help-23

[devout peak]

im stuck here aswell

[covert yoke]

I don't know that I quite understand

C and D are on the same leg

So the probability is going to be pretty low more likely than not

As it requires both C and D fail and then just one of A and B

Whereas for the overall component to fail at least one of (A and B) and at least one of (C and D) needs to fail

[safe radish]

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[devout peak]

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[woven mica]

uhhhh

[woven mica]

do i just add f'(x) and g'(x)?

that seems to easy

i have calculus trust issues

[vale oriole]

yes, when u derive u(x), u should get u'(x)=f'(x)+g'(x)

[woven mica]

oh ok, that's really it?

[vale oriole]

yes

[woven mica]

well alr, thank you

.close

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[versed isle]

0.6320

[lean otter]

Thanks.

How did you do it?

That is wrong your answer.

[versed isle]

no. of women who talked between 600-799 = 316, divided by total students, 500

[versed isle]

ahh my bad :(

didnt read the mid part

[versed isle]

alright lemme try this properly

[lean otter]

Thanks.

[versed isle]

0.8226

[lean otter]

Thanks. But can you explain how you did it again?

[versed isle]

alirght

the question is talking about a person being selected at "random"

so do not get confused and assume all the persons are being taken as the total events

later on it is seen that we know about the selected person that they talk about "600-799 minutes"

this makes that, the person is being chosen from the "600-799" minutes category

specifically a women is asked from that

hence divide the no. of women present in that category by the total no. of people in that category

[lean otter]

So I do 600-799?

I am so confused.

[versed isle]

no, u look at the table

under the 600-799 minutes category

do u see the no. of men and women being specified?

[lean otter]

Yes.

[versed isle]

we need to select a women from the total no. of people in that category

the total no. of people are = men + women = 25 + 116 = 141

the total no. of women = 116

hence ans is 116/141= 0.8226

[lean otter]

How did you get 116?

Bru.

[gleaming coral]

ok

[lean otter]

How do I solve this?

[craggy holly]

Have you tried 1/61?

[lean otter]

1/61m

?*

[craggy holly]

whats m?

[lean otter]

I meant ?*

[craggy holly]

ya

[lean otter]

.*

[craggy holly]

I mean, have u tried it

[lean otter]

What does 1/61 mean?

[craggy holly]

Is there a limit to how many answers u can give?

[lean otter]

Yes...

Like 3 attempts.

[craggy holly]

Well, there are 61 men on average that talked less than 600 minutes, so my theory was that we are picking a person at random from those 61 men.

[lean otter]

Bro I tried 1/61 and it says it is wrong.

[craggy holly]

Did you type it as "1/61"?

[lean otter]

Yes.

[craggy holly]

Okay, idk how they are judging ur answers - based on fractions or decimals. Don't give them any more answers. Or have u reached ur limit?

of answers

[lean otter]

Just decimals.

4 decimal places.

[craggy holly]

Well, ya if they are doing to least 4 decimals then it's not gonna take 1/61. But, don't give them an answer yet. I don't want u to get punished for wrong answer.

[lean otter]

Okay.

[craggy holly]

is there a time limit?

[lean otter]

No.

.closw

.close*

.close

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[craggy holly]

what was the answer?

[lean otter]

0.5598

6*

0.5596* I figured it out.

It was 61/109.

[craggy holly]

ya, but didn't they say that's incorrect?

Or was that cross just there?

[lean otter]

System glitch had to refresh.

[lean otter]

System glitch.

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[sage tapir]

anyone available to help me with my pre calc review

[icy lance]

anything specific

[sage tapir]

yes should i send image?

[gleaming coral]

ye

[sage tapir]

one second please

how do i tell if it is restricted

[icy lance]

it has to be restricted if its not one-one
ie if multiple x values can result in the same y value

like with y=x^2 you get the same y value at -2 and 2
so we have to restrict it to make it one-one, for y=x^2 we'd either do x>=0 or x<=0 generally

then you can find its inverse

[sage tapir]

ok when graphing where or how do i get the points for each graph

[icy lance]

the points?

[sage tapir]

yes

[icy lance]

im asking what you mean by that

[sage tapir]

how do i sketch the graph or find the points to sketch the graph

[icy lance]

its sufficient to just graph one really
once you have restricted your domain you can graph it through whatever methods you normally do, finding intercepts and asymptotes, or just plugging some points in

if those are points (x,y)
then the inverse equivalent is the point (y,x)

since its a reflection in y=x

so you can draw that too

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[finite ivy]

pls help

[twin hamlet]

The bm was re-delated to the intersection with the dc

[twin hamlet]

.

[finite ivy]

wdym?

[twin hamlet]

?

[twin hamlet]

What u got from it?

[finite ivy]

olympiad

[twin hamlet]

?

Huh

[twin hamlet]

My question is: what did you get after this action?

[eternal carbon]

consider midpoint N on BC

[finite ivy]

oh sorry i dont speak english very well

[finite ivy]

i dont get this srry

[finite ivy]

oooooh

[twin hamlet]

What did you get from that?

