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[keen bane]

What would be the span here in this case

[round plover]

you've found x = 2A+B, y = A+C and z = 2A + B

A, B, and C are free variables

so you can just parametrise the span in terms of them: {(x,y,z) : A,B,C in R}

expand the expressions for x,y,z in that set and clean it up a little

[keen bane]

But the span isn't r2

This is a plane corrct5

[round plover]

yes

what do you think the span is

[keen bane]

We get b in terms of x and y

But for a just canc3ls out

And u for c you can only get in terms of y

[keen bane]

So I'm a
Bit confused here

[round plover]

I have no idea what you're doing

[keen bane]

My work

I substituted to see if you can get x and y in terms of a b and c

So there is an x and y such that you can use any coeff

[round plover]

you have this

X=Z so they are not independent

Y has a C term in it, so it is independent from X and Z

so your span is 2 dimensional

[keen bane]

Is this because x and z are not independent

[round plover]

yes

[keen bane]

If x was independent

That would make it r3

Right

A plane in r3

[keen bane]

And can I say vector x = R2

For span instead of equations

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[pale perch]

thats not valid because its a specific case. You have only proven one case of it being true

rather than all

[hard atlas]

not at all what they want you to do

why only x=1? why not something like x=27?

so?

27 is aswell

yes but if x=27 then it might

or maybe x=8164926

so x+y and x-y are both integers

so you have a product of integers which gives 1

x-y doesn't have to be positive

which integers multiply to 1

yes

and?

that does not mean that x-y has to be positive

take 3-5

3 and 5 are both positive

yes this is one half of the proof by contradiction

which integers multiply to 1

that's what it is asking

[hard atlas]

can you just humor me and answer this

there is also another pair

ok

so these are our two options for what the left side could be

yes

and now we use that x,y are positive

and can't satisfy these equations

and that's where we get the contradiction

yes

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[lunar venture]

In argand diagrams, what does locus of points actually mean?

[lunar venture]

All possible values of an unknown complex number? (I.e z)

For example this question

I would take the perpendicular bisector of the line segment joining (3,0) and (0,-1) so I can find all possible values z might be?

But this doesn’t really make sense would it?

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[karmic epoch]

Im stuck if any of this is right and how to do the rest ,rotate

[novel ridge]

Can you rotate this? Just say ,rotate

https://tenor.com/view/the-rock-upside-down-how-gif-22996439

[karmic epoch]

,rotate

[karmic epoch]

,rotate

[novel ridge]

once more?

[novel ridge]

yay! lol

[karmic epoch]

haha. sorry

[novel ridge]

Thanks lol... No worries 🙂

So I'm assuming that you're talking about the last problem with the graph? @karmic epoch

[karmic epoch]

yes. well sort of. im talking about the last two and im unsure about the answers on 7 and 8

[novel ridge]

Okay so if AC is 34... and if AC = AB + BC, then that means AB + BC = 34

Because you're just adding those two distances

So (2x+5) + (x-1) = 35

Can you solve x from here?

@karmic epoch

[karmic epoch]

yeah sorry. im having trouble understanding already.

[novel ridge]

AC is your total distance and they said it was 34...

AB's distance is 2x+5
BC's distance is x-1

Would you agree that adding the distances AB and BC would get you AC?

[karmic epoch]

one second

oh yes. i would agree with that

[novel ridge]

Okay so AB + BC = AC, right?
Remember what we said about AB's and BC's distance?

[karmic epoch]

yes that is right

[novel ridge]

So if AC's distance is 34, would you agree that since
AB + BC = AC
AB + BC = 34
(2x+5) + (x-1) = 34?

[karmic epoch]

yes because they are both the same distance

[novel ridge]

Well, no they aren't the same distance but they do both add up to equal the total distance

Once you find x, you'll know that they aren't equal distances 🙂

[karmic epoch]

oh yeah thats what i meant. if you add them together they are the same distance?

[novel ridge]

Yes if you add them up, they will give you the total distance of 34

So now you just have to solve for x

(2x+5) + (x-1) = 34...

Can you solve for x? If not I have an idea of how I can explain further

[karmic epoch]

the closest number i can think of for x can be is 9

[novel ridge]

Can you tell me how you got that?

(It's okay to say that you just guessed lol)

[karmic epoch]

so 9 plus 9 (the 2x) plus 5 is 23, then add another 9 and minus 1 i just assumed id be 34

but its not

[novel ridge]

Okay so just chose x values until you got 34... Understood...

So what if I told you that you can do this...

(2x+5) + (x-1) = 34

3x + 4 = 34

3x = 30

x = 10

Soooo no you're not right

[karmic epoch]

ohhhh

[novel ridge]

You were really close actually... But if you just guess what x is by not doing this math, it will take a very long time to solve x lol

[karmic epoch]

i wasnt thinking to use 10 because its so easy to multiply. i forgot about the other numbers

that was an awesome explenation

i actually got that

[novel ridge]

Awesome 🙂

[karmic epoch]

so BC would be 9?

