help-13

[flat kayak]

I think it's 8t^2-6/t-1

that'

that's the answer I got

[dire geode]

!show

[cedar kiln]

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

[dire geode]

oh you're just missing parentheses

use

[flat kayak]

is that all?

[cedar kiln]

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[flat kayak]

.close

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[still pecan]

Is that the correct interpretation?

[dire geode]

no

either you wrote it down wrong or you need to use chain rule

[storm mural]

if numerator is one differential and denominator is product of 2 the answer will be 0 so the left can’t be correct

oh maybe it’s dhdh/dxdy

[dire geode]

d^2 h would be correct

[still pecan]

Yes my bad, I was meant to write d^2 h

[dire geode]

yea that's the definition of mixed partial derivatives

[cedar kiln]

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[storm ivy]

i can’t seem to get it right, the answer should be x between 3 and 5

[cedar kiln]

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[cursive trout]

How would you find the most amount of unique rationals summed together, in the form of 1/x, that sum to 1, However your x has to be between 2-2023?

[cedar kiln]

@cursive trout Has your question been resolved?

[cedar kiln]

@cursive trout Has your question been resolved?

[crystal raptor]

interesting, where is this problem from?

[cursive trout]

this is from some competition my school is currently a part of i could send the screenshot of the question that we had submitted into teams

just pulled it out

[brave aspen]

I dont know how to find the answer other than just randomly guessing xD. Curious what kind of number theory stuff someone else may be able to suggest

That being said i found a set of 3 such numbers

The other part i have no clue someone ping me too if you know how to solve it

[cursive trout]

i found 11 doing something basic

but apparently theres a lot more

these were my values of x in the set

2,4,8,16,32,64,128,256,512,768,1536

i know people got atleast into the 40's in terms of number of numbers

and apparently you could get up to 579 i think it was said

[cedar kiln]

@cursive trout Has your question been resolved?

[fair geyser]

just as i got the idea how it should be bruteforced, i;m too tired

[mint prism]

due to 6 being a perfect number, 1/2 + 1/3 + 1/4 work where 2,3,6 are all divisors of 6 other than one

similarly

,w 1/2 + 1/4 + 1/8 + 1/16 + 1/31 + 1/62 + 1/124 + 1/248 + 1/496

[wraith dagger]

Results provided by WolframAlpha

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

Requested by yenn#4559

[mint prism]

will add up to one, as 496 is the second perfect number

maybe you could do sth with these

but doesnt seem like its in the heart of the problem

[cursive trout]

i see

i'll give it a shot anyway

thanks a lot

[cedar kiln]

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[sly cape]

An ethical hacker, also referred to as a white hat hacker, is an information security expert who systematically attempts to penetrate a computer system, network, application or other computing resource on behalf of its owners. The purpose of ethical hacking is to evaluate the security of and identify vulnerabilities in systems, networks or system infrastructure. It includes finding and attempting to exploit any vulnerabilities to determine whether unauthorized access or other malicious activities are possible.
An ethical hacker is required to randomly guess the correct pin code that consists of the number 0 through 9 that must be entered in the correct order to access a company system.

1. What is the probability that the ethical hacker will guess the pin code correctly on the first try?
2. There are many variations of this guess. Assuming the primary variation allows the ethical hacker to guess correctly if the four-digit in the number are selected in any order as long as they are the same four digits as set by the IT security of the company.
   For example, if the ethical hacker picks four digits making the number 2376, then the he will guest it right if 2376, 3726, 6327, 7632, and so forth, are entered. Consider the following four different versions of his presumptions.
   (a) All four digits are unique (e.g. 1234)
   (b) Exactly one of the digits appears twice (e.g. 2334, 8185)
   (c) Two digits each appear twice (e.g. 1212, 8855)
   (d) One digit appears three times (e.g. 2226, 8188)
   Find the probability that the ethical hacker will successfully guess the accurate pin code in the first try for each of these four situations.
   Show the necessary steps and explanation for the four presumptions stated above.

someone help me answer this please

[sly cape]

<@&286206848099549185>

.close

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[vast yacht]

how do i find the total distance? i've tried like three times and it's all come out wrong

I though you took the distance traveled between t=2 and t=6 and add it to the 3200 but it said it was wrong

[cedar kiln]

@vast yacht Has your question been resolved?

[vast yacht]

@cold briar

[steel heart]

its <@&286206848099549185>

[vast yacht]

thank you

so how do I find the total distance that the particle travels

[steel heart]

is it displacement or distance?