@finite ivy

[finite ivy]

amm two congruent triangles?

[twin hamlet]

Continue

[finite ivy]

ahhh

i think i solve it

no wait

[finite ivy]

oc i get that but am

that is a isosceles triangle right?

[twin hamlet]

Let me remind you that the median drawn to the hypotenuse in a right triangle is equal to half of the hypotenuse

[finite ivy]

oooooh

ooooooh

well thats all i guess

thank you so much

[twin hamlet]

Yes

Close channel pls

Write: .close

[finite ivy]

.close

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[icy crystal]

so the problem here is to simplify $\frac{5+7i}{2-6i}$ I got $\frac{8}{10}$+$\frac{11i}{10}$ and I simplifed the first fraction to $\frac{4}{5} but this is incorrect I know I'm close but still I think I messed up the simplification process

[icy crystal]

why did It clump up the last part like that? lol

[plucky elk]

missing $

#latex-testing

[icy crystal]

?did I do something?

[hearty egret]

just add the missing $

[twilit moat]

show your working?

[plucky elk]

#latex-testing is where you test your latex

[twilit moat]

what part do you think you made a mistake on

[icy crystal]

oh nevermind

[flat frigate]

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

[icy crystal]

how do I delete the message?

.close

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[icy crystal]

.reopen

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✅

[icy crystal]

I originally got 8/10 for the first fractional component but nevertheless I screwed up simpifying somewhere

[versed wave]

show your working?

[icy crystal]

I can take a picture but it's super blurry?

[versed wave]

what works works

[icy crystal]

so the first step was to foil both sides against the denoms complex conjugate

$\frac{5+7i}{2-6i}$

[flat frigate]

Protestant Protocologist

[icy crystal]

times

$\frac{2+6i}{2+6i}$

[flat frigate]

Protestant Protocologist

[icy crystal]

and I got

$\frac{10+3014i+42i^2}{4+12i-12i-36i^2}$

[flat frigate]

Protestant Protocologist

[icy crystal]

after the foiling process

[versed wave]

wtf

[icy crystal]

this then became $\frac{10+44i+42i^2}{4-36i^2}$

[versed wave]

\frac

[flat frigate]

Protestant Protocologist

[icy crystal]

and after knowing that i^2=-1

[versed wave]

quick note

[icy crystal]

yeah?

[versed wave]

the product of two complex numbers conjugate with each other is the modulo squared of any of them

[icy crystal]

ah okay I didn't know that

[versed wave]

$z\cdot\overline z =\lvert z\rvert^2$

[icy crystal]

$\frac{10+44i-42}{4+36}$

[flat frigate]

Protestant Protocologist

[versed wave]

the arithmetic so far is fine, continue

[versed wave]

in fact i dont think your simplification is wrong

but rather ||you forgot something||

[icy crystal]

$\frac{-32+44i}{40}$

[versed wave]

remove a {

[flat frigate]

Protestant Protocologist

[icy crystal]

yeah

and then that became

$\frac{8}{10}$ + $\frac{11i}{10}$

[flat frigate]

Protestant Protocologist

[icy crystal]

and I simplified 8/10 to 4/5

[versed wave]

and this is where you went wrong

[icy crystal]

okay

[versed wave]

can you spot the error?

[icy crystal]

let me rework it from there...

WAIT

negative

[versed wave]

[icy crystal]

wait can I just leave it as -8/10 or do I need to simplify it into -4/5?

well that would change the equal denominators wouldn't it?

WHELP

I got it right but I accidentally pressed 1

what does it mean when a problem asks me to "Plot a number"?

just the slope "3"?

Yes

Cartesian

Oh that worked

huh

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[split badge]

bit of an odd ball

[split badge]

I have a minecraft server im in with a banking system

My money is at roughly 57,000 and every hour i earn 12% interest

my max capacity is 2 million

How many hours will it take before I reach 2million

and just to specify

i earn 12% of my TOTAL money in my bank every hour

so this hour im at 57k but next hour itll be higher so the total amount ill earn from interest will be higher

I just idk how to calculate this

[lean otter]

Do you want to know the math behind it or just want to know it

Also money is credited as soon as one hour lapses right?

Or is it continuous

[split badge]

so

my total money just went up due to a "loan"

I have 157k and I need to get up to 2 million

The money is recieved as soon as that timer hits, and the next cycle INCLUDES the new money

@lean otter

I dont need to know the math but I would like to

[lean otter]

I need help

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[latent eagle]

can someone explain to me in absolute dumb fuck terms why this converges "slowly"

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[woven quartz]

So doing a lesson for Calc 1 and realizing Ive forgotten a lot of (nonderivative) equations I now need.

Anything else I should add for rates of change I should know?

[glossy dock]

i mean as long as you understand the basic concept, some common scenarios, implicit differentiation, logarithmic differentiation, memorized the trig derivatives, and probably also l'hopital's rule then ur probs fine

[woven quartz]

Well so like for all of these, I wasn't given any of the non derivative equations.