[novel ridge]

Yes, that's correct

Because 10 - 1 is 9 when you put x back into the equation of x - 1

[karmic epoch]

okay perfect. Thank you. do you have any idea on how to do the bottom problems

[novel ridge]

Okay so the bottom problems... Let's just start by drawing the dots and drawing a straight line that connects those dots

[karmic epoch]

okay and i do that...

at

-1 and -4

[novel ridge]

Well I just mean making a dot a (-1,-4) and (3,2) and drawing a line that connects those dots

[karmic epoch]

i dont know where the numbers go on the box exactly

[novel ridge]

Gotcha. So each box is 1 step.... So moving left will make your x coordinate go negative

Moving to the right will make your x coordinate go positive

Moving up will make your y coordinate go positive, moving down will make your y coordinate go negative

[karmic epoch]

okay i understand that now

[novel ridge]

So from S, let's travel in the x direction first towards T... So you're just moving to the right...

How far is it from -1 to 3?

[karmic epoch]

i think id be 4 right

[novel ridge]

Correct, so what is half way?

[karmic epoch]

2?

[novel ridge]

Yes so what is half way between -1 and 3?

[karmic epoch]

i think 2 again?

[novel ridge]

Not quite... what is -1 + 2?

[karmic epoch]

1?

[novel ridge]

Yes so on the x direction, 1 is going to be your half way point....

Now let's do it again for the Y direction

How far is -4 from 2?

[karmic epoch]

5

[novel ridge]

Almost

2 - (-4) = ?

[karmic epoch]

oh shoot

i went to the x

[novel ridge]

All good lol, we're going upwards now

[karmic epoch]

2 - -4 is -2?

i think

or -6

[novel ridge]

2 - -4 is just 6

Subtracting negative numbers is like adding the positive of that negative number

But anyways... so yes, -4 to 2 is 6 units in the positive Y direction, upwards

What is half way of that?

[karmic epoch]

are you counting the middle to be a point) but i think id be -1

[novel ridge]

Yes, spot on

So you told me that in the x direction, you half way point is at x = 1 and now you told me that you half way point in the y direction is at y = -1... So what is your coordinate for your half way point?

[karmic epoch]

1, -1 im pretty sure

[novel ridge]

Yes that's correct 🙂

[karmic epoch]

ahh awesome

[novel ridge]

Alright last question...

The length of the line... I'll give you a hint on this one... I would use Pythagorean Theorem.

Since you told me that the total x distance between -1 to 3 is 4 units away and that the y distance is 6 units away... You can sort of see a triangle with this....

[karmic epoch]

i think you can yes

[karmic epoch]

oh yea yeah

[novel ridge]

Using Pythagoreon's Theorem, what do you think that distance is?

[karmic epoch]

12?

[novel ridge]

Nope... Do you know Pythagoreon's Theorem?

[karmic epoch]

i know its something smart... can you give a quick introduction. i think ill remember it

[novel ridge]

Yes so Pythagoreon's Theorem basically just says that you can find a diagonal line's distance, provided that you know the distance of two straight lines that make a right angle...

Have you gone over this in class?

I just want to make sure you're only using the things you have learned. Otherwise, I can help you find the length of the line a different way

[karmic epoch]

i think i understand that. im positive weve learned it

[novel ridge]

Okay in that case, let's say that, from my picture A = 4, B = 6 and C is the length of that line.

To find C, you need to take A and square it so 4^2 = 16, take B and square it so 6^2 = 36, and C is the square root of A^2 plus B^2.

^^^ This is not the answer, just an example

You will need to use a calculator for C. I don't think that's a perfect square number

[karmic epoch]

im slowly dying in my chair. but i think i get that. do we need to find C in our problem?

[novel ridge]

Yes

We're almost done lol

Once you find that, you're good to go 🙂

[karmic epoch]

thats where is confuses me. i didnt think there was a C to solve

[novel ridge]

C will just tell you the distance of that line

or ST, according to your paper

[karmic epoch]

so C is the question mark

[novel ridge]

Yes

[karmic epoch]

okay cool

[novel ridge]

I think you got it from here. I'm gonna eat 🙂

[karmic epoch]

okay sounds good... i

i dont think i do

i wanted to eat aswell

but Thank you

[novel ridge]

Answer is the square root of 52 lol

Whichi is about 7.211 if you used a calculator

[karmic epoch]

i just got 42

[novel ridge]

16 + 36 is 52 lol

[karmic epoch]

gosh dangit. Oh yeah

i did that wrong

[novel ridge]

LOL no worries man. All good and all done

[karmic epoch]

Thank you so much

i literally

[novel ridge]

You're welcome 🙂

Good luck with your class!

[karmic epoch]

could not have done that

[novel ridge]

No problem 🙂

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[hollow kestrel]

I'm a TA for Pre-Cal, I'm fairly confident x = 4 is not a relative extrema but I wanted to confirm before I mark anything.

[livid tapir]

Yeah not a relative extrema

[hollow kestrel]

ight cool, i helped the student out and i saw some others with different answers and just wanted to make sure

ty!

.close

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[quick ridge]

Confused on the top problem

[waxen hawk]

Find the value of f(1) first

[quick ridge]

Ok

Is this correct?

[waxen hawk]

Yes

Now you have the value of f(1), could you find the f[f(1)] by that

[quick ridge]

So I plug 4, 0<1<2 into the first equation at the top for x?

