

[cosmic charm]

ok so for 36 is q3

cos is + or - in q3

[silent zenith]

can some one help in my channel pls

[thorny agate]

negative

[cosmic charm]

tan?

[thorny agate]

Positive

[cosmic charm]

and what is negative multiply positive

[thorny agate]

Negative

[cosmic charm]

so 36 is negative

[thorny agate]

Ahhh I get it

What would it be for 37 then, when it can be in any quadrant?

[cosmic charm]

do you know what sec is?

[thorny agate]

Because if cos and sec are positive they can only be in 3 and 1

[thorny agate]

yes, the reciprocal of cos

[cosmic charm]

and if you multiply it by cos?

what do you get

[thorny agate]

1

[cosmic charm]

1 is positive right

[thorny agate]

Indeed

[cosmic charm]

so for any quadrants

sec * cos always gives 1

1 is positive

so sec * cos is positive

[thorny agate]

Okay got it

Alright I’ve got 38 and 39 done too

Positive and negative, right?

[cosmic charm]

yeah

[thorny agate]

Alright then im good on thise

On a similar topic tho, is there a way to find the reference angle without consulting one of those big circle charts?

Nvm im@being dumb

Thank you for your help!

.close

[pearl pond]

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[pearl pond]

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[naive trellis]

Hello, looking for a bit of help with calculus if anyone would be willing to support. Thank you

[naive trellis]

I am trying to find the derivitive of this function

using the product rule I get this answer, however im lead to believe the x^-3/5 addition is incorrect

[gilded temple]

u dont need product rule

take 6 outside the derivative coz its constant and differentiate x^-3/5

[robust oar]

it should be -2/5

[gilded temple]

btw d/dx (c) = 0

[naive trellis]

Really? So derivitive of a constant is 0?

Not 1

I see. That makes much more sense

[silent zenith]

magnus pls help me

[naive trellis]

@gilded temple Thank you

.close

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[nocturne cosmos]

For g, how do you solve this without a graphing calculator?

[nocturne cosmos]

For h, how do you solve it without doing synthetic/long division over and over again?

Don't have any work since I don't know how to get started.

[pearl pond]

@nocturne cosmos Has your question been resolved?

[nocturne cosmos]

<@&286206848099549185>

[pearl pond]

@nocturne cosmos Has your question been resolved?

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[marble obsidian]

I found out that its continuous within the range as well as it is differentiable

what's next?

evaluate both f(-1) and f(2)?

[lilac quartz]

equate $f’(c)= \frac{f(b)-f(a)}{b-a}$

[jolly parrot]

.doc

[marble obsidian]

that's my f'(c) right?

[lilac quartz]

well g(x)

[marble obsidian]

g(x)?

g'(c)?

[lilac quartz]

The question is given in g(x)

find g’(c)

[marble obsidian]

using this and plug in values right?

I got 7/18 from it

but is that just my g'(c)?

[lilac quartz]

your g’(c) must be an expression in c

it should be a value

[marble obsidian]

but I thought we defined g'(c) as a constant

from the equation

so is it just matching it up with g'(x)?

[lilac quartz]

what’s g(x)?

can you write it here?

[marble obsidian]

21/(x+7)^2

[lilac quartz]

so what’s f’(c)?

[marble obsidian]

7/18

[lilac quartz]

how?

we didn’t get to find a c yet?

[marble obsidian]

we defined it here no?

we know all the values

[lilac quartz]

just replace x by c

and then solve the result equation you get

$\frac{21}{(c+7)^2} = \frac{f(b)-f(a)}{b-a}$

[jolly parrot]

.doc

[lilac quartz]

find the right side value and solve for c

[marble obsidian]

alr did

[lilac quartz]

essentially you are looking for all c that has this property

[marble obsidian]

it was 7/18

[lilac quartz]

does -7/18 solves this?

[marble obsidian]

positive 7/18

[lilac quartz]

Then yes, it could that

[marble obsidian]

it's just finding c after that

then

k thxs

[lilac quartz]

yes, that’s all

Check your calculations again

[marble obsidian]

?

should be right

21/(c+7)^2 = 7/18

which crossmultiplies and solves to be c = 3sqrt(6)-7 and -3sqrt(6)-7

[lilac quartz]

yes

those are the possible values for c

[pearl pond]

@marble obsidian Has your question been resolved?