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[tiny tusk]

What's the point of taking out the sqr p^4 off the bottom

And how would I know I have to do that

[hard crest]

like separating it into √33 • √p⁴?

[tiny tusk]

Yeah

It's so you can divide it from the top too right,m

By when you do that do you just divide it from one of the radicals on the top or both? Just one right?

[hard crest]

yeah I mean it's not strictly necessary to write it out like that

both of them

distributive property

[tiny tusk]

Ohhh ok thanks

I'm cooked

Whatever tho

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[harsh eagle]

How would I prove that the equation equals 2/sin theta?

[rustic goblet]

the first thing I would do is rewrite the LHS as one fraction

[harsh eagle]

as in "sin theta + 1 + cos theta/ 1 + cos theta + sin theta"?

[strange trout]

omd i forgot all my trig identities

[harsh eagle]

I though cross canceling initaially but it didn't get 2/sin theta

[wispy monolith]

Do you know how to perform operations on fraction?

Like addition subtraction

[harsh eagle]

yes, they need to have the same denominator

[wispy monolith]

You need to use something called butterfly method

a/b+c/d = (ad+bc)/(bd)```

[harsh eagle]

give me 1 min, I'll try it rn

[wispy monolith]

[viscid badger]

HELLO?

anyone here?

[harsh eagle]

yes

[viscid badger]

can u help me with a question?

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[harsh eagle]

if you go to open channels, someone can assist you 🙂

[viscid badger]

o

[harsh eagle]

I stopped at sin^2 theta + 2 + cos^2 theta/ 1 + cos theta/ sin theta. What would be the next step?

[wispy monolith]

Show your work

[wispy monolith]

(1+costheta)(1+costheta) ≠ 2+cos²theta

[harsh eagle]

https://tenor.com/view/paper-gif-21548222

ahhh

[harsh eagle]

(1+costheta)^2?

or 1^2 + cos^2 theta

[wispy monolith]

Yes

[wispy monolith]

No

1²+cos²theta+2costheta

[harsh eagle]

I see, you can then convert the cos^2 theta and sin^2 theta into 1 with the Pythagorean Identities

Thank you for the help! I can now see how the equation can equal 2/ sin theta

[wispy monolith]

Great

[harsh eagle]

.close

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[harsh eagle]

.close

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[zinc cedar]

I understand that f(0)>0 because c is positive but why is f(x) >0 for all values of x?

[median vigil]

if it has no real real roots then it can't change signs. so either it is always positive or always negative

[zinc cedar]

Oo ok thnx got it

.close

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[odd lava]

How do i solve this?

[hasty marsh]

i don't think u can integrate this

[odd lava]

do i just say its = 0

[hasty marsh]

i think so

[hasty marsh]

its an odd function

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[lean otter]

im not getting how the sinu able to get out of that

absolute function

[plucky elk]

Did you see that

[lean otter]

no no i meant

like | sin u |

how u get sin u out of it

[plucky elk]

tan(u/2) has the same sign chart as sin(u)

[lean otter]

so both sides just gonna cancel out the negative?

thats it?

[plucky elk]

Verify it by using half angle or other trig identities

"cancel out the negative" isn't really all that's happening

[lean otter]

sure

[lean otter]

yeah figured

imma use trig identities

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@lean otter Has your question been resolved?

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[shrewd bluff]

If i have a system of 3 equations

with 3 unknowns

[shrewd bluff]

and theyre all involving cos

is it always possible to solve them

[charred jetty]

What's the question format?

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[lean otter]

(2x-4)^3=1(2x)^3(-4)^0+3(2x)^2(-4)+3(2x)(-4)^2+1(2x)^0(-4)^3
\
=1(8x^3)(-1)+3(4x^2)(-4)+3(2x)(-16)+1(1)(-64)
\
=-8x^3-48x^2-96x-64

[lean otter]

Oopsies

$(2x-4)^3=1(2x)^3(-4)^0+3(2x)^2(-4)+3(2x)(-4)^2+1(2x)^0(-4)^3
\
=1(8x^3)(-1)+3(4x^2)(-4)+3(2x)(-16)+1(1)(-64)
\
=-8x^3-48x^2-96x-64$

[flat frigate]

grimm.fromblacksouls.real.1784

[lean otter]

sorry for th shiddy formatting

where’d i go wrong

[hasty wagon]

(-4)⁰=1 , not -1

and (-4)²=16, not -16

[lean otter]

Sorry

what’s the difference between (-x)^0 and -x^0 though

why do they give different answers

[dapper star]

calculator?

its like -(x^0)

unles you keep brackets

[hasty wagon]

you can generally treat x^0 = x/x for x≠0

[lean otter]

and for 0/0 its undefined ? Or i heard somewhere its both positive and negative infinity or smth or you can just write it out as 1

[lean otter]

ohh no.i know i just wantedt o know why exactly they have different answers withthe bracketsa nd without the brackets. Sorry for not clarifying

[vapid glen]

we can sidestep this whole dividing by 0 thing by thinking of it as x^{n+0} = x^n * x^0 so we ought to have x^0=1 so we have x^n = x^n*1.