[vast yacht]

it's distance

[steel heart]

do you know all real roots of this eqn

[vast yacht]

wdym by that

[balmy tide]

I am here to help

[vast yacht]

the only thing I'm confused about is how to find the total distance - it says that subtracting the distance travelled backwards between t=2 to t=6 and then adding that to the position at 16 doesn't work

hello thank you for helping

I'm just stumped on what else to do

[steel heart]

wait i am preparing a graph

[vast yacht]

I'm also trying not to graph it with desmos

is there any strictly numerical way I can solve it

[balmy tide]

Umm why do you need graph

[balmy tide]

What is dv/dt?

What is ds/dt

Do you know it

[vast yacht]

no

[balmy tide]

If you know differentiation you can solve it

[vast yacht]

I know differentiation I think

[balmy tide]

What is s/t

Yeah

[vast yacht]

is that multiplying the power by the constant and moving the power down 1

right?

[balmy tide]

You know differentiation of s wrt to t is speed right?

[vast yacht]

I am lost now

no

[steel heart]

its velocity @balmy tide

[balmy tide]

Yup you are right

[balmy tide]

Ok think slow and say what you know

Differentiate the eqn and let's see if you know differentiation correctly

Just do it

[steel heart]

Distance = $(2) + |s(6)-s(2)| + (s(16) - s(2) + |s(6)-s(2)|)$

[balmy tide]

You are right

[wraith dagger]

Pro_Hecker

[balmy tide]

Your problem is solved

S'(t) is the eqn of velocity

[steel heart]

is this right?

[vast yacht]

yes but I'm not trying to find velocity

[steel heart]

this

[balmy tide]

Yeah

You want when particle stops right?

[vast yacht]

I want the total distance

the particle stops at 2 and 6 I used the quadratic formula

[balmy tide]

Oh ok

Listen carefully

[vast yacht]

ok

[balmy tide]

When particle stops it reversed direction?

[vast yacht]

yes

[wraith dagger]

Pro_Hecker

Distance = $s(2) + |s(6)-s(2)|  + (s(16) - s(2) + |s(6)-s(2)|)$

[balmy tide]

Good after 6 sec it again reversed?

[vast yacht]

yes

[balmy tide]

Great

[steel heart]

Distance = $s(2) + |s(6)-s(2)| + (s(16) - s(2) + |s(6)-s(2)|)$

[wraith dagger]

Pro_Hecker

[balmy tide]

Now the particle travels in opposite direction between 2 to 6

[steel heart]

😦

[vast yacht]

correct

[balmy tide]

So total distance travelled is

[vast yacht]

segment from 0-2 plus 2-6 plus 6-16?

[wraith dagger]

Pro_Hecker

[balmy tide]

Distance travelled in 2 seconds+distance travelled in6-16-distance travelled between 2-6

[balmy tide]

Yup

But minus

The 2-6

[vast yacht]

minus 2-6? but isn't it total distance?

would it not count?

[balmy tide]

Distance is scalar remember it

[vast yacht]

ok

[steel heart]

is my internet not working?

or am I being promptly ignored 😦

[balmy tide]

Yh they are asking total displacement I guess

[balmy tide]

Sorry man your internet is working alright

[vast yacht]

ohh now I see why I was confused

[steel heart]

total distance and total displacement are different

[vast yacht]

yeah sorry pro hecker I just wanted to focus on one thing at a time

[balmy tide]

Yess

[vast yacht]

I didn't know the difference

hmm that was a bit misleading

[balmy tide]

Yup you are right

[balmy tide]

This is distance

[balmy tide]

This is displacement

[vast yacht]

oh ok

but if the problem's wording was distance?

[balmy tide]

Yes

It is distance

[vast yacht]

so I would solve by adding them all?

[balmy tide]

If the answer is displacement then question is wrong

[vast yacht]

and not subtracting the segment?

[balmy tide]

Yeah

[balmy tide]

Yeah

[vast yacht]

ok thank you

[balmy tide]

Yup

Welcome

You are great at kinematics

All the best

[vast yacht]

so how would I plug the numbers into the equation

[balmy tide]

🫡adios

[vast yacht]

for example how do i find distance travelled from 0-2

putting in 2 just gets me 0

[balmy tide]

Put 2

[vast yacht]

WAIT

yeah I put 2 into the velocity equation and it gave me 0

6(2)^2-48(2)+72=0

[balmy tide]

🤦

Baba put it in the distance equation

[vast yacht]

oops what was my mistake

[balmy tide]

Not velocity

[vast yacht]

ohh s(t) not s'(t)?