[marble obsidian]

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[midnight haven]

Can someone explain this: https://cdn.discordapp.com/attachments/890523039786229770/1219216151742251059/IMG_2330.jpg?ex=660a7e86&is=65f80986&hm=52c9dcba8ab1f1bd4acd72744a89a92c62a73d2cad1943b2363fc1c9e181e327&

[last summit]

why is your picture watermarked

, rotate

[last summit]

What is your doubt?

[midnight haven]

I just dont really know what it means

Thanks for helping btw

Like if i had

For example

X^5-5x^4+5x^3+5x^2-6x

[midnight haven]

How would i solve it with this

[last summit]

did you just make that up randomly

[midnight haven]

No its a question in my paper

[last summit]

I see

so your paper is being tricky because

in general a 5th degree polynomial will not just have roots you can find easily

but this one (and likely any that are given to you) will

[midnight haven]

Oh

[last summit]

for instance, those terms all share a common factor of x

do you agree?

[midnight haven]

Yes

We can take x out

[last summit]

mhm

also by looking at it

you can kind of think

hey maybe x=1 works

and try that out

and if that works i mean you may aswell try x=-1

because if x=1 is a root

(x-1) is a factor

[midnight haven]

The end result should equal 0 right

If i sub 1 in

Oops

[cosmic charm]

basically you use the rational root theorem and try to find one root first

which is list the positive and negative factors of the constant term, and see which one makes the function equal to 0

[midnight haven]

Oh i see

So if i take out z

X

The constant would be 6

And i just use the positive and negative factors of 6?

[cosmic charm]

yeah

[midnight haven]

How would i go further though

[cosmic charm]

what are the factors of 6

[midnight haven]

1, 2, 3, 6

[cosmic charm]

and -1, -2, -3, and -6

[midnight haven]

Oh yes

[cosmic charm]

now you put all 8 of these numbers in and see which one gives 0

[midnight haven]

Ohh

[cosmic charm]

then that is one of your factors

then use long division

[midnight haven]

Wouldn’t that take a while though

Is there a faster way

[cosmic charm]

idk calculator i suppose

[midnight haven]

Hm ok

Thanks

I get it now

.close

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[stoic siren]

give me a hint please

(idk calculus)

(havent been taught it yet)

[vivid grove]

So what have you attempted? You need to know calculus for this question

[stoic siren]

no other way?

cause i havent been taught calculus

as for what I have attempted, i havent done anything, just made them equal to each other and that gave me x^2-mx-b = 0 but that isnt anything i think

[vivid grove]

So we know two things

1. x^2 -mx - b = 0

For the point of intersection of the parabola and line

We don't want there to be two solutions as that would be mean the line is a secant and not a tangent

I will let you think about it

[stoic siren]

discriminant

and we want one solution

[vivid grove]

bingo

[stoic siren]

so its =0

delta = 0

[vivid grove]

yep

[stoic siren]

$\delta b^2-4ac= 0$

[jolly parrot]

pixel

[stoic siren]

woah

uh

i meant like

the triangle thing

yk what i mean

$(-m)^2-4(1)(-b)=0$

[jolly parrot]

pixel

[stoic siren]

$m^2+4b=0$

[jolly parrot]

pixel

[stoic siren]

wait

cant we use this

so that like

y + 2 = m(x - 1)

[vivid grove]

There might be multiple ways. I think the path we are on is good.

[stoic siren]

ok lets continue doing what we were doing

cause im curious on what to do next

[vivid grove]

So we can get rid of the m in mx + b

[stoic siren]

why

[vivid grove]

Using substitution

[stoic siren]

elaborate

we havent found m

to get rid of it

oh wait

wait yeah we havent found it

[vivid grove]

m^2 +4b =0 => m = +/- sqrt(-4b)

[stoic siren]

yeah but we dont know what b is

[vivid grove]

Think about it for bit. What information do we have.