[lean otter]

it took a lil bit to understand but thank you very much!!!

[dapper star]

I don't know how calculators' function but, -1^2 can be interpreted as -(1^2) and also (-1)^2

[lean otter]

Thx to

all.t hx every1

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[thin swan]

How do I solve this differential equation?

[thin swan]

i think that f(x) = e^(x^2) or something along that line but i cant figure out how to get the - 2x

[median vigil]

note that this is linear & separable, which may suggest 2 potential strategies

[thin swan]

the 2 strategies being?

[median vigil]

first order linear odes are solvable by integrating factor, and separable odes are solvable by separation of variables

[thin swan]

👍 (in case anyone sees this and has the same question the answer i got was e^(x^2) + 1)

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[wise oyster]

Why does 150 n have -

I don’t understand why

<@&286206848099549185>

[thin swan]

because its to the left

it doesnt really matter where the negative is, its just that its in the opposite direction of the other two

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[lean otter]

how to find instantaneous rate at point P

i know like u make secants keep getting closer to the P

then match with a tangent line (?)

get the slope

like i took P(10, 80) and G(9, 60)

but the answer is 16 not 20

lmao

😭

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

@lean otter Has your question been resolved?

[charred niche]

Was there any additional information in the question?

Like the graph seems to be that of a parabola

[lean otter]

yeah

let me share

[charred niche]

So if i assume fn. to be y=kt² solving the given points i would get k=4/5

[lean otter]

well its suppose to be done just graphically!

[vital oracle]

slope=tan(theta)

[lean otter]

know about that formula

but im thinking like u firsst hace secant slowly making it towards

tangent

[charred niche]

You can do graphically but i doubt that you will get exact value

[lean otter]

yeah no need to be

exact

just approx

[charred niche]

Have you estimated slope of PQ4?

I think Q4 would be somewhere round (9.5,72)

Or 71

[vital oracle]

is it 20

[charred niche]

We cant find the exact but it would be somehere round 18

[lean otter]

(9.5, 72) (10, 80)

8/0.5 = 16

[charred niche]

Take y=71

If 72 then 16

[lean otter]

so its approx. 16

which is the right answer too

[charred niche]

Yes and btw the function is 4/5 t²

And you could that way too

Since its a free fall

Its supposed to be a parabola

Since it is passing through origin there would not be any constant

So y=kt² then putting (5,20) you could find k

Then just differentiate and put t=10 to get exact slope

[lean otter]

ohhhh

thank you

🙏🏻

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[compact flame]

no you got it wrong

x -> infty

let x = 1/k

where k -> 0

what do you get

@modern bloom

[astral goblet]

1

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[stable kestrel]

what to do?

[vagrant ice]

x = -7, 7 should be the only solutions

IDK if you are allowed calculus but you can use it to show the function is increasing for x > 0

and decreasing for x < 0

[stable kestrel]

yes

okay

[vagrant ice]

cause note that $6 - \sqrt{35} = \frac{6^2 - \sqrt{35}^2}{6 + \sqrt{35}}$

[flat frigate]

south's secret twin brother

[vagrant ice]

that means that if f(x) is the LHS, f(x) = f(-x) so it's an even function

[stable kestrel]

What is lhs

[toxic stratus]

its a quadratic

use usual quadratic solving methods

[stable kestrel]

okay thank you

[vagrant ice]

no need cause you can get x = 7 by inspection

the main part is showing that this is an even function and that it is increasing for x > 0 and decreasing for x < 0

[toxic stratus]

i mean if you want to solve everything by inspection then ok...

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[flat frigate]

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

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[lean otter]

Can anyone provide me and explain Section Formula

[lean otter]

Section Formula in Geometry
It's proof

[craggy sedge]

,w section formula

[flat frigate]

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[stable flint]

like you need its proof?

[lean otter]

Yup

[stable flint]

if so last time till i remember

to prove it they assumed the lines to be hypotenuse

and the point which divided it, from that a perpendicular was drawn which resulted in two similar right angle triangles

[lean otter]

K
I got it.
Thanks

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[verbal ether]

Hello! i want a help realed to proof of trigononmetry

[safe radish]

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/

[verbal ether]

I was doing trigonometry proofs and i practised them enough
but still i am lacking to develop that kind of approach to solve it immediately .

Can anyone help me ?

[random hazel]

..

[thin bridge]

Its recommended that you ask clear specific questions as opposed to vague/broad/open ended ones

[lean otter]

yes

Here help is provided on a specific question
For advice you may go to other channels

[digital sparrow]

!da2a

[safe radish]

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/

[lean otter]

Bro just solve them by ur known.
Think of the step that how to begin solution.
Then look at what is easy to solve either LHS or RHS of the given identity.

[verbal ether]

Nah! i mean what should be approach for harder q le me send one

[digital sparrow]

Bro ask the question

@verbal ether

[verbal ether]

wait le me take its pic

This

@digital sparrow btw sry for this tag

[digital sparrow]

Hey np

Tell me how you would get sin⁶x?