[balmy tide]

Yup

You want distance not velocity

[vast yacht]

ok I see

[balmy tide]

Yup

[vast yacht]

and then the other numbers i plug in are the difference between the starting point and ending point?

[balmy tide]

Yes

So can I leave brother

[vast yacht]

ok great thank you

[balmy tide]

?

Welcome

[balmy tide]

?

[vast yacht]

I got 416, but that doesn't look right

shouldn't it be bigger than 3200?

[balmy tide]

Why?

[vast yacht]

because it's total distance, and the position at 16 is less than the total because the function doubles back

that was my original thought, that I could add the segment between 2 and 6 to 3200 but that was wrong

[balmy tide]

Yes

[balmy tide]

Did you add the magnitude or vector

?

[vast yacht]

I added no magnitude or vector

or is it 3232

[balmy tide]

Umm i mean like absolute value?

[vast yacht]

because the distance between 2-6 is s(4) = 32?

and 32 + 3200 is 3232?

[balmy tide]

Noooo

[vast yacht]

how come

[balmy tide]

Put s4

Is wrong

Put s(6)

[vast yacht]

288

no wait

[balmy tide]

Minus s(2

)

[vast yacht]

ohh

s(6) minus s(2)?

but both s(6) and s(2) are 0

[balmy tide]

S(6)-s(2)not s(6_2)

[vast yacht]

?

[balmy tide]

That's velocity

[vast yacht]

no I'm using s(t) not s'(t)

[balmy tide]

Umm

Brother

I think you have to redo the entire question

[vast yacht]

everything else is right... in the pinned message, all the other answers are highlighted in green

[balmy tide]

If you get s(6) as zero

[vast yacht]

which means they're correct

the only one left is the last one

[balmy tide]

Ok

[balmy tide]

Is s(16) wrong?

Can you recheck if s(6)

[vast yacht]

2(6)^3-24(6)^2+72(6)=0
432-864+432=0

nope s(16) is 3200

2(6)^3-24(6)^2+72(6)=0

2(216)-24(36)+432=0

432-864+432=0

the math checks out...this is very strange

[balmy tide]

Wait a second

I wil help you

Ok

Here

S(6) is zero that means

The particle went to orgin again

[vast yacht]

ohh so it was situational not a mistake

[balmy tide]

Yup

Now do jt

It

[vast yacht]

so I do add s(2) and s(6)

64 + 0

and then add that to the 3200?

so it will be to 16, but including how it doubled back?

[balmy tide]

Wait

I will explain clearly

Find S(2)

[vast yacht]

s(2) is 64

[balmy tide]

Now that is the distance travelled

Back again

To reach orgin

[vast yacht]

yes

[balmy tide]

128

[vast yacht]

yes

[balmy tide]

Is the distance travelled till s(6)

Now add it to s(16)

[vast yacht]

3200+128 = 3328

[balmy tide]

Yup

Great that is the answer

[vast yacht]

thank you! that was very kind of you to go so slowly for me

[balmy tide]

Thanks for being careful

[vast yacht]

have a good night!

[balmy tide]

Welcome

Goodniht

[vast yacht]

.close

[cedar kiln]

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[steel heart]

is it night?

[balmy tide]

Nope

[cedar kiln]

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[vast yacht]

it is 1 am for me

[vast yacht]

oop

.close

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[balmy tide]

😂

You can dm me yk @vast yacht

[cedar kiln]

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[hushed badger]

hi can someone help me find the critical points for this equation?

how would i simplify it?

hello zfn we meet again

[storm mural]

set it equal to 0? you can make every power an integer once you do that

[hushed badger]

hmm

wait i dont see it

lemme work it out

wait why can i make every power an integer

[hushed badger]

dont get this part

[storm mural]

get rid of the fractions by multiplying both sides by denominator

[hushed badger]

ok nvm i see what u mean

[storm mural]

there's actually 2 critical points here

*3

[hushed badger]

do i find it with these

oh wait that doesnt make sense

[storm mural]

when you move something to the denominator the power switches sign

[hushed badger]

x = 1 and x = -1 are two

[storm mural]

jsyk

[hushed badger]

cough

oops

right

[storm mural]

might wanna fix this

[hushed badger]

these are two of the roots right

oh wait

the errors all the way there

lemme check

oh crap

[storm mural]

yeah

that also simplifies

[hushed badger]

wait u said there were 3 roots

cant seem to find the third one

[storm mural]

numerator simplifies

[hushed badger]

o ye

[storm mural]

this is -1

I think

[hushed badger]

wait

hm

[storm mural]

^

[hushed badger]

wait why lol

to be sure ur referring to this?