[stoic siren]

so we have the straight line(s) passing through (1, -2)

[vivid grove]

yep

[stoic siren]

$m = \pm \sqrt{-4b}$

[jolly parrot]

pixel

[stoic siren]

uhh

$x^2-mx-b=0$

[jolly parrot]

pixel

[stoic siren]

and then the two general equations which are y= x^2 and y=mx+b

[vivid grove]

Yes and y=mx+b is what we want to find and now we know m

[stoic siren]

so we substitute value of 'm' into y=mx+b

so it would be

$y=x(\pm \sqrt{-4b}) -b$

[jolly parrot]

pixel

[vivid grove]

Yep

[stoic siren]

mm

now we make it equal to x^2?

[vivid grove]

No that is for finding the intersection point

[stoic siren]

oh right

we still need to find 'b'

right?

[vivid grove]

yep

[stoic siren]

ok

so b is y-intercept

[vivid grove]

yes

[stoic siren]

mhmm

oh wait

we can just sub in the coordinate now

only one unknown variable

[vivid grove]

yep

[stoic siren]

which is b

(1, -2)

$-2=\pm \sqrt{-4b} - b$

[jolly parrot]

pixel

[stoic siren]

wait we have a negative surd i just realised

that means b < or = 0

[vivid grove]

yep

[stoic siren]

$b \leq 0$

[jolly parrot]

pixel

[stoic siren]

ok so how would we solve now

to get rid of square root

we can square everything

but is that efficient

[vivid grove]

yep

do that

[stoic siren]

$4 = 4b - b^2$

$\pm -4b = - - 4b = 4b$

\pm just becomes 4b

[vivid grove]

slowdown

[stoic siren]

wait no

[jolly parrot]

pixel

[vivid grove]

It is probably best we add b to both sides therefore we don't have to deal the plus/minus sign

b-2 = +/- sqrt(-4b)

[stoic siren]

ok right one sec

[vivid grove]

b^2 -4b + 4 = -4b

[stoic siren]

$-b-2= \pm \sqrt{-4b} \ \text{square everything} \ (b-2)^2 = \pm \sqrt{-4b}^2 \ \ b^2-4b+4=-4b$

[vivid grove]

yep

[jolly parrot]

pixel

[stoic siren]

ok yes

so i dont need pm?

plus minus

cause we squared it

right?

[vivid grove]

yep

[stoic siren]

ok

$b^2 + 4 = 0 \ b = \pm 2$

[jolly parrot]

pixel

[vivid grove]

No

[stoic siren]

oh?

[vivid grove]

Something is either wrong with our calculation or the question

We can't take the square root of a negative number

[rapid oak]

4 + 4 does not equal 0 ngl

[stoic siren]

where did u get 4+4 tho

[rapid oak]

From the plus or minus

2 squared is 4

[stoic siren]

oh right yea

when testing it

it doesnt work

which means we did it wrong

[rapid oak]

Ok

[vivid grove]

When we did mx+b, mx-b was written instead mx+b

[vivid grove]

This post over here

[stoic siren]

i fixed that already

by taking it to the other side

making it

-b

oh wait

(-b-2)^2

[rapid oak]

@jolly parrot idiot

[vivid grove]

Yep that would be the issue

[stoic siren]

okay yeah

i see where we went wrong

[rapid oak]

@jolly parrot your dum

[stoic siren]

$(-b-2)^2 = \pm \sqrt{-4b}^2 \ \ b^2+4b+4 = -4b$

[jolly parrot]

pixel

[vivid grove]

yep that looks good

[stoic siren]

$b^2+8b+4=0$

[jolly parrot]

pixel

[stoic siren]

solve for 'b'?

[vivid grove]

Yep

[stoic siren]

quadratic formula

ugly number

i cant be asked simplifying surd

can i just give decimal

and round up to 2 decimal points

b = -0.54
b = -7.46

[vivid grove]

Yep rounding is good

You can also give exact answer. Itis probably best in this case otherwise use wavy equal sign to show your answer is an approximate

[stoic siren]

yeah

what would we do now

we got two b values

so two equations

right?

two straight line equations

[vivid grove]

We can find m using the formula we have

[stoic siren]

yeah

$m = \pm \sqrt{-4b} \ \$ let b $\approx -0.54 \ m= \pm \sqrt{-4(-0.54)} \ m\approx \pm 1.47$

how would we take the plus minus into consideration

thats gonna give us so many different m values

[vivid grove]

I think checking with desmos could be good idea, i.e. said I would expected one solution not two

[jolly parrot]

pixel

[vivid grove]

One thing to watch out for is the error propagating as you are using the rounded version b.