Btw 11th?

[verbal ether]

nope 10th

[verbal ether]

may be by taking out 2 (in power)
like 3*2

[digital sparrow]

Idk what that means

But if you are saying cubing sin²x + cos²x

Then yeah

[lean otter]

You remember the sum of cubes identity?

Like a³ + b³?

You are in 10th, you must?

Where did you go man?

[verbal ether]

oh! thxs now i solved it

by cube and at last by square property

[digital sparrow]

Kk nice

Btw this prob isn't 10th grade stuff

[lean otter]

Depends on where you are from, it may be thrown into a 10th grade book since it doesn't require any 11th grade trig Identities.

[digital sparrow]

CBSE?

[digital sparrow]

I guess yeah but still no use

10th trig for me was all about finding lengths of stuff in right triangles rarely isosceles

[lean otter]

State, if I remember correctly we had the same problem in 10th grade, it's in Loney too, I wonder where all these questions finally come from.

[digital sparrow]

Ohh

Nice

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[random hazel]

!done @verbal ether

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[full hazel]

Why is the coefficient of t in the parametric equations 1 and -2 respectively? Shouldn’t it be -2 and 4, the same as the derivative at t = 1?

[safe radish]

@full hazel Has your question been resolved?

[plucky elk]

because x(t=0) = -1 and y(0) = 3

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[rare crest]

I dont understand the + - +

[patent flame]

the function x^2+2x-15 is strictly positive when x<-5 and x>5, and is strictly negative when -5<x<3, hence the + - +

[full hazel]

x(t=0) is 0 though. and y(t=0) is 2

[rare crest]

Is this wrong?

[safe radish]

@rare crest Has your question been resolved?

[rare crest]

<@&286206848099549185>

[atomic salmon]

yes

[rare crest]

Why?

[merry cove]

you should have changed the sign on second line

no didnt say anything

[atomic salmon]

are u trying to find range of x?

[rare crest]

Thanks!

[atomic salmon]

U don't know wavy curve method ?

[rare crest]

No

[atomic salmon]

ahh thats why u didnt understood this question

like it requires a bit of wavy curve method

[rare crest]

Okayyy

[stable flint]

it can be done with common sense tho (without wavy curve)

keep any 3 values of x for:
x ≤ -1
-1 ≤ x ≤ 4
x ≥ 4

and if it comes positive and -ve

and then you can rewrite in the form of answer easily

[fallen fulcrum]

I saw the question, people helped before me. But thought I’d try it anyway, is this right method?

And the wavy curve method, is that similar to splitting the middle term?

[rare crest]

I lack that in maths

[stable flint]

na I didn't meant that way

nor with that intention

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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[rare crest]

Lol issakay!

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[crimson shore]

Prove that (n-1)n(n+1)(n²+1) is divisible by 16, knowing that n is odd

[honest perch]

uneven = odd

any ideas?

[crimson shore]

Are you happy now?

[honest perch]

i was always happy

[crimson shore]

To replace n with 2K+1

[honest perch]

yes you can do that

[crimson shore]

Idk what to do after

[honest perch]

what do u get

[crimson shore]

2K(2K+1)(2K+2)(4K²+4K+1)

I can only prove the divisibility by 4

[honest perch]

should be 4k+2

[crimson shore]

Where?

[honest perch]

it's 2K(2K+1)(2K+2)(4K²+4K+2)

[crimson shore]

Ah thanks

Didn't see that

Still

[honest perch]

well take out all the 2s

[crimson shore]

I can only prove divisibility by 8

[honest perch]

8 * K(2K+1)(K+1)(2K²+2K+1)

need to show K(2K+1)(K+1)(2K²+2K+1) is even

[crimson shore]

Should I do on cases

?

[honest perch]

you can

[honest perch]

look at K(K+1) in particular

[crimson shore]

Ah

Consecutive

[honest perch]

yeah cuz the other 2 factors are always odd

[crimson shore]

Numbers

[honest perch]

yes

[crimson shore]

Thanks

[honest perch]

wl

[crimson shore]

.close

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[solar hazel]

what the fuck

[honest perch]

sorry it was just the song you were listening to

[solar hazel]

well i know that much

i wasn't even here

[crystal hatch]

.close

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[near tendon]

For c i got x>1.25

Is that right?

If not im stuck

[safe radish]

@near tendon Has your question been resolved?

[near tendon]

why

it says

a=1.25

a>2

cant u have negative

values of a

tho

yh

one sec

is

like

a=1.25 or a>2

well

it says

{a:a,,1.25} U {a:a>2}

idk set notation tho

thats what i was thinking

nah he dont like me

.close

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[remote folio]

help

[severe pond]

if it’s not divisible by 4 it’s tens digit must be odd

and if its divisible by 2 and 5 then it’s divisible by 10 so it’s one’s digit is zero

and if it’s not divisible by 25 then it can’t be 50

[remote folio]

so the answer is

[severe pond]

you tell me

!nosols

[safe radish]

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

[severe pond]

i’m not just going to tell you the answer

i essentially already did

i told you more than enough

[safe radish]

@remote folio Has your question been resolved?