[storm mural]

yes

[hushed badger]

wait sorry im confused af rn

why is that

[storm mural]

because you cross multiplied
(kind of)

you multiplied top and bottom by (x-1)^4/7

[hushed badger]

uh i multiplied the left top with (x+1)^3/7 and the right top by (x-1)^4/7

[storm mural]

exactly

[hushed badger]

oh wait

OH

I THOUGHT U MEANT THE POWER

oops

[storm mural]

sorry

[hushed badger]

no i should have interpreted it better lol

so now we have this

[storm mural]

exponent rules on numerator

[hushed badger]

okay

hm

1/7, 1, -1

[storm mural]

indeed

[hushed badger]

okay i got it

tysm

[cedar kiln]

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[primal solar]

how to do part b ples help

[livid hound]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@primal solar Has your question been resolved?

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[crimson obsidian]

if A is a square matrix and |A|=0 then can I say that A is nilpotent matrix? If not then provide a counter example

[tropic oxide]

do you know what a nilpotent matrix is

[crimson obsidian]

Nilpotent matrix is a matrix that satisfies the equation A^K=O such that k is natural number

[tropic oxide]

wording could have been better, but yes.

[crimson obsidian]

So

[tropic oxide]

so can you say that a matrix whose determinant is 0 is necessarily nilpotent?

[crimson obsidian]

I don't know tbh

I wanted to know if converse exists

[crimson obsidian]

That's why I asked for a counter example

[full gust]

consider the matrix [1,1][1,1]

[tropic oxide]

^

i was gonna go for something more "minimal" but that works

[crimson obsidian]

I see

Thanks for example. Let me verify it

[crimson obsidian]

Wow that's a interesting example. It makes a pattern of A^n=[2^(n-1), 2^(n-1)][2^(n-1), 2^(n-1)]

Thanks to the example, I now know that A matrix with Determinant=0 isn't necessarily a nilpotent matrix

.close

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[wary hamlet]

how do i solve 9b?

[cedar kiln]

@wary hamlet Has your question been resolved?

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[austere plume]

How do I answer 8b and c

[quartz frost]

you know that the lines intersect at (p,-2)

thus they both share the point (p,-2)

[austere plume]

Oh yea

I forgot about that

What about c

I tried this to find k

But for some reason it don’t work

[quartz frost]

suppose we have p sorted out
thus we know a point on line l2
if we sub the value of that point into l2, we will get k

[austere plume]

Oh yea

Thanks for the help

[austere plume]

I know this isn’t the best way to find k but shouldn’t this also work tho

[quartz frost]

,rotate

[quartz frost]

in line 4 you forgot the =0

in line 5 im confused what you did

the logic i used was:
sub in the point (0,-4) into l1 to get the complete equation for l1
rearrange l2 to for y because i prefer it that way
sub point (p,-2) into l1 to get p, since everything else in l1 is known
sub point (p,-2) with the now known p into l2 to get k

[cedar kiln]

@austere plume Has your question been resolved?

[austere plume]

I used this formula b^2-4ac=0

Since it would have gotten rid of x

So I can solve for k

[cedar kiln]

@austere plume Has your question been resolved?

[cedar kiln]

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[brittle field]

not sure how they got c and d

I see that at x = 0 , the y = 0

[versed kayak]

calculate area under curve

[shell smelt]

And g(2) + sin(1 radian) = g(3)

Assuming that's cosine graph after x = 2

[versed kayak]

yo dont confuse him lmao

[shell smelt]

He asked d too

[brittle field]

start from the beginning

[shell smelt]

0-1: 1 * 2 / 2

1-2: 2 * 1

[brittle field]

not sure how they got g(1) either, shouldnt it be y = 2

[shell smelt]

They are the area under the curves

[shell smelt]

It's a triangle

With sides with 1 and 2

1*2/2 = 1

[brittle field]

?

[shell smelt]

Yes

For 2-3, it's(probably) a cosine graph

Since it starts from top and goes down

And the integral of cosine is sine

[brittle field]

ok so its asking for the area. the area of g(0) = 0 because the function starts at 0 (?). g(1) = 1 because that is the area under the function between x = 0 and x = 1 ? (howd they get that?). g(2) = 3 because that is the area under the function between x = 1 and x = 2 ? (again how?) and so forth

[shell smelt]

A triangle, a rectangle, a curve

[brittle field]

am i right in saying g(1) is asking for area between x = 0 and x = 1 ?

so i gotta use a triangle formula?

and then rectangle

and then somehow curve ?

