help-26

[uncut crow]

3.) Check sign of f'' to find concavity

[tropic zephyr]

If I plug in pi/4 i get a negative number

[uncut crow]

[tropic zephyr]

Something between pi/2 and 3pi/2 can be pi right? if so -8cos(pi)=(-8) (-1) =8 positive

[uncut crow]

yes

You've got the idea. The concavity would just keep switching

, w inflection points of x+8cos(x)

[thorny flame]

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[tropic zephyr]

I got it's concave down on (0, pi/2) and concave up (pi/2,2pi)

Damn counted it wrong for concave down

Do I need to include interval of a number bigger than 2pi

[uncut crow]

No

[tropic zephyr]

i figured it out I did a mistake somewhere, thanks for the help

.close

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[regal stratus]

helloo

[regal stratus]

4. A car travelling in a straight path is recorded to have traveled 8m after 1s, 20m in 2s, 32m in 3s, and so on. How long will it take for the car to travel 1380m?

I am confused since my answer is 115.33 seconds and it seems wrong

[hasty smelt]

well the car is going at a constant speed

[regal stratus]

12m /s

[hasty smelt]

yes

[regal stratus]

how do i get the answer then

1380/12?

[hasty smelt]

yes

[regal stratus]

115

but theres

the 8

1372/12?

[hasty smelt]

well it started going at a constant speed from 8m

so yes

[regal stratus]

114.33?

[hasty smelt]

+1 second from the 8m

so 115.33

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[regal stratus]

.closer

.close

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[errant thistle]

quick question abt this one here? What does "all of its interior diagonals" of a rectangular box refer to ?

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[deft venture]

The obtuse-angled triangle ABC has sides of length a, b and c opposite the angles A, B and C respectively. Prove that a^3cosA + b^3cosB + c^3cosC < abc.

[topaz sinew]

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

[deft venture]

Step 1

[pulsar sun]

hmm

lets see...well by AM-GM $\frac{a+b+c}{3} > {abc}^{\frac{1}{3}}$

[deft venture]

Oh right I completely disregarded am/gm

[thorny flame]

ItzKraken

[pulsar sun]

you could also make a substitution for the cosine of the angles using cosine law, but that might get a bit too complex

[deft venture]

I could try that

[pulsar sun]

the substitution would be: \
\
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

[thorny flame]

ItzKraken

[deft venture]

$a^4(b^2+c^2-a^2)+b^4(a^2+c^2-b^2)+c^4(a^2+b^2-c^2)<2a^2b^2c^2$

[thorny flame]

es126

[topaz sinew]

@deft venture Has your question been resolved?

[pulsar sun]

where does the 4th power come from..?

u would have

wait.. ohh

hmm i dont see anything

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[lofty prairie]

can i solve the equation alebraically

[supple mist]

they're equal

[drifting swift]

algebraically as opposed to what else?

[supple mist]

^

[drifting swift]

rotten notation tbh

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[final lance]

How do i know if i do just 0.243 or 1 - 0.243

[topaz sinew]

@final lance Has your question been resolved?

[lofty prairie]

i got 1/6 vs 1/8 algebrically

[drifting swift]

your channel timed out and now somebody else is here, open a new one.

and show your work there.

[topaz sinew]

@final lance Has your question been resolved?

[final lance]

<@&286206848099549185>

[neon iron]

Quality A is greater

[lofty prairie]

hey no c it is

[neon iron]

Correct answer?

Or according to you?

[topaz sinew]

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[final lance]

<@&286206848099549185>

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[eternal spoke]

i need to prove that a function of degree 4 can have a maximum of 4 zeros

[eternal spoke]

and idk how

[drifting swift]

do you know the relationship between zeros and factors?

you can prove the more general stmt that a polynomial of degree n has no more than n roots.

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[timber hull]

Basically does something approaching zero dominate?

[keen raptor]

f(z)?

Do you mean for the variable to match with the n?