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[spiral saddle]

\textbf{Exercise 8.-} Define a linear transformation that satisfies the stated conditions.

\begin{enumerate}
\item[a)] ( f : \mathbb{R}^2 \to \mathbb{R}^2 ) such that ( \operatorname{Ker}(f) = { \mathbf{x} \in \mathbb{R}^2 \mid x_1 + 2x_2 = 0 } ), ( \operatorname{Im}(f) = \langle (1, 0) \rangle )
\item[b)] ( f : \mathbb{R}^4 \to \mathbb{R}^4 ) such that ( \operatorname{Ker}(f) = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 + x_2 + x_4 = x_2 + x_3 = 0 } )
\item[c)] ( f : \mathbb{R}^3 \to \mathbb{R}^4 ) such that ( (1, 1, 2) \in \operatorname{Ker}(f) ), ( \operatorname{Im}(f) = \langle (1, 0, 1, 1), (2, 1, 0, 1) \rangle )
\item[d)] ( f : \mathbb{R}^4 \to \mathbb{R}^2 ) such that ( (1, 0, 1, 3) \in \operatorname{Ker}(f) ) and ( f ) is an epimorphism
\item[e)] ( f : \mathbb{R}^4 \to \mathbb{R}^4 ) such that ( \operatorname{Ker}(f) = \operatorname{Im}(f) = \langle (2, 1, -1, 0), (0, 1, 0, 1) \rangle )
\item[f)] ( f : \mathbb{R}^3 \to \mathbb{R}^3 ) such that ( f ) is not a monomorphism and ( (1, 1, 1) \in \operatorname{Im}(f) )
\item[g)] ( f : \mathbb{R}^4 \to \mathbb{R}^4 ) such that ( \operatorname{Ker}(f) = \operatorname{Im}(f) ) and ( f(3, 2, 1, -1) = f(-1, 2, 0, 1) \neq 0 )
\end{enumerate}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[rustic goblet]

[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

for a)

Ker(f) = <(1,2)>

maybe we can treat f as a multiplication of matrices

and then

use the vectors from the kernel

since the kernel is solving for Ax = 0

A(1,2) = (0,0)

also

(mxn)*(2x1) = (2x1)

==> n = 2

m = 2 aswell

A is a 2x2 matrix

lets call it $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

A(1,2) = (0,0)

aka

,, \begin{pmatrix} a & b \ c & d \end{pmatrix} \cdot \begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix}0 \ 0 \end{pmatrix}

we get a system

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,, \implies \begin{cases} a + 2b = 0 \implies a = -2b\ c + 2d = 0 \implies c = -2d\end{cases}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

so $A = \begin{pmatrix} a & b \ c & d \end{pmatrix} = \begin{pmatrix} -2b & b \ -2d & d\end{pmatrix}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

ok and the image is

Im(f) = <(1,0)>

I think maybe the columns of the matrix representation of f must be scalar multiples of the vector in <(1,0)>

so $\begin{pmatrix}a \ c \end{pmatrix} = \alpha \begin{pmatrix}1 \ 0 \end{pmatrix} \ \begin{pmatrix} b \ d \end{pmatrix} = \beta \begin{pmatrix} 1 \ 0 \end{pmatrix}$ with $\alpha, \beta \in \mathbb{R}$

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

we get a system

,, \begin{cases} a = \alpha \ c = 0 \ b = \beta \ d = 0 \end{cases} \implies A = \begin{pmatrix} \alpha & \beta \ 0 & 0 \end{pmatrix}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

and Ker(f) = <(1,2)>

f(1,2) = (0,0)

f(x,y) = A * x

f(1,2) = A * (1,2) = (0,0)

,, \begin{pmatrix} \alpha & \beta \ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \ 2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

a + 2b = 0 ==> a = -2b ==> (a,b) = (-2b , b)

,, \implies A = \begin{pmatrix} -2\beta & \beta \ 0 & 0 \end{pmatrix} = \beta \begin{pmatrix} -2 & 1 \ 0 & 0 \end{pmatrix}

[flat frigate]

938c2cc0dcc05f2b68c4287040cfcf71

[spiral saddle]

,w {{-2,1},{0,0}} * {{1},{2}}

[flat frigate]

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[spiral saddle]

should be correct I will continue with b) now

f : R⁴ -> R⁴ s.t. Ker(f) = { x in R⁴ / x1 + x2 + x4 = x2 + x3 = 0}

i) x1 + x2 + x4 = 0 ==> x1 = -x2 - x4
ii) x2 + x3 = 0 ==> x2 = -x3

(x1,x2,x3,x4)=(-x2-x4,x2,-x2,x4)

(x1,x2,x3,x4) = x2(-1,1,-1,0) + x4(-1,0,0,1)