[vivid grove]

So looks like the -m is the answer

[stoic siren]

yeah thats right

cause it passes through (1, -2)

and then the second question

is

hold on

lemme cook

$m = \pm \sqrt{-4b} \ \$ let b $\approx -7.46 \ m\approx \pm \sqrt{-4(-7.46)} \ m\approx \pm 5.46$

[jolly parrot]

pixel

[stoic siren]

+m is answer for this one

so the two equations are

[vivid grove]

I think what happened is when we applied the quadratic formula the first time when doing x^2 - mx -b = 0 a second solution got introduce as a result.

[vivid grove]

Yep

[stoic siren]

god finally

that took sso long to solve

iff something like this comes up in an exam

im most likely gonna get cooked

which means ill get it wrong

like you did an amazing job in hinting me in the right direction

like seriously damn, you never even gave me the answer

you made me think

[vivid grove]

I don't think it will come. To many steps

[stoic siren]

and i got it right but made a lot of mistakes along the way

[stoic siren]

yeah hopefully

thing is with us

we have questions called

complex unfamiliar

the questions are literally what they're called

they're complex

and they're unfamiliar

and these questions are usually worth like

lots of marks

so like

I really wanna be able to answer them

its the only thing i struggle with

so a question like this one would've prob be a complex unfamiliar

[vivid grove]

I see

[stoic siren]

anyway

i appreciate your help

like seriously thank you so much

you did really well explaining each step

all your messages were really concise and made sense to me

so thank you

not everyones help is like that

sometimes they use big ahh words that I've never heard of

🙏

i will now be heading to sleep

this is my revision for exam this thursday

[vivid grove]

good luck on your test

[stoic siren]

got a couple more exercises to go through 💪

tomorrows job

[stoic siren]

thanks champ

have a great rest of your day/night/evening/afternoon.

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[mild shuttle]

I need help matching figures with a corresponding area equation i have an image

[dark sparrow]

how do you calculate the area of a square

oh ok your just a troll

[shut mason]

this is simple 😭 use the formula of a = a^2 and figure it out

[shut mason]

probably pre-algebra math

[mild shuttle]

Im not trolling i just didnt know how to do this. Im in a geometry class

High school

[mint flax]

Do you know how to find areas of squares and rectangles?

[mild shuttle]

It looks confusing to me until its explained in a easy way then i get the fundamentals

[mild shuttle]

Add the sides? I havent done that since middle school to be honest our school gives easy work

[mint flax]

No

[mild shuttle]

I know it seems like im trolling or its easy but i promise im just clueless 😭

[mint flax]

Area of square = (side)²

And area of rectangle = length x breadth

Using this find the areas of the figures

[mild shuttle]

breadth?

[mint flax]

The smaller side

[mild shuttle]

Ohh

I can use that for this? It shows one side on two of them

[mint flax]

Yeah those are squares buddy

[mild shuttle]

So A= 2x ^ would be what i use and the ^ is supposed to represent squared

[mint flax]

For first one?

No

The two also has to be multiplied

[mild shuttle]

Each side is the same so they are all 2x right

What would i multiply the 2 by

I know all the sides are same and it shows all 90 angles are same

[mint flax]

For area=side², so you have to do (2x)²

[mild shuttle]

Oh yeah thats what i meant i couldnt find the squared thing to put on the equation i meant that though

You multiply 2 by the 2

B would go to #1 and A to #2 right

[mint flax]

C to #1

[mild shuttle]

Oh that makes sense after multiplying the 2

So awnsers are C,A,B Correct

Cause #2 x squared is x

Thanks for the help i just needed a refresh

[pearl pond]

@mild shuttle Has your question been resolved?

[pearl pond]

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[dusk zodiac]

I am having some trouble in analytical geometry. The question is: If a line 3x+y-9=0 cuts a segment line joining the points (1,3) and (2,7). Find at what ratio does the line cuts it.

[pearl pond]

@dusk zodiac Has your question been resolved?

[dusk zodiac]

<@&286206848099549185>

[pastel flax]

first find the line that joins (1,3) and (2,7)
do you know how to do it?