[shell smelt]

Yes

[shell smelt]

Let's think about curve at the end

Do you get the first parts?

[brittle field]

almost, I use triangle area formula to find g(1)?

and rectangle area is just base * height

[shell smelt]

Yep

[brittle field]

wait a min, g(2) is area of rectangle between x = 1 and x = 2 , its height is 2 , that would be 1 * 2 = 2 . not 3

because its base is a width of 1

and its height is 2

I must be missing something

[cedar kiln]

@brittle field Has your question been resolved?

[cedar kiln]

@brittle field Has your question been resolved?

[brittle field]

<@&286206848099549185>

[shell smelt]

g(2) = area under the curve from 0 to 2

not 1 to 2

[brittle field]

ok that makes more sense

now how do i find area of next part

between x = 2 and x = 3 (I will add the areas of the previous rectangle and triangle)

because g(2) means the area of the triangle + the area of rectangle

[granite knoll]

looks like a sinusoidal function w/ period 4 maybe?

[brittle field]

not sure what that means, I don't think my teacher used words like that

[granite knoll]

e.g. if you shift sin(x) over 2 to the right, increase the amplitude, and change the period (so it wouldn't be 2pi)

[brittle field]

is there another way you could describe how to find the area here?

[granite knoll]

does the function from x=2 to x=5 look like sin(x) or cos(x) to you?

I guess cos(x)

[granite knoll]

yea so it looks like if you were to shift cos(x) to "start" at x=2 instead of x=0, right?

and the period would be 4 instead of 2pi. and the amplitude is 2 instead of 1

[brittle field]

is it easier to separate them? finding area between x = 2 and x = 3 . height = 2 and width = 1. does period = height? and amplitude = width?

since its asking for g(3)

[granite knoll]

no... amplitude is height. the amplitude of cos(x) is 1. the function you're given goes up to 2

the period of cos(x) is 2pi. for your function it'd be 4

[brittle field]

I don't get it. the height/amplitude is 2 and the width/period is 1

[brittle field]

.close

[cedar kiln]

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[austere plume]

How do I answer 6B

[shell smelt]

,rotate

[storm mural]

expand out (a + 4i)^2 + 2*(a + 4i)

separate into real and imaginary parts

[austere plume]

I did

[shell smelt]

z^2 + 2z can be written as z(z + 2). Replace z with a + 4i you get (a + 4i)(a + 2 + 4i)
If this expression is real then total imaginary part should be 0 according to the question

[austere plume]

I tried to solve a^2-2a-16 but it’s wrong

[shell smelt]

Which is 4ai + 8i + 4ai = (8a + 8)i

[austere plume]

Wait so basically

I had to solve the imaginary part?

[shell smelt]

This might be easier to work with though

[shell smelt]

Yes

[austere plume]

Oh ok

[shell smelt]

The question states that it's real

So it shouldn't have an imaginary part

[austere plume]

I didn’t know what I meant by that

[shell smelt]

or Im(z^2+2z) = 0

[austere plume]

Ok that makes sense

[shell smelt]

Alright you can close with .close

[austere plume]

.close

[shell smelt]

If solved

[austere plume]

.close

[cedar kiln]

Channel closed

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[austere plume]

Thank you for the help

[shell smelt]

My pleasure

[cedar kiln]

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[hardy sable]

hello, i found this in the studying documents, it shows that the axioms 2 and 3 for a vector space to be subspace are merged like cU, and U+V became cU+V does this really work

[crimson sedge]

yes

[hardy sable]

how does it work

[crimson sedge]

substitute B=0 to get aA in S

and substitute a=1 to get A+B in S

[cedar kiln]

@hardy sable Has your question been resolved?

[cedar kiln]

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[junior timber]

ZQ

i labeled everything

[valid tendon]

Do you know where to start?

[junior timber]

uhhh kinda no

all o know that its below 30

[valid tendon]

Since you want to find ZQ, so that the shaded and non shaded regions r equal

I’d suggest solving for the areas of both regions

and setting them equal to each other

and solve for ZQ

[junior timber]

okk

@valid tendon wait can u tell me if this way works

[valid tendon]

sure

[junior timber]

so i turn point O (the center)ZQY is a half a trapezoid

and i use that to solve for base a

and then i divide by 2 which give me ZQ

[valid tendon]

I’m confused atm

[junior timber]

about what

[valid tendon]

what is base a?