Or I plug that into the original f(x) equation

[waxen hawk]

Emm

It is important to understand what you’re dealing with

F(x) is like a function

It is a function

[quick ridge]

No yes I understand that

[waxen hawk]

So you put x=1 in f(x) in order to find f(1)

Now you got f(1)=4

just put this result into f(x) again

[quick ridge]

Yes, with the 0<1<2

[quick ridge]

Ok cool sometimes the words/equations get to me

[waxen hawk]

It is fine, I as well got confused when first encountered to this type of question

[waxen hawk]

Is that f[f(1)]=12

[quick ridge]

Yes

[waxen hawk]

It is kinda blurry and sketchy

[quick ridge]

Sorry

I’m writing small to conserve space

[waxen hawk]

It is not consistent with my answer

Could you show the calculation process

[quick ridge]

Oh

13

My correction

I forgot to add the 1

[waxen hawk]

Yes

[quick ridge]

So with 13 I proceed to plug it into the other equations?

[waxen hawk]

I think f[f(1)]=13 is the final answer

The question asks for its value, isn’t it

[quick ridge]

Oh ok. It has a bracket with other equations no?

[waxen hawk]

I couldn’t see it, the question says find f[f(1)] for...

[waxen hawk]

Ohh

It took me some time to understand the question

However, it asks you to find the f[g(x)] and g[f(x)]

You just simply put the function into another function

Then you will get the answer

[quick ridge]

So sorry, I’m referring to the top one, number 15

[waxen hawk]

I thought you have already done that question

As you find its value, which is the question asks for

[quick ridge]

Oh, 13 was it?

[waxen hawk]

It is, is there anything makes you feel uncertain about your answer

[quick ridge]

What was the reason for the other equations?

[waxen hawk]

Other equations, could you be specific

[quick ridge]

The ones in the bracket including 3x+1, 0<x<2

[waxen hawk]

It literally just indicates you that f(x)=3x+1 when x is in between 0 and 2. Like it tells you what the graph of f(x) looks like in that range

[quick ridge]

Oh ok sorry for the trouble 👍and thank you for your time

[waxen hawk]

It is ok, ure welcome

[quick ridge]

And for reference, the way to do these is by putting the g(x) function into the f(x) one

[waxen hawk]

Yes exactly

[quick ridge]

Sounds good, thanks a ton

Currently sitting at a D in pre calc, I was advanced in math as a kid but online algebra 1 messed me up😂thanks for the explanation

[waxen hawk]

It is all good, get more practice you will soon obtain a A

[quick ridge]

Aiming for a 80 by the quarters end. Trying to make realistic goals

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[subtle knot]

,rotate

[subtle knot]

Need help on 23 B

It’s supposed to = 1.93% but I keep getting the wrong percentage

[fathom drum]

Can you show your work

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[waxen magnet]

in step 4.. why didn't we multiply 5 in 4? my answer is 3 - 8f(-5x + 20)

[waxen magnet]

(please mention me if anyone of you ansnwered)

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<@&286206848099549185>

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[outer shadow]

im extremely confused..

[outer shadow]

what is a counterexample

and how do i provide one

[dull maple]

,w counterexample

[twin meteor]

Results provided by WolframAlpha

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

Requested by .enemagneto

[dull maple]

You give an example which refutes the claim.

Like if i say that all primes are odd.

Counter example - 2 is a prime.

[plush sage]

yes

[outer shadow]

im a bit confused

😭

[plush sage]

you are to prove the statements false, just as you said in the example

[outer shadow]

wait are these geometry proofs

so is #1 m<1 = 45 and m<2 = 45

because 45+45 is 90

complementary angle is 90 degrees

RIGJT

[plush sage]

You got it

[outer shadow]

omg okay

so then

for b

i dont kneo

[plush sage]

I think you can put any other value for l and b other than the given

for example if you put l=14, b=2, you get perimeter 32 as well

same with 13,1 also

[outer shadow]

okay

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[jovial shadow]

Hello there

[jovial shadow]

I need help finding the 2 zeros of this question

Can someone please help as I keep getting the wrong answer

The actual answers are supposed to be 4 and 8 according to the answer key

[lusty python]

You are going to want to move all the terms to the left and make them all equal to zero

and solve from there

for each x term, you want to make it equal to 0 and solve

[jovial shadow]

Can you please give me a step by step explanation because the quadratic I get isn’t giving me the right answers

🙏 please

[bright yew]

show your work. what quadratic did you get

[jovial shadow]

When I plug these numbers into the quadratic formula I don’t get the right answer which means I messed up somewhere before

[bright yew]

your mistake was $(\sqrt{a} - \sqrt{b})^2 \neq a - b$

[twin meteor]

nebula40

[jovial shadow]

Oh ok thanks

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[split swallow]

∫ x^2+2x-3)/(x^3-4x) dx
lim 0 - 4 ∫x√x^2 + 9 dx
∫4dx/x^4-1 need help with these 3. im not familiar with partial fractions yet

[rugged orchid]

Try factorise

[obtuse radish]

x^3 - 4x

how to

[rugged orchid]

What ideas do you have?

[split swallow]

(x^3-4x) = x(x^2 - 4)?

or i could just factorize it again so x(x-2)(x+2)

[rugged orchid]

Looks good

What about the top?

[split swallow]

you mean the x^2 + 2x - 3?