[dusk zodiac]

yea i know how to do it

[pearl pond]

@dusk zodiac Has your question been resolved?

[pastel flax]

write the equation of that line and find the intesection point of them and then use section formula

[dusk zodiac]

thanks man!

.close

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[raw whale]

Hey everyone, having trouble finding a tutor so I came here.
The question is
Given the function f(x)= 5x - 6
If f(x) = 5w + 9 what is x?

[snow sail]

can you explain the question a little bit more

is w a function of x?

so its 5x-6 = 5w(x) + 9

[terse cedar]

x = w

is it that simple ? :p

[raw whale]

I got x = w+3

Hold on

[terse cedar]

ye

[snow sail]

sure

[terse cedar]

u're right

[raw whale]

It is not there are two equations

[snow sail]

two?

[pearl pond]

@raw whale Has your question been resolved?

[raw whale]

Well theres the given function and the one that says to solve for x

I thought I had to make then equal to each other

<@&286206848099549185>

[snow sail]

yea, to be honest, its not really clear

i think you have the right idea though

like, f(x) = 5x-6

but,,, it just doesnt really make sense

i almost wonder if its a typo

theres no other w on this worksheet?

[snow sail]

this problem, if im interpreting it right, and if it make sense, would be

i take it back actually, wondering if its just a typo now

[raw whale]

No there isnt

I got it just now nvm

[snow sail]

trying to think of what it should be

okay

[raw whale]

Thanks anyways guys!

Ill close up I guess

.close

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[weak flare]

why did we assume that t will be after 5

bc that equation can only be formed if we assume that

[wooden merlin]

Then they are changing constantly.

[wooden merlin]

Is that a formula? x=1/2 a t^2?

[weak flare]

just equating area under the graph

[wooden merlin]

I guess you could just set up an equation and solve for it.

[weak flare]

bc distance=area under graph for a speed-time graph

[wooden merlin]

$$A=B$$ $$50+20t=60+12t$$ $$t=\frac54$$

[jolly parrot]

Good

[wooden merlin]

This equation tells when the distance is equal after 5s, if you want it included, t=t-5) would probably work (as what the solution were doing).

[pearl pond]

@weak flare Has your question been resolved?

[weak flare]

I see

[pearl pond]

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[wind lagoon]

Hello there. Let $[a,b]\subset\mathbb R$ and $f$ a function from $\mathbb R\to\mathbb R$. Under which conditions on $f$ does it hold that $$f^{-1}([a,b])=[f^{-1}(a),f^{-1}(b)]?$$

[jolly parrot]

Philip

[wind lagoon]

I guess f needs to be bijective for one thing. But are there any more requirements?

[autumn narwhal]

It doesn't

Consider something like this

[gritty onyx]

Consider, as a counterexample, a function that sends 4 to -1, 5 to 1, -1 to 4, 1 to 5, and for all other numbers its the identity function. This is bijective, but f([4,5]) is not [f(4), f(5)]

[keen rain]

@wind lagoon Injectivity & Continuity

[wind lagoon]

ok, interesting 👍

[keen rain]

Injectivity to ensure existence of inverse

Continuity to ensure the inverse of an interval strictly aligns with the interval of inverses

[cinder flower]

i think there's a bit more nuance there

[gritty onyx]

I think it might need to be fully bijective in order to insure invertible over all of reals

[keen rain]

if [a,b] can be chosen even if f^-1(a) or f^-1(b) wouldn't be defined then yes, bijective

[cinder flower]

to even have an f^{-1} you need f to be bijective

what you want instead of f^{-1}(a) is like

inverse image of {a} under f has one element

random example: take f(x) = abs(x) and let a=-1 and b=0. then f^{-1}([a,b]) = {0}, but [f^{-1}(a), f^{-1}(b)] doesn't even make sense

[keen rain]

only if f is unbound in range

ah yh f:R->R

[keen rain]

but f(x) doesn't have an inverse

[wind lagoon]

ok, the reason for the question; I was watching this video and there is this part where the author writes f^{-1}([a,b))=[f^{-1}(a),f^{-1}(b)), hence I was wondering on what grounds this is justified

[cinder flower]

yea

[cinder flower]

but the fact can hold for functions that don't have inverses

[keen rain]

yes due to the blue requirements

they include

bijectivity

as well as continuity

[wind lagoon]

👍

[keen rain]

due to differentiability, monotony and surjectivity

[keen rain]

are the requirements from before somewhat clear why they must hold?