Ker(f) = <(-1,1,-1,0),(-1,0,0,1)>

[spiral saddle]

okay so we found the kernel

maybe we can use the rank nullity theorem to find the dimension of Im(f)

dim(Im(f)) + dim(ker(f)) = dim(domain)

f : R⁴ -> R⁴ s.t. Ker(f) = { x in R⁴ / x1 + x2 + x4 = x2 + x3 = 0}

dim(Im(f)) + 2 = dim(R⁴) ==> dim(Im(f)) = 2

Im(f) = <(?,?,?,?),(?,?,?,?)>

we are trying to find two vectors that are linearly independent in the image, that are not present in the kernel

let Ker(f) = <(-1,1,-1,0),(-1,0,0,1)> = <v1,v2>

v1 = (-1,1,-1,0)
v2 = (-1,0,0,1)

no I think

there is a better approach to finding the linearly independent vectors for the image that are not present in the kernel

[flat frigate]

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[spiral saddle]

I think is simpler if we find the orthogonal complement of the kernel
instead of doing mambo jambo with the rref colspace rref rowspace pivots

[spiral saddle]

,w rref {{-1,1,-1,0,0},{-1,0,0,1,0}}

[flat frigate]

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[safe radish]

@spiral saddle Has your question been resolved?

[spiral saddle]

OKAY SO

synthetictaylor helped me a bit

and gave key hints about the problem

like

i) x1 + x2 + x4 = 0 ==> x1 = -x2 - x4
ii) x2 + x3 = 0 ==> x2 = -x3

like

we can use the first two canonical vectors of R⁴ for the basis of the image of f

those two vectors are orthogonal with the vectors in the kernel

and when we got the basis of the image of f we can form the first two columns of the matrix representation of the linear transformation by placing the column vectors of the basis of the image of f

Im(f) = <(1,0,0,0),(0,1,0,0)>

and the other two columns of the matrix representation of f are a linear combination of the vectors from the image

x1 = (1,0,0,0)^t
x2 = (0,1,0,0)^t
x3 = -(0,1,0,0)^t
x4 = -(0,1,0,0)^t - (1,0,0,0)^t

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@spiral saddle Has your question been resolved?

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[chrome plank]

i get the idea of it, and did a bunch of scratch work. its just that im unsure how to label the constants

[chrome plank]

im assuming i shouldnt use the same C in the second part of the inductive step?

or is it fine

[rich elm]

as long as u dont confuse urself its probably fine

label it C_{1} or C_{2} if need be

or like C_{k} and C_{k+1}

whatever is intuitive

[chrome plank]

ok cool. i think i got the basic idea for this down

[chrome plank]

i got it down to this. im unsure how to deal with the C since ik i can just multiply the j+1 and -1 into the sum then add and subtract a j+1 term to get it into the form of the "conclusion" thing

if i do that arent i left with a C(j+1)? do i need to be careful ab that?

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@chrome plank Has your question been resolved?

[safe radish]

@chrome plank Has your question been resolved?

[chrome plank]

I'll just ask for help tomorrow ig

.close

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[fathom jewel]

And did you progress?

[lethal vortex]

did a whole lotta nothin

[rich elm]

show your process of sub-ing

[rich elm]

ur sub is dubious

[lethal vortex]

wtfs dubious mean

[rich elm]

not reliable according to oxford dictionary

im just trying to say u need to redo ur sub

[rich elm]

$dx \neq ln|u|du$

[flat frigate]

Dootud

[rich elm]

even then, u didnt even sub in ur differential

[lethal vortex]

oh yea i forgot about that

[lethal vortex]

what is dx = then

[fathom jewel]

you differentiate both sides, not integrate

[lethal vortex]

RUH

bruh

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[lean otter]

How do i integrate this

I feel like some kind of trigonometric substitution might work but i can't quite figure it out

[indigo valve]

put x = a sint or a cost

[rich elm]

I would multiply top and bottom by sqrt x and separate the integrals into two fractions

But amitts suggestion will also work

It’s just oreference

U will end up doing a trig sub going down my suggestion either way

[lean otter]

alright ill try again with that

[indigo valve]

yed its pretty nice way

[lean otter]

so im gonna be using the sin2x and (1-cos2x) formulas after this i suppose?

[indigo valve]

right

[lean otter]

but that

hold up ill send a picture of it

im having a brainfade moment right now i feel

would x = a sin^2(t) have worked

[indigo valve]

wait actually yeah

sorry

[lean otter]

oh yea i got it

no worries

thanks for the help

.close

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[steep epoch]

How do I do this ive been stuck on this for a couple days please..

[vagrant ice]

which part do you need help with?

[river oriole]

what part

[steep epoch]

i dont know how to find the number of terms

or the sum

[river oriole]

whats the formula for Tn

[steep epoch]

im guessing Tn is the same as Un for me?

[vagrant ice]

okay, so split it into two geometric series

2, 9/2, 81/4......
-3, -27/4, -243/16.......

the common ratio is 9/4 for both sequences

ah so if you figure out what x is in 3^x = 177147

you just divide by 2

to find the what the number of terms is for -1 - 2.25 - .......