[rugged orchid]

Yeah

Does it cancel with the bottom if you factorise?

[split swallow]

https://www.youtube.com/watch?v=6rXByMcuAyI&t=348s i usually based on this one

technically i dont factorize the top

[rugged orchid]

Well, usually you need to do some guesswork when solving integrals

[split swallow]

only the denominator

[rugged orchid]

As a general approach I would try to see if the denominator and numerator cancels

If it does our lives might be a lot easier

If it doesn’t we can try partial fraction decomposition

[split swallow]

thats basically need XD

[rugged orchid]

Well I mean, this is the kind of line of thought I’d try for integrals

It doesn’t always work

But it’s good to just check there aren’t easier paths

[rugged orchid]

The answer to this was no anyway

[split swallow]

shouldve answer it straightforward mb

[rugged orchid]

Eh integrals are usually educated guesswork

Anyhow, what did you try for partial fraction decomposition?

[split swallow]

this one (x^2 + 2x - 2)/(x^3 - 4x) = A/x + B/(x - 2) + C/(x + 2)

but i have to multiply it by both sides

[rugged orchid]

What do you mean

You can just do A(x-2)(x+2) + Bx(x+2) + Cx(x-2) = numerator

[split swallow]

to clear the denominators

[rugged orchid]

Yeah this

[split swallow]

i was gonna send the answer but ok AHAHAH

but the next part is very confusing to me

[rugged orchid]

Ok let’s see it

What is confusing about the next part

[split swallow]

A(x-2)(x+2) + Bx(x+2) + Cx(x-2) about getting the value of a b and c

[rugged orchid]

So expand it

Then match coefficients

Of the numerator

[split swallow]

isnt there another way though

[rugged orchid]

What’s wrong with this way

[split swallow]

well actually the teacher taught us kinda like this one

[rugged orchid]

Mhmm clever

That is a bit confusing to explain indeed

[split swallow]

and this is the part where i got stuck literally HAHAHA

i mean sure i can do partial fraction decomposition but after that then yeah

[rugged orchid]

So we have A(x-2)(x+2) + Bx(x+2) + Cx(x-2) = x^2 + 2x - 2

[split swallow]

wait sorry

its actually x^2 + 2x - 2

[rugged orchid]

Oh

[split swallow]

my fault for the correction

[rugged orchid]

That’s a lot harder

[split swallow]

according to this part i need to make some terms zero

[rugged orchid]

Yeah let me think about how to explain this

Ok so

We have 2 expressions

And we know they are equal

Agree?

[split swallow]

yes

[rugged orchid]

Those being these 2 expressions

Ok so we want to find out what A B and C are

Now fortunately for us, we know exactly what the value should be if we plugged in particular x’s

So let’s say I want to find what A(x-2)(x+2) + Bx(x+2) + Cx(x-2) is when x = 1

Aha we can simply put x = 1 into x^2 + 2x - 2

Do you agree?

[split swallow]

yes i agree

[rugged orchid]

Ok if I plug x = 1 into this it doesn’t do shit

It’s useless

Oh but if I chose x carefully, something useful happens!

Let’s say I chose x = 0 (you will see why)

Then I have A(0-2)(0+2) + B(0)(0+2) + C(0)(0-2) = 0^2 + 2(0) - 2

Do you agree?

[split swallow]

yes

[rugged orchid]

Aha but the 2nd and 3rd terms are just * 0

Which is just 0

So then what we really have is A(-2)(2) = -2

Yeah?

[split swallow]

wait got confused a little bit

its not gonna be -4?

or simple just -2?

[rugged orchid]

Look at the right hand side

0^2 + 2*0 - 2

That’s just 0 + 0 - 2 = -2

[split swallow]

ohhh

now i see

[rugged orchid]

And I chose x = 0 here

[rugged orchid]

Precisely because of this

I can then choose some other “special” values of x that eliminates the other terms

So I can in a way, “isolate” the A, B, and C

Do you think you can do the rest on your own?

[split swallow]

hold on im not familiar with it yet i apologize for that

[rugged orchid]

You can find A from this step

Yes?

[split swallow]

yes i can

it will become -4A = -2

[rugged orchid]

And then?

[split swallow]

ummm

A = -2/-4

then A = 1/2?

[rugged orchid]

Yes

Ok so now we have So we have (1/2)(x-2)(x+2) + Bx(x+2) + Cx(x-2) = x^2 + 2x - 2

Do you understand why I chose x = 0

[split swallow]

yes

[rugged orchid]

So what other numbers for x should you try

[split swallow]

and also the value of b and c is now 0

[rugged orchid]

Nope

We don’t know that

[split swallow]

no no i mean

[rugged orchid]

We don’t know what B and C is

[split swallow]

from A(-2)(2) B(0) C(0)

[rugged orchid]

So I chose x = 0

[split swallow]

but yeah since we already got the value of a next is b and c

[rugged orchid]

Can you suggest another value

[split swallow]

x = 2

[rugged orchid]

Ok do the steps I did

[split swallow]

wait

[rugged orchid]

Let’s see what we get

[split swallow]

A(2-2)(2+2) + B(2)(2+2) + C(2)(2-2) = 2^2 + 2(2 ) - 2

[rugged orchid]

And then?