[wind lagoon]

yeah, so continuity and bijective, or?

[keen rain]

yep but I meant is it clear to you why these two

[wind lagoon]

hmm, well, bijective and continuous, then it is strictly monotone, right?

[cinder flower]

the reason i brought it up was because whether f needed to be bijective was under contention

[keen rain]

but f(x)=abs(x) illustrated the need for injectivity

which was at the start

surjectivity would be due to the requirement f:R->R

if f has to strictly cover R for domain and range

however we could also take a non-surjective function

e.g. sigmoid

[cinder flower]

it is not needed if you loosen up what f^{-1}(a) means

[keen rain]

which would fulfill the requirements

but doesn't have R as range

[keen rain]

I consider it under the definition in analysis

[cinder flower]

yea to me f^{-1}(a) is the inverse of f evaluated at a too

and i wouldn't write that if f wasn't bijective

[keen rain]

hm but any sigmoid f: R -> R would not be surjective, yet has an inverse by restricting the range

maybe the definitions on it can differ

[keen rain]

yes, but the crucial part is the continuity

because if f were discontinuous at any x

then you can find an interval

such that the given statement doesn't hold

because that interval will have a gap in its inverse interval

[cinder flower]

you could drop the bijective requirement and require f^{-1}({a}) and f^{-1}({b}) have one element and some other condition. that's what redstone's example was like

[keen rain]

ic

but since [a,b] is arbitrary

nowhere on f^-1(x) we'd be allowed to have two values

[keen rain]

nywys apprec the clarification 🐛

[pearl pond]

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[smoky saffron]

Am i doing it wrong

[gritty onyx]

You cut off part of the problem

[smoky saffron]

?

[gritty onyx]

Your acceleration should have two terms, one in t to the 1/2 power and one in t to the 3/2 power. Your work will be more clear if you simplify v(t) and a(t)

[smoky saffron]

oh

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

How can I get the derivative of this expression?

[midnight haven]

I've tried using the chain rule twice, once on the outer function and once on the inner exponential, but that didn't seem to work

Here’s my work:

[nocturne plover]

why is the exponential in the root changed from e^-5t^2 to e^25t^2?

[midnight haven]

i assumed i could convert it to 25t^2 since squaring -5t results in 25t^2

unless that's only with parenthesis?

[nocturne plover]

but you're not squaring the -5

yes if it was (-5t)^2

then it would become 25t^2

[midnight haven]

ah i see

[nocturne plover]

but atm it's just -5 t^2

[midnight haven]

i'll try it without changing the exponential

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[glossy roost]

this look right ?

[meager trellis]

well the exact answer would have pi in it because it's a cylinder

[glossy roost]

49pi then

wait no

4/49pi

[meager trellis]

that's also what i got

and that isn't 2.5 m/min

[glossy roost]

im tripping then

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[calm quiver]

Someone help with c

[slate swift]

Lmao did you even solve the previous question

[calm quiver]

BRO HOW DID YOU APPEAR

[prime bramble]

what're you trying to do? prove an identity?

[slate swift]

Is there a problem?. This server is for everyone to learn.

[calm quiver]

Yes

Precisely

[calm quiver]

Yeah

No I was just saying

Like

Your method was making no sense lol no offense

ING got us tho

You can learn with me

[slate swift]

Bruv I was trying to help you after I had solved it, have some patience

[calm quiver]

Oh my fault cuh

[slate swift]

I was just trying to make sure you understand each step

[prime bramble]

okay, what have we tried?

how far did you get?

[calm quiver]

So

I got this far

[prime bramble]

okay

that's good so far

do you see anything that you can do with the numerator?

perhaps, a common factor?

[calm quiver]

Hmmm

No because the cos and sine are one term

[prime bramble]

you have two terms in the numerator though

cos^2 and cos^2 * sin^2

[calm quiver]

Right

[prime bramble]

right

[calm quiver]

So how do I take a common factor out

[prime bramble]

what's common to both terms?

[calm quiver]

I can’t take out cos

Uhh

Cos