[river oriole]

is U your terms?

ill just put it this way, whats the formula for the nth term?

[steep epoch]

yea and n is term number

[steep epoch]

divide what by 2

[vagrant ice]

can you solve 3^x = 177147 first

[versed pendant]

11, idon't think you need to divid by two

[steep epoch]

Un = u1 + (n - 1) d

[river oriole]

this is gp

thats for ap

[steep epoch]

oh my fault

Un = u1 x r^n-1

[vagrant ice]

I mean it helps to combine everything into 1 geometric series

[river oriole]

should bracket the (n-1) but yea

[vagrant ice]

but yeah like that's the 12th term, cause the first term has 3^0, the second term has 3^1 and so on

[versed pendant]

oki

[river oriole]

you have a first term and the common ratio from your part i)

so you solve for n

[vagrant ice]

so if you have the combined geometric series -1 - 2.25 - .....
you're doing it up to the 6th term

yeah the key is to group the terms 2 - 3, 9/2 - 27/4 and so on once you know the number of terms

[steep epoch]

i dont know how

[river oriole]

divide both sides by the first term

and then use log

[vagrant ice]

did you read what I wrote above?

[steep epoch]

yea

[vagrant ice]

so......

[steep epoch]

are you refuring to his

[river oriole]

i think im confusing him lmao sorry

[vagrant ice]

yes, once you find x = 11, then that's the 12th term in the original series

[steep epoch]

nah you werent your good

[vagrant ice]

my strategy is to combine 2, 9/2, 81/4...... and -3, -27/4, -243/16....... by adding both series together

so that you only have to do one calculation

so the first term of the combined series is 2 - 3

the second term would be 9/2 - 27/4

[steep epoch]

so your saying to add or subtract each 2 terms

.close

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[spare crow]

how do I divide imaginary numbers?

[rich elm]

divide by what tho

u just divide as u normally would

[proper bramble]

Multiply by conjugate of denominator

[spare crow]

I think I overcomplicated things a lot

I got this problem where 4x+11 / 3x+2 in Z

so 3x+2 | 4x+11

[proper bramble]

Just polynomial division

[spare crow]

so 3x+2 | 24x+66

so now what....

[proper bramble]

Same with real numbers

[proper bramble]

Wait what exactly is the problem

[spare crow]

.

[proper bramble]

Yeah but what are u supposed to do

[spare crow]

erm... for now this...

find x

[proper bramble]

But there’s no equation

[spare crow]

mhm

x in N

[vagrant ice]

keep reducing using the Euclidean algorithm, so gcd(3x + 2, 4x + 11) = gcd(3x + 2, x + 9) and so on

then next do 3x + 2 - 3(x + 9)

[spare crow]

omfg yes

[vagrant ice]

and then you get like x + 9 is a factor of 28 idk

[spare crow]

I had something like that in mind but not exactly

thanks

[vagrant ice]

npnp!

[spare crow]

<3

.close

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[spare crow]

wait what

is this true if 3x+2 > x+9

?

cuz like. OH YE IT IS TRUE

thx

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[wheat gyro]

kind of unsure what to do here

[vagrant ice]

if k people fill up each room, then (n - k) people can go in k rooms, so there are (k choose (n - k)) ways to arrange the people I think

[safe radish]

@wheat gyro Has your question been resolved?

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[compact flame]

I don't know where to begin

[compact flame]

<@&286206848099549185>

[digital sparrow]

I think c

Does subset include that Pi can be the whole A?

@compact flame

[digital sparrow]

!15m

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[compact flame]

really?

[compact flame]

No it says the correct answers are A and D

[plucky elk]

what does replacing the elements of P1 mean

[compact flame]

I don't think so

[compact flame]

Yeah that's what I'm thinking

So vague

[digital sparrow]

Sorry I read 18 as 48 💀

[digital sparrow]

Yeah that is why we subtract 2

1*

I think I know

[plucky elk]

If A = {1,2,3}, P1 = {1,2}. How is A reconstructed?

[digital sparrow]

Ways of choosing Pi = 2^m - 1

[compact flame]

Replacing as in

[digital sparrow]

(Binomial Theorem)

[compact flame]

Re-place-ing??

shifting them around in A

😭 idk sets man

[plucky elk]

like P1 = {a, b} ?

[digital sparrow]

I think now new elements would be added

[plucky elk]

So A = {a, b, 3} ?

[digital sparrow]

A will become {3, 4, 5}

[compact flame]

This makes sense

but there's one problem

[digital sparrow]

🤓

[compact flame]

Is there a condition that the elements replaced with cannot be the same elements in A?

[digital sparrow]

Read the union and Sigma part of it

[compact flame]

there's no sigma

[digital sparrow]

It ensures that all Pi are distinct and have no elements in common

[compact flame]

Where?

[digital sparrow]

That is basically a Sigma + Union/Intersection

[plucky elk]

that's not what sigma means

[digital sparrow]

Yeah

I mean you get it lol

Instead of summing all elements

We are taking union