[split swallow]

A(0) + B(2)(4) + C(0) = 4 + 2

[rugged orchid]

What

2^2 + 2(2) - 2

Oh

You simplified a bit

[split swallow]

yeah i did XD

[rugged orchid]

That’s a weird thing to do, simplifying only part of it lol

It looked wrong but it’s correct

[split swallow]

i though you were gonna say im wrong XD

[rugged orchid]

My bad

Nah it’s good

So what’s next

[split swallow]

A(0) + B(8) + C(0) = 8

[rugged orchid]

Not 8

[split swallow]

wait so basically B = 1?

oh

its actualyl 6

asklnjaksnasasdsa

[rugged orchid]

Yes

So you’ve found A and B now

Do the same thing for C then it should be easy sailing

[split swallow]

B8 = 6

[split swallow]

wait whats the exact value of b? 6/8?

[rugged orchid]

Yes

[split swallow]

i can still simplify it right?

like make it 3/4 or something

[rugged orchid]

Sure

[vocal sleet]

@split swallow Has your question been resolved?

[split swallow]

wait is it possible ? i see isee

i will try x = -2 for c

[vocal sleet]

@split swallow Has your question been resolved?

[vocal sleet]

@split swallow Has your question been resolved?

[split swallow]

.close

[vocal sleet]

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[tropic anchor]

can someone check my work (the one written in pencil)

[paper depot]

,rccw

[paper depot]

yours is in pencil and rejected?

[tropic anchor]

no

[paper depot]

i see an X next to it

in red ink

[tropic anchor]

the pencil is what i corrected

[paper depot]

right ok

yeah, your pencil-work looks fine to me barring arithmetic mistakes you might have made that i missed

let's ask WA

,w expand (x + 3/x^2)^6

[tropic anchor]

WA?

[twin meteor]

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Requested by ann_dec

[paper depot]

wolfram alpha

ok yeah WA agrees with you

the constant term is indeed 135

[tropic anchor]

i didnt know u could do that

[tropic anchor]

thanks

.close

[vocal sleet]

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[tropic anchor]

Find the values of the constant k for which the equation (2k-1)x²+6x+k+1=0 has real roots

[merry python]

!status

[vocal sleet]

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

[tropic anchor]

I'm getting x<= -5/2 but the answer says its x>=-5/2

wait

oops i wrote x instead of k

but yeah

[merry python]

How did you conclude to this?

[tropic anchor]

I factored it

[merry python]

and then?

2x^2+x-10=(x-2)(2x+5)

[tropic anchor]

and then i tried to find x

[merry python]

well, have you ever solved quadratic inequalities before?

[tropic anchor]

Yeah but we only did it 1 lesson so maybe I didn't get it

How do u do it

[merry python]

Okay, so product of two factors is less than zero

so you can conclude that they are of the opposite sign

Like if (x-2) is positive or =0 then, (2x+5) must be negative or =0 and vice-versa

so as to satisfy the inequality

[tropic anchor]

ohh ,is thst always the case

[merry python]

no

That is just a trick that works here

usually you can graph

[tropic anchor]

wdym it's less than 0?

[merry python]

the method is WAVY-CURVE METHOD

[merry python]

well, x<=0 is read as "x less than or equal to zero"

[tropic anchor]

what's product of 2 factors

[merry python]

Okay, let me guide you the same through wavy curve method

mark the critical points on a number line

i.e. set (x-2)=0 and (2x+5)=0 and solve for x

[tropic anchor]

I'm kinda confused

[merry python]

Like this

[vocal sleet]

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.

[ashen rose]

draw a curve

[merry python]

Now, choose a value for x greater than 2 and check whether (x-2)(2x+5) is positive or negative

[tropic anchor]

positive

[merry python]

Okay

Now choose a number between -5/2 and 2 and

do the same

[tropic anchor]

wdym? if i choose -1/2 its negative if i choose 1 its positive?

[merry python]

are you sure it is positive at x=1?

[tropic anchor]

I don't get it sorry

[merry python]

at any value between -5/2 and 2, the value of (x-2)(2x+5) will be negative

[tropic anchor]

The other person said draw a curve so is it bc the curve is below the line

[merry python]

well, that method is actually better but the issue is it has some extra rules to follow

That works well and fine if the linear factors have an odd power

[ashen rose]

and then since it says ≤ 0 it want the lower part of the curve

[tropic anchor]

Im getting more confused

[ashen rose]

you can just use the curve method

i think youll undertand that better

[merry python]

well, relate that to a normal cartesian plane. The line is y=0. so anything below it would represent negative and anything above it would represent positive

The idea is to cut the critical points whilst going in a curve

[tropic anchor]

ok, but what was the other method you said the one without drawing a curve

[merry python]

well, lets finish this one first

Okay so you need (x-2)(2x+5)<=0.
Notice the less than equal to zero. That would correspond to the area/interval below the line.
AND the area below the line is covered by the interval [-5/2,2]

[ashen rose]

if youre drawing a curve ^

[tropic anchor]

why is it less than 0

[merry python]

That's what the inequality was right

basically anything and everything you did below that box is wrong

[tropic anchor]

I only get it when u drew the curve

what if u don't use the curve then how do i solve it

[merry python]