[junior timber]

ZQ * 2

[valid tendon]

And the “half trapezoid”?

[junior timber]

OZQY * 2

is the trapezoid

QZYO is the half trapezoid

and O is the center point

[valid tendon]

if it’s half a trapezoid then it’s a triangle, if so how does it have 4 angles?

or am I missing something?

[junior timber]

how would it be a triangle?

see

no triangle

[valid tendon]

oh

still

cutting a trapezoid wouldn’t technically make both of the parts equal to eachother

[junior timber]

okk

[valid tendon]

anyways

i just solved it

You have to solve for the shaded areas and non shaded areas

e

lets see the white square is 30*30 which is 900

the white triangle on the left is 10*30/2 is 150

[junior timber]

okk

[valid tendon]

Then the triangle above that is 900 - 245 = 555,

but how would you solve for the last non-shaded triangle?

[junior timber]

which small triangle

the ZQ one?

[valid tendon]

this one

[junior timber]

idk how to find it

so the height is 30

[valid tendon]

yeah.

you know that half the length of the square is 30

and the length of the base + ZQ = 30

[junior timber]

yea

[valid tendon]

what is the length of the base?

[junior timber]

wait

ok nvm

[valid tendon]

you got it?

[junior timber]

but how do we find the base

[valid tendon]

subtract ZQ

[junior timber]

yea but u dont know what that is

so how

because if i found the small base

i might as well just so 30 - the small triangle

i get ZQ

[valid tendon]

The question wants you to find the area of the shaded and non shaded regions

[junior timber]

ohh

[valid tendon]

and then let them be equal to eachother

[junior timber]

right

[valid tendon]

it's gonna result into like a equation

where ZQ is like the "x"

[junior timber]

yup

here wait

ill try to find all the other area first

[valid tendon]

alright.

tell if you have any difficulties

[junior timber]

OKK

ok

so

[valid tendon]

yes?

[junior timber]

one second

let me write it down

[valid tendon]

Ah

goodluck

[junior timber]

here

[valid tendon]

555*

[junior timber]

oh right oops

sorry me

*mb

now what

[valid tendon]

find the area of that white triangle

[junior timber]

how tho

[valid tendon]

like i said earlier, the base + ZQ = 30

[junior timber]

but we need a length of one fo them

[valid tendon]

hm?

[junior timber]

like base or ZQ

[valid tendon]

you're solving the area in terms of ZQ.

[junior timber]

6ea

yea

[valid tendon]

Anyways solving for the white triangle

the base = 30 - ZQ

[junior timber]

yes

[valid tendon]

and length is 30

so the area is?

30 * (30 - ZQ) /2 = ?

[junior timber]

450-15x

[valid tendon]

yeah.

then what's shaded trapezoid?

if you know the square is 900 and triangle is 450-15x

[junior timber]

so u do

900-(450-15x)

[valid tendon]

mhm

[junior timber]

its 450+15x

[valid tendon]

yeah.

now sum up the shaded areas

and non-shaded

[junior timber]

wait is 450+15x the shaded trapezoid right

[valid tendon]

correct

[junior timber]

okk

and 450-15x is the small one

[valid tendon]

yeah

[junior timber]

okk

so after i add it

then what

[valid tendon]

whats the total area of the shaded region?

and non-shaded

[junior timber]

i got

1545 +15x

and 2055-15x

[valid tendon]

nice.

[valid tendon]

the question wants them to be equal to each other

so

1545+15x = 2055-15x

do some algebra magic 🪄

[junior timber]

x = 17

[valid tendon]

yeahhhhhh

congrats

🎉

[junior timber]

YAYY

[valid tendon]

got any more?

[junior timber]

yea

like 4 more

[valid tendon]

[junior timber]

do u wanna help

[valid tendon]

well, if you want me to.

[junior timber]

tytyty

[valid tendon]

oh my

[junior timber]

these are 3 of them

i did 12 alr

[valid tendon]

ic

ADF and EFB seem to be quarter of a circle

[junior timber]

yea

[valid tendon]

you got q 10 then?

[junior timber]

nope

because it looks a little longer then a quarter of a circle

[valid tendon]

ah

[valid tendon]

But considering the lack of information such as what E is assigned for, I'd guess it's just quarter circles

[junior timber]

okk

[valid tendon]

and seeing that the first shaded area is the quarter circle and the other is the outerpart which is area of square - quarter circle

the sum of that is just the square

[junior timber]

okk

[valid tendon]

so what's the area of the square?