[timber hull]

sorry yes

e.g. f(n) = a^n - b^n or something

[keen raptor]

Consider f(n) = 1/((i/3)^n)

[timber hull]

ah okay fair point

how do we characterise these sort of "side cases"

surely in general if one part approaches zero

then the whole thing approaches zero

[keen raptor]

I wouldn't really consider this a side case

If one part approaches zero and the other part approaches some other finite number then it would approach zero

[timber hull]

right

[keen raptor]

But if it's unbounded then all bets are off

[timber hull]

right okay, tyvm

.close

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[molten spoke]

This is more of a generic question but does $$\left( \frac{\partial V}{\partial T} \right)_P = \frac{1}{\left( \frac{\partial T}{\partial V} \right)_P}$$ since rearranging for V is harder than rearranging for T in the question ive been given

[thorny flame]

I can't believe you've done this

[topaz sinew]

@molten spoke Has your question been resolved?

[molten spoke]

<@&286206848099549185>

[uncut crow]

https://math.stackexchange.com/questions/1744014/partial-derivatives-inverse-question

This should help

[molten spoke]

thanks

ok so ig that means thats wrong

this is the full question:

ive done the first and last one but idk how to do the middle one unless im missing something rlly obvious

[uncut crow]

I would multiply through by $V^2$ and then expand out the lhs

[thorny flame]

Xenophon

[uncut crow]

Take the partial wrt to T of both sides and if I remember correctly, things work out

Checking, I got about half way through and I am 90% sure you can solve for the middle partial

gotta go though, gl

[molten spoke]

thanks ill try it now

[topaz sinew]

@molten spoke Has your question been resolved?

[molten spoke]

wait so when doing it w partials on both sides is it similar to implicit differentiation

[topaz sinew]

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[molten spoke]

.reopen

[topaz sinew]

✅

[topaz sinew]

@molten spoke Has your question been resolved?

[uncut crow]

yes

[molten spoke]

ok

[molten spoke]

yh im still stuck

its cuz i got some rlly complicated fraction which seems wrong

[uncut crow]

How do you know it's wrong?

[molten spoke]

its cuz i dont know how it would ever cancel out with the other two

it might be that i got like one term wrong im just gonna check over it again

[uncut crow]

What fraction did you get?

This video covers many examples of this sort of thing

https://www.youtube.com/watch?v=nIydwSyhWXo

[molten spoke]

ive been working on it ive managed to get to this $$\frac{2a(V-b)^2 - V^3 RT}{2a(V-b)^2 + V^3 RT - 2aV(V-b)}$$ but it should all cancel to -1

[thorny flame]

I can't believe you've done this

[molten spoke]

oh my fucking god ive just realised smthn

i wrote the question down with P-a/v^2 and not P+a/v^2

smthings actually wrong w me

finally i got -1 thanks sm for the help

.close

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[pure coral]

I got the right side figured out, but the left side is causing some issues

[tough cargo]

ur close

[pure coral]

Should I take the derivative of the first set of parenthesis? Do I Do chain rule again, because it has an exponent?

[tough cargo]

yes and yes need to do chain rule again somewhere

[pure coral]

This doesn't look right

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@pure coral Has your question been resolved?

[pure coral]

Is this right

.close

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[broken flint]

Have I displayed this correctly ?

[broken flint]

So I am trying to see if that is what they are asking about as far as the “hint” goes

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[next vortex]

Whats the answer for this question

<@&286206848099549185>

This is what I gotten but I believe I’m wrong

[next vortex]

.

[fossil coyote]

Ok, a couple things

[next vortex]

?

[fossil coyote]

1. Don’t ping helpers right when you ask the question
2. Don’t assume coordinates in space unless the line is concurrent to the gridlines
3. (0,10) isn’t even on line A

[next vortex]

okay

[fossil coyote]

Let’s start with line B since you marked (0,10) and (4,0) already

[next vortex]

how do I solve A since the only thing I can mark in line A is 2.2 ,4.5 since that’s where it intersects

[fossil coyote]

Find lattice points (points where the x- and y-coordinates are integers) that are on line A

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[icy pilot]

why is

[icy pilot]

log3(0) undefined

[restive inlet]

$3^{\what} = 0$

[thorny flame]

ℝam()n()v

[distant wren]

There is no exponential number that can get your output to be zero. Your number will always be extremely small but never zero. It’s like cutting a chocolate bar in half 10000 times. There will always be something left

[icy pilot]

I thought undefined was dividing by zero

so you can get undefined thru multiple ways

what kind of rule do they use for taking it dividing by zero and make it multiplying by 1/2

and second question, using change of base formula, how does the 1/2 translate into 2 in the bottom?