By marking if (x-2)(2x+5) is positive/negative in the given interval

for example for x less than -5/2, (x-2)(2x+5) would be positive

and for x between -5/2 and 2, (x-2)(2x+5) would be negative

This is a bit more tedious

[tropic anchor]

"for x less than -5/2" does that mean substituting x with a number less than -5/2

[merry python]

YES

[tropic anchor]

Ok if i say x os -3

2×-3+5= -1

Its negative

[merry python]

,w (x-2)(2x+5) at x=-3

[twin meteor]

Results provided by WolframAlpha

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[tropic anchor]

Ohhh u multiply

The factors

[merry python]

so we need the negative part (since <=0). You can conclude that x must be between -5/2 and 2 or equal to them

Both the methods are almost same. The wavy-curve method is a bit easier but tricky at the same time for higher degree polynomials

[tropic anchor]

wait but if x is 2

[merry python]

,w (x-2)(2x+5) at x=2

[twin meteor]

Results provided by WolframAlpha

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[tropic anchor]

ohh nvm

I think i understand now

thank u

[merry python]

welcome

[tropic anchor]

.close

[vocal sleet]

Channel closed

Closed by @tropic anchor

Use .reopen if this was a mistake.

[tropic anchor]

qait

i have another quest

question

[merry python]

.reopen

[vocal sleet]

✅

[tropic anchor]

when it says find the values of the constant is it always asking for a range like -5/2<=x<=2

[merry python]

well, there could be cases where only one value would work

But in general a range is asked

[tropic anchor]

how do I know if its asking for a range or a value

[merry python]

You need to solve the question for that

[tropic anchor]

Okk

Thanks

.close

[vocal sleet]

Channel closed

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[merry python]

consider

for example x^2+kx+(k-1)

for it to have non real or equal roots

[tropic anchor]

.reopen

[merry python]

you have discriminant=<0

[vocal sleet]

✅

[merry python]

k^2-4k+4=<0

(k-2)^2=<0

Solving this you get k=2

only one solution!

[tropic anchor]

Did u take the square root

To get 2

[merry python]

square of anything real is >=0

so, (k-2) must be equal to zero

to satisfy =<0

[tropic anchor]

but its greater thsn or equal to

why is k=2

[merry python]

non real or equal roots or basically less than 2 real roots

so discriminant must be less than or equal to zero

I guess this stuff is of higher level than what you've studied so far

[tropic anchor]

Ohh nvm

I understand

[merry python]

Oh well

I am going afk

so do you have any more doubts?

[tropic anchor]

nope

thanks

.close

[vocal sleet]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crisp agate]

This question as a while mind boggles me I need help

[crisp agate]

why is 1-e^-3t = 1-0?

why is arctan(infiity) = pi/2?

how did they get ln(2/5) to be the answer?

[proven garden]

,w graph e^x

[twin meteor]

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[proven garden]

it approaches 0 for -inf

or think of it like 1/e^inf

[crisp agate]

good point ok

what about arctan

[proven garden]

what is tan(pi/2)?

[crisp agate]

tan(pi/2) = 1/0 no?

[proven garden]

so it diverges

so arctan(inf)=pi/2

[crisp agate]

huh wdym by diverges

wait

[proven garden]

1/0 is infinity

[crisp agate]

we can just graph arctan as well which is just a horizontal tan graph right

[proven garden]

yes

[crisp agate]

so as x approaches infinity, y approaches po/2

ohhh

ok

and then just the last one

i know if the degrees equal one another you can just divide coefficients for the end behavior

but is there another way to do the question if the degrees weren't equal one another?

[proven garden]

?

it's just L'hopitals law

[crisp agate]

im stupid

i get it 💀 i didnt know you could just

take the Ln out of a lim function

and evaluate thel imit inside

[proven garden]

I think you can do that for all continuous functions

[crisp agate]

ill be sue to keep it in mind

thanks!

[proven garden]

no problem

[crisp agate]

.close

[vocal sleet]

Channel closed

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[crisp agate]

.reopen

[vocal sleet]

✅

[crisp agate]

.close

[vocal sleet]

Channel closed

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[crisp agate]

.reopen

[vocal sleet]

✅

[crisp agate]

@proven garden how would I perform L'H on the j direction of the vector?

im struggling specifically on the sin^2(t)

i end up with 1/sin(x)cos(x)

[worn wadi]

yeah its correct

[crisp agate]

do i just evaluate it as infinity???

[silk comet]

you might have to apply it twice

[crisp agate]

so product rule sin(x)cos(x)?

[silk comet]

should be t/(sin(t)cos(t))

[crisp agate]

yeyes mbmb

[silk comet]

not 1 in the numerator

[worn wadi]

oh yeah ^ mb

[crisp agate]

typo

[worn wadi]

t/cos(t)sin(t)

[crisp agate]

still 0/0 L'H again

so apply product rule for the denominator

?