[junior timber]

1*1

[valid tendon]

which is?

[junior timber]

1

[valid tendon]

yes.

[junior timber]

so the raidus is 1 too

[valid tendon]

wait wait

what question r u on

...

[junior timber]

10

[valid tendon]

i mean you only needed the area of the shaded part

so 1.

[junior timber]

??

oh ok

yea

true

[valid tendon]

LOL

[junior timber]

ok 11 now

[valid tendon]

this one is tricky e

havent found the solution yet

[junior timber]

so the radius has to be less than 1

[valid tendon]

ohhh i solved it

damn that was tricky

@junior timber

[junior timber]

yea

how did u do it

[valid tendon]

So

wait

you see this?

[junior timber]

ye

[valid tendon]

alright

[junior timber]

ohh

the radius

[valid tendon]

so

you solve the area of the triangle first.

which is?

[junior timber]

1

[valid tendon]

yeah

alright

now using some pythagorean theorem magic 🪄

what is BC?

[junior timber]

2.24

[valid tendon]

No, just the exact value

square root of....?

[junior timber]

oh

5

[valid tendon]

alright

now using some ez area of triangle magic

[junior timber]

huh

[valid tendon]

radius is the height, square root of 5 is the base

[junior timber]

ohh

[valid tendon]

and you know the triangle's area is 1

what is r??????

i know u can do it

[junior timber]

0.89

[valid tendon]

bro...

exact value 🤣

decimals is a nono

[junior timber]

2*1/√5

[valid tendon]

remove *1 and ur good

but

do you know how to rationalize fractions?

[junior timber]

no

[valid tendon]

oh

uhm

[junior timber]

wait so

[valid tendon]

hm?

[junior timber]

2/√5

is what u mean?

[valid tendon]

yes

multiply top and bottom by sqrt(5)

[junior timber]

okk

[valid tendon]

can u do that for me?

[junior timber]

top and bottom?

whats tje top

and whats the bottom

[valid tendon]

denominator and numerator

mb

[junior timber]

wait so

2*√5

[valid tendon]

almost there

[junior timber]

and √5*√5'

]right?

[valid tendon]

what is it equal to ?

sqrt(5) * sqrt(5)

[junior timber]

so

2√5 and 5

[valid tendon]

yeah

2√5/5 yeah?

[junior timber]

yea

[valid tendon]

that's the answer

[valid tendon]

you just rationalized a fraction

[junior timber]

okk

so nw what

[valid tendon]

that's the radius

u got ur answer 👍

for q 11

question 13 now?

[junior timber]

wait so 2√5/5 is the radius?

[valid tendon]

yes

[junior timber]

wait

how did we get 2 again

[valid tendon]

r*sqrt(5)/2 = 1

r*sqrt(5) = 2

[junior timber]

ohh ok

[valid tendon]

r= 2/sqrt(5)

= 2*sqrt5)/5

got it?

[junior timber]

yae

[valid tendon]

make sure to show that r is perpendicular to BC

[junior timber]

okk

13

[valid tendon]

alr

[junior timber]

theres a ratio

[valid tendon]

pretty confused bout the question rn

[junior timber]

same

idk how u can solve for thath

[valid tendon]

ah i know

First and foremost

solve for the area of the rectangle.

[junior timber]

how

[valid tendon]

well, whats the base and height of the rectangle?

[junior timber]

yea

but theres a ratio

[valid tendon]

hm?

[junior timber]

ad it doesnt give us what DA or CB is

[valid tendon]

where?

[junior timber]

question 13

the first pat

*part

[valid tendon]

it's a circle

[junior timber]

Dm : MC = 2 : 1

i know its a circle what abt it tho

[valid tendon]

e

so it's a ratio huh

[junior timber]

yes

[valid tendon]

thought mc = 2

lel

oh well doesn't matter

it's still solvable

so anyways

let's solve for the area of the rectangle.

@junior timber

If AN=a and NB=b, what is AB? or rather the base?

[junior timber]

uhh

a + b

= AB

[valid tendon]

yes

[junior timber]

which is the base

[valid tendon]

👍

[junior timber]

which is also 2:1

[valid tendon]

wait wait

[junior timber]

well if 2+1 would be the base

[valid tendon]

lets do the ratio later

[junior timber]

but since its ratios

we dont know

yea

[valid tendon]

a+b is the base.

so what's the height?

if ADN is a arc/quarter circle

[junior timber]

DA would be the height and raidus

[valid tendon]

yea which is

?

what's the radius?