[molten raven]

x*1/2=x/2

[icy pilot]

why can't you just divide by 1-2Ln5

[topaz sinew]

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@icy pilot Has your question been resolved?

[topaz sinew]

@icy pilot Has your question been resolved?

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[thorny flame]

QuasiStar 超新星

QuasiStar 超新星

[long stirrup]

no

p^4

sure

[craggy haven]

use q for another prime, p^2 p looks strange lol
but yes

[long stirrup]

eyah

i think so, you just find what feels smallest

[craggy haven]

seems right

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[gray basin]

How do I apply de Movire’s on this?

[drifting swift]

write (1+i)/sqrt(3) in polar form first

[gray basin]

ok

ok did it

[icy sky]

Are you going to show us?

[gray basin]

yes

[icy sky]

It might be more helpful to put it in e^(i theta) form

[gray basin]

ok will do that

so

is that fine?

[icy sky]

Sure

[cinder bloom]

correct

[icy sky]

Now raise that to the power 12

[gray basin]

ok

is this ok?

[icy sky]

Sure

Make sure you simplify all the way

[gray basin]

ok cool

thanks

[icy sky]

Youre welcome!

[gray basin]

.close

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[shut violet]

@vernal vale

[vernal vale]

yo

[shut violet]

Continuing the question from last channel

So basically I’ve found the points of intersection

Hold on I’ll send my working

[vernal vale]

okay

we can use one more property idk if its gonna make it that much easier

$\int _{-a}^a = 2\int _0 ^a$ here

[thorny flame]

jan Niku

[vernal vale]

because its symmetric

[shut violet]

Yeah you are right but let’s just do it this way

[vernal vale]

okay

so you have the integrand $$9-x^2-(kx^2-x^2)$$

[thorny flame]

jan Niku

[vernal vale]

which is $$9-kx^2$$

[thorny flame]

jan Niku

[vernal vale]

antideriv $$9x-\frac{kx^3}{3}$$

[thorny flame]

jan Niku

[shut violet]

I’ve already integrated the whole thing

I’m just stuck on the last part

[vernal vale]

im catching up lol

[shut violet]

Oh ok all good, lol

[vernal vale]

oh man this just sucks with both bounds

can we do the 2 thing?

itll greatly simplify your life

[shut violet]

I’d rather not

If you don’t mind

[vernal vale]

bleh alright

simplify both terms at once then

$9 \qty( \frac{\pm 3}{\sqrt k }) - \frac k3 \qty( \frac{\pm 3}{\sqrt k} )^3$

[thorny flame]

jan Niku

[vernal vale]

this is $\frac{\pm 27}{\sqrt k} - \frac{\pm 9}{\sqrt k}$

[thorny flame]

jan Niku

[vernal vale]

or $\frac{\pm 27 \mp 9}{\sqrt k}$

[thorny flame]

jan Niku

[shut violet]

Wait hold on

[vernal vale]

which is $\frac{\pm 18}{\sqrt k}$

[thorny flame]

jan Niku

[vernal vale]

which step?

[shut violet]

This part

$$\left[9\left(\frac{3}{\sqrt{k}}\right)-\frac{k\left(\frac{3}{\sqrt{k}}\right)}{3}^3\right]-\left[9\left(-\frac{3}{\sqrt{k}}\right)-\frac{k\left(-\frac{3}{\sqrt{k}}\right)}{3}^3\right]=24$$

[thorny flame]

Lex1729

[vernal vale]

right, im simplifying the upper and lower bounds

so your left bracket expression is 18/sqrt(k)

and your right is -18/sqrt(k)

[shut violet]

Would the left hand side simplify to this?

$$\left[\frac{27}{\sqrt{k}}-\frac{9k}{k\sqrt{k}}\right]$$

[thorny flame]

Lex1729

[vernal vale]

the entire left hand side?

[shut violet]

Yeah

[vernal vale]

no

for one its not fully simplified but also youre missing a factor of 2

[shut violet]

No I mean just the left hand side

the left bracket

[vernal vale]

the left hand side of the equation?

oh left bracket

[shut violet]

not the full side of the left just the left bracket

[vernal vale]

yes

but you should fully simplify it

this is 18/sqrt(k)

simplify the second fraction

[shut violet]

But I can't simplify this any further

[vernal vale]

you can

[shut violet]

How does it become 18?