[silk comet]

yeah

if you use known limits you can just do
[\lim_{t \to 0} \frac{t^2}{\sin^2(t)} = \left(\lim_{t \to 0} \frac{t}{\sin(t)}\right)^2]

[crisp agate]

whhops didnt mean to click that

[twin meteor]

tushar

[crisp agate]

wha

[silk comet]

sin(x)/x as x approaches 0 is a common limit that many people learn

L'H works too

[crisp agate]

i get

cos^2(x) - sin^2(x)

after product rule

ok i have a general matj question

if i can do this then i can solve now i think

[silk comet]

yes (for the denominator)

now you can evaluate

[crisp agate]

is cos^2(x) = ( cos(x) )^

^2

[worn wadi]

yes

[silk comet]

yes

[crisp agate]

ok so then its just 0/1

0

[silk comet]

no

numerator is not 0

it was t before

[crisp agate]

1*

1/1

[silk comet]

yes

[crisp agate]

1

[silk comet]

yes

[silk comet]

this also gives you 1^2 = 1

[crisp agate]

so u can just plug n square ur finding?

like les say

cos^2(pi/3)

cos(pi/3) = 1/2

so 1/4

[silk comet]

yes

[worn wadi]

using what tushar mention should be easier istead of using Lhopital

[crisp agate]

i c i c

[worn wadi]

since you should already be familar with sinx/x when x approaches 0 = 1

you dont have to spend your time differentiating using L'H

[crisp agate]

that makes sense, i'm just trying rn to practice derivatives too since the exam today is gonna use them

killing 2 birds w/ one stone

[worn wadi]

ultimately its your choice do what you feel most comfortable with

goodluck on your exams!

[crisp agate]

ty 😭

ill probably be back soon here to ask another question 🫡

.close

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[vast shale]

is this correct ?

[lapis marten]

Shouldn't the argument be -3/4pi?

Also kuyaroka please respond meaningfully

[vast shale]

why -3/4pi

oh yeah

it's -3/4pi

wait ichange

[quaint mulch]

Nice nasus pfp

[vast shale]

thanks ncie nunu pp

the best character is yuumi

[obtuse sierra]

#chill

[vast shale]

so this is right or no

what's this bro

[unkempt oasis]

) How can you add a 5N and 4 N forces acting simultaneously on a body to produce a resultant of 7 N force?

[lapis marten]

What did you get for cis(-21pi/4)?

[vast shale]

-3pi/4 *7

[unkempt oasis]

Didnot get it?

[lapis marten]

Yes, but what was your next step?

[vast shale]

idk

[unkempt oasis]

Bro I need to prove by vector law

[lapis marten]

Also, just noticed, shouldn't the 7 be distributed over the modulus by raising the modulus to the seventh power instead of multiplying

[vast shale]

so what i should do

[obtuse sierra]

<@&268886789983436800> not listening

[vast shale]

i need to do 21sqrt(2)^7

[lapis marten]

(3sqrt2)^7

is the new modulus

[vast shale]

yes 3sqrt2^7

3(sqrt(2))^7*cis(-21pi/4) @lapis marten ?

[lapis marten]

You did 7 * (3sqrt2)

[vast shale]

yeah

its ^7

[lapis marten]

The three should also be inside the brackets

so (3 * sqrt(2))^7

[vast shale]

yeah

[lapis marten]

Then it's correct

[vast shale]

so it's (3*sqrt(2))^7 cis(-21pi/4) ?

[lapis marten]

Yes

[vast shale]

thanks brother

.close

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[vocal sleet]

Please don't occupy multiple help channels.

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[steel fjord]

Here we will look at bit strings where 11 does not occur.
Let ai be the number of bit strings of length i in which 11 does not occur. We have a0 = 1
(there is only one bit string of length 0), a_1 = 2 and a2 = 3.

show that an = a(n−1) + a_(n−2) for
all n ≥ 2 by induction

[steel fjord]

Can anyone show me how to do that proof?

[cyan shadow]

do you know how induction proofs work?

[steel fjord]

yes

but I think I have to do it by strong induction, which I am not good at

cause I tried by weak

that is where I got

@cyan shadow you know how to solve it?

<@&286206848099549185>

[vocal sleet]

@steel fjord Has your question been resolved?

[cursive turret]

look at the strings with length n and not containing 11. if the string ends with 1 it must have a 0 at n-1. if it ends with 0 it could be any appropirate string of length n-1.

[steel fjord]

yeh

but how do i proove it mathematicly

[cursive turret]

hmm, why do you think thaz my argumentation is not mathematically?

[steel fjord]

im just used to weak induction

is what you did induction?

[cursive turret]

yes, of course. it is the induction step.

[steel fjord]

ok, but where does the n-2 come from?

Is that if the strings ends with 1?

[cursive turret]

from the first part. if the stirng ends with 1, it has to have 0 at n-1 so it is an n-2 string with 01 added.

[steel fjord]

ok

kind of understand

[vocal sleet]

@steel fjord Has your question been resolved?

[vocal sleet]

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[vocal sleet]

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[upper zodiac]

for the trianlge on the right my values are 7for bc and cd

and ab = 2x from special triangles

so i get 14

but its not correct

$$why is 14sqrt2$$

[twin meteor]

puckmyseen

[upper zodiac]

$$14sqrt 2 $$

[twin meteor]

puckmyseen

[upper zodiac]

the correct one

and not 14

.close

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[abstract prawn]

Need Help with this.

[formal gazelle]

Hey I’m gonna try to help u today ;)

!status

[vocal sleet]

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

[abstract prawn]

For Number one I got X = 4

You have to go to another channel.

?