[junior timber]

a

which is also AN

[valid tendon]

correct

so what's the area?

[junior timber]

a+b * a

[valid tendon]

u forgot parantheses/brackets but yeah

alright

[junior timber]

oh oop

[valid tendon]

now that we know the area of the rectangle

lets find out the areas of the other figures remaining

like the big quarter circle. or ADN

if radius = a, whats the area?

@junior timber u having a hardtime?

[junior timber]

uhh

wait area of the circle?

[valid tendon]

yes

the larger one

Since it’s a quarter of a circle

or rather said one fourth of a circle

[junior timber]

okk so

(pi*a^2)/4

[valid tendon]

yeah.

remember taht

So, what’s the area of the smaller circle

with the radius=b?

[junior timber]

(pi*b^2)/4

?

[valid tendon]

yes!

Alright

the last one

So, what’s CK?

[junior timber]

idk

[valid tendon]

What is CB?

if AD = a?

[junior timber]

CB is a

[valid tendon]

What is CK + KB?

look at the figure

@junior timber u good?

[junior timber]

sorry

[valid tendon]

u having a hard time?

[junior timber]

kinda

[junior timber]

a

[valid tendon]

yeah

it’s a

lets solve for CK now

CK + KB = a

what should you do?

[junior timber]

wait does KB = b?

[valid tendon]

Yeah

lol

forgot bout that

[junior timber]

ok so CK = a-b

[valid tendon]

right nice.

[valid tendon]

Alright, MC now

Since DM:MC = 2:1

we know that DC or AB = a+b yeah?

type if ur here

[junior timber]

yea

[valid tendon]

But what’s DM + MC?

[junior timber]

a+b

[valid tendon]

nice

Since DM:MC = 2:1

we know DM is twice as much as MC

so DM = 2*MC

[valid tendon]

So then this expression becomes 3*MC

[junior timber]

oh ok

[valid tendon]

Then what is MC?

[junior timber]

wait MC * 3 = what?

[valid tendon]

If DM+MC = a+b

what is 3*MC?

[valid tendon]

^

[junior timber]

uhh

DM * Dm/2

[valid tendon]

DM + MC = 2*MC + MC

= 3*MC

= a+b

[junior timber]

okk

[valid tendon]

what is MC then?

[junior timber]

MC = a+b/3

[valid tendon]

nice

solve for the triangle MCK

[junior timber]

its (a-b) * (a+b/3) / 2

[valid tendon]

yes

so

how do we solve for the shaded area?

when we have all the figures areas

[junior timber]

u take the area of the rectangle

[valid tendon]

mhm

[junior timber]

and substact the two quarter circles

and subtract the triangle

[valid tendon]

yeah

[junior timber]

wait whats the rectngle area again

[valid tendon]

Uh

a(a+b)

U alright?

[junior timber]

yea

[valid tendon]

got ur answer?

[junior timber]

not yet

[valid tendon]

ah

[junior timber]

i got this

wait its this right

[valid tendon]

Oh my

[junior timber]

yea idk

help please

[valid tendon]

wait

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

((a-b)(a+b)/3)/2 = (a^2 - b^2)/6

[junior timber]

huh

[valid tendon]

nvm

[valid tendon]

the pi*b^2/4 should be positive and

((a-b)*(a+b)/3)/2 should be positive too

that should be good enough

[junior timber]

oh right

[valid tendon]

maybe simplify ((a-b)(a+b)/3)/2 to (a^2-b^2)/6

and ur good to go

[junior timber]

wait so simplify what u wrote?

[valid tendon]

this but turn all the terms in the parentheses to positive

[junior timber]

i did

so its that long answer?

[valid tendon]

and rewrite ((a-b)(a+b)/3)/2 to (a^2-b^2)/6

[valid tendon]

i mean yeah

[junior timber]

okk

[valid tendon]

this is using difference of squares

[junior timber]

okk

[valid tendon]

So what’s the end result lookin like?

u done?

[cedar kiln]

@junior timber Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

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• Be polite and have a nice day!

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[crimson sedge]

How to write an equation for this

[gritty viper]

what's the first time cos(x) hits 0?

[crimson sedge]

?

[gritty viper]

your graph is 3cos(bx), right

if 3cos(bx) = 0, what is bx

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Idk 🤷‍♂️