[vernal vale]

$\frac{9k}{k\sqrt k} = \frac kk \frac{9}{\sqrt k}$ right?

[thorny flame]

jan Niku

[shut violet]

Oh I see now

[vernal vale]

yea

now your denominators match

and the whole thing just combines super easy

[shut violet]

Okay so now we have this right?

$$\left[\frac{18k}{k\sqrt{k}}\right]-\left[-\frac{27}{\sqrt{k}}-\frac{-9k}{k\sqrt{k}}\right]=24$$

[thorny flame]

Lex1729

[vernal vale]

you didnt fully simplify

$\frac{18k}{k\sqrt k} = \frac{18}{\sqrt k}$

[thorny flame]

jan Niku

[shut violet]

$$\frac{18k}{k\sqrt{k}}+\frac{27k}{k\sqrt{k}}+\frac{9k}{k\sqrt{k}}=24$$

[thorny flame]

Lex1729

[vernal vale]

okay

sure

[shut violet]

$$\frac{54}{k\sqrt{k}}=24$$

[thorny flame]

Lex1729

[shut violet]

So is everything correct until here?

[vernal vale]

no

[shut violet]

Where did I go wrong?

[vernal vale]

do this more carefully

you have powers of k on the top and bottom

so simplify these fractions

[shut violet]

can I cancel out the k in the numerator and denominator?

[vernal vale]

of course

[thorny flame]

Lex1729
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[shut violet]

$$\frac{54}{\sqrt{k}}=24$$

[thorny flame]

Lex1729

[shut violet]

How is this?

[vernal vale]

should be able to get a k here

wait a second

something bad happened

[shut violet]

What do you mean?

[vernal vale]

i think there is a sign error here

[vernal vale]

you lost a minus

[vernal vale]

this work says that the contribution from each end should be 18/sqrt(k)

but youre getting 36/sqrt(k) from your lower bound

[shut violet]

How did I loose a minus?

[vernal vale]

right

[shut violet]

that right part becomes a positive right?

[vernal vale]

look at the right set of brackets

those terms have different sign

one is -

one is - -

[vernal vale]

here theyve suddenly become both positive

[shut violet]

yeah because they both have two negative signs?

[vernal vale]

they do not

distribute the - across the brackets

then the first term has 2 and the second term has 3

so the first term ends up positive but the second is negative

[shut violet]

Oh, I didn't know I had to distribute the - across the brackets

do I always have to do that?

[vernal vale]

yes absolutely

distribution always goes across brackets

[shut violet]

So for example if I had this:

$$\left(5x^2+3x-9\right)-\left(9x^2+3x+5\right)$$

[thorny flame]

Lex1729

[shut violet]

it would become this?

$$5x^2+3x-9+9x^2-3x-5$$

[thorny flame]

Lex1729

[shut violet]

wait hold on

If I had this:

$$\left(5x^2+3x-9\right)-\left(-9x^2+3x+5\right)$$

[thorny flame]

Lex1729

[shut violet]

Would it become this?

[vernal vale]

yea

[shut violet]

Oh ok

But I mean what if there are no brackets

[vernal vale]

well this is an integration problem right

so there are going to be brackets

[shut violet]

Are there any exceptions where I don't need to distribute the negative sign?

[vernal vale]

the distributive property is just fundamental

thats all this is

[vernal vale]

i can't think of anything off the top of my head

[shut violet]

$$\frac{36}{\sqrt{k}}=24$$

[thorny flame]

Lex1729

[vernal vale]

yes

[shut violet]

Okay finally

[vernal vale]

you wanna know whats really funny

[shut violet]

For the moment we have all been waiting for

[vernal vale]

well

[shut violet]

k = 2.25

[vernal vale]

i wont spoil it

[vernal vale]

[shut violet]

?

[vernal vale]

https://www.desmos.com/calculator/mz5erlz7gt

the bounds actually ARE -2 and 2

its unclear to me if the problem intended for you to take this as true

i mean if you could take it as true, then you could solve the problem with algebra

and the instructions tell you that you must use calculus

so i'm guessing its just a funny coincidence

[shut violet]

😩

Well at least I got the answer in the end

Thanks so much for your help btw 👍

[vernal vale]

thats a challenging problem

glad u got it

[shut violet]

Yeah, it is. Only thanks to the glorious Jan Niku

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[final lance]

Why do you not do continuity correction when you are approximating with the central limit theorem

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[final lance]

Why do you not do continuity correction when you are approximating with the central limit theorem

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[final lance]

<@&286206848099549185>

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[final lance]

<@&286206848099549185>

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[elfin thicket]

is this allowed?

(imaginary/complex numbers)

prove that the left hand side = the right hand side

[elfin thicket]

basically can i split tthe complex fraction on the left in the way i did

[mint crescent]

Seems fine

[elfin thicket]

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[sand moth]

how do i find the value of "a" if the given is only foci and conjugate axis

[sand moth]

foci is (-3,-1) (7, -1) CA:6

i found the "b" because i divide it by 2

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[cobalt finch]

63. [5 marks]
    (a) Find the set of values of k for which the following system of equations has
    no solution.
    x + 2y − 3z = k
    3x + y + 2z = 4
    5x + 7z = 5

[frosty axle]

using the third relation, try to get two equations

(remove z)

[cobalt finch]

wdym

[acoustic tangle]

Try finding out what x + 2y - 3z is equal to only using the last two equations

And simply set k to be some number different from that so that there is a contradiction disallowing the system to have any solutions

[cobalt finch]

ive got to 22x +14y=7k15 and 11x+7y=13

thus they r parrelel right

so i say taht

k cannot equal to whatever makes them parrele

so two of the planes meet

and the other plane is parrelel to one of them

@acoustic tangle sry for ping

[acoustic tangle]

I said using the last two equations, where did you get k from?

[cobalt finch]

the first equation...

[acoustic tangle]

Anyway there's an easier way

[cobalt finch]

which is

[acoustic tangle]

Take the second equation, multiply it by 2, and subtract the third equation from the result

What do you get?

[cobalt finch]

first equation?

[acoustic tangle]

You should get that x + 2y - 3z = 3

So simply pick k to be any number other than 3 and that's it

[cobalt finch]

ohh k

btw @acoustic tangle sry for ping

is there a consistent way to find taht

multiplyinf

or justby the eye

[acoustic tangle]

I guess you could do substitution

[cobalt finch]

for lets say that there is no solution

how would you change it for it be the easiest

[acoustic tangle]

Wdym?

[cobalt finch]

like in this

what would the easiest way to solve it be

[clear lodge]

rank of the matrix and augmented matrix

[cobalt finch]

liek gaussian

[acoustic tangle]

Oh I was genuinely puzzled until I realised it says that there's no unique solution rahter than no solutiuon at all

[cobalt finch]

naa i got it

now

add the two last

and compare the fact that they are multiples

but what about this

what does it mean by consisten

[clear lodge]

consistent means it has solution(s)

[cobalt finch]

k

and general solution would be whattho

[clear lodge]

it can be determined, when the solution is unique, and indetermined, when it's not unique

[acoustic tangle]

Express the form that the solutions have

[clear lodge]

or in other words: solve it for every lambda

[acoustic tangle]

I think part ii is referring to the answer of part i rather than every lambda though

[clear lodge]

i mean, i havent solved it.
But i'd read this as:
a) it's never determinate. Prove it.
bi) It's consistent for some lambda. Prove it.
bii) For every lambda where it's consistent, solve it

[acoustic tangle]

Yeah

[clear lodge]

actually doing it, there's only one value of lambda for which it is consistent

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[distant wren]

Hi

[tough cargo]

Always try proving. Even if it's false tryna prove it will give u an idea of why

Gl I believe in u

[distant wren]

Should I start with:

B= Ax (by definition of divides)

C = ay (by definition of divides)

Aww ty

[tough cargo]

Por que c = by?

Oh nvm

[wary tulip]

maybe looking for a counterexample will give an idea why it’s true

[distant wren]

Meant a. Sorry

[tough cargo]

Could work that way too tbh

[distant wren]

And then a|b^2+ c would turn into
(Ax^2) + ay?

[odd jacinth]

be careful

[tough cargo]

close

[distant wren]

Did I miss a step?

Ok I did lol

[odd jacinth]

if b = ax
b² = ?

[distant wren]

A^2x^2?

[odd jacinth]

yeah

[distant wren]

A| a^2 * x^2 + ay?

[odd jacinth]