[formal gazelle]

<@&268886789983436800> blade is being annoying in all channels

[blissful sentinel]

thanks

[formal gazelle]

Ok send me the answers u got

[abstract prawn]

Of course.

One second.

[formal gazelle]

Sure take ur time

[glass sorrel]

lol what did he do

[formal gazelle]

Spamming and toxic lol

[glass sorrel]

lol

[formal gazelle]

@abstract prawn ?

[abstract prawn]

Yeah?

[formal gazelle]

Oh my messages lagged sorry

[abstract prawn]

That’s fine

You can see the picture, right?

[formal gazelle]

Yep

Ok let’s go through the first knee

One

The question is

4x - 6 + 2x = 18

[formal gazelle]

Can u show me how u did it?

The answer is correct but I wanna make sure ur doing all the steps and don’t lose marks in the exam for ur process :)

[abstract prawn]

First I combined the x terms: 4x + 2x = 6x.
Now, rewrote the equation: 6x - 6 = 18.
Next, I add 6 to both sides of the equation:
6x = 24.
Lastly, divide both sides of the equation
by 6 to solve for x: x = 4.
Therefore, the solution to the equation is
X = 4.

[formal gazelle]

Yep perfect

The first three are correct lemme see the last one

[abstract prawn]

You mean how I got number 4?

[formal gazelle]

No the questions 1 to 3 are correct

Oh wait all ur questions are correct

👍

Nice

[abstract prawn]

Wait, I still have more.

Let me show you.

[formal gazelle]

Oh Alr send me pics lol

[abstract prawn]

Of course

One second

[formal gazelle]

||also are these the questions from online quizzes of schoology || lol

Might be a a few mins late I. Responding gotto switch classes in school

[abstract prawn]

No, it’s Homework.

Lol

[formal gazelle]

Aight send pics

[abstract prawn]

Got it

[formal gazelle]

All of them are perfect

:)

Are there more?

[abstract prawn]

Nope, that’s it.

Are you sure it’s all good?

[formal gazelle]

I think lemme check 1 last time

[abstract prawn]

Alright.

Thanks

[formal gazelle]

Yep

[formal gazelle]

All correct

[abstract prawn]

🙏

.solved

[formal gazelle]

Good luck lol

.close is the one haha

[abstract prawn]

My bad lol

.close

[vocal sleet]

Channel closed

Closed by @abstract prawn

Use .reopen if this was a mistake.

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cursive fern]

can someone help me with exercise D?

[cursive fern]

I know I have to use 1/2 mv^2 and mgh but I dont know how to get the right answer

[vocal sleet]

@cursive fern Has your question been resolved?

[young blaze]

So you have $\displaystyle Mgh_0 = \frac{1}{2} M (L \dot{\theta}_{\max})^2$. Can you write $h_0$ in terms of $L, \theta_0$

[twin meteor]

adzetto

[cursive fern]

My friend is actually asking the question and he has an exam tomorrow

I passed that exam last year

Would you be able to provide some steps to the answer so I can help him out? Would be much appreciated

[young blaze]

Well, firstly, set up the conservation of energy equation. The initial kinetic energy $K_i$ is zero because the pendulum starts from rest. The final kinetic energy $K_f $ at the lowest point is $\frac{1}{2} M (L \dot{\theta}{\max})^2$. Put $h_0$ in terms of $L, \theta_0$ . Solve for $\dot{\theta}{\max}$.

[twin meteor]

adzetto

[cursive fern]

Thank you very much @young blaze!

I hope he will manage

.close

[vocal sleet]

Channel closed

Closed by @cursive fern

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic otter]

Hi, I would like my work to be checked

[civic otter]

This is the given info (the first sentence is just "given are the events A and B")

[civic otter]

This is my solution to questions (a) and (b)

[civic otter]

And this is my solution to questions (c) and (d)

[vocal sleet]

@civic otter Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[opal dragon]

!help

[opal dragon]

Someone please run me through this question

[sly sierra]

similar triangles?

[opal dragon]

Yea

[sly sierra]

what do you know about them

[opal dragon]

I know that similar triangles both have two angles that are equal

[sly sierra]

yes

and hence three angles that are equal

how about their side lengths, what can you say about those?

[opal dragon]

so similar triangles pretty much have 3 equal angles

Thats a yes right

[opal dragon]

Are the sidelengths equal too then

[sly sierra]

yep

[sly sierra]

not necessarily equal, but proportionate

the side lengths of one triangle are scaled versions of the side lengths of the other triangle (with the same scale factor for all three sides)

[opal dragon]

Alright

[sly sierra]

in your figure, you have enough info to work out what the scale factor is

[opal dragon]

so it's an enlargement sf

[sly sierra]

yeah the bigger triangle is an enlarged version of the smaller one

in some cases the triangles can be moved around as well, but that won't affect whether they're similar or not

[opal dragon]

So the bigger triangles got a base of 9cm and one side is 16cm

[sly sierra]

right

[opal dragon]

we know that the smaller triangle has a side of 12cm so it means we are doing a reduction scale factor

[sly sierra]

yes correct

and what is the scale factor?

[opal dragon]

so rsf = 12/16 = 3/4

[sly sierra]

yep

so now you can use that to find x

[opal dragon]

3/4 * 9 = 27/4

[sly sierra]

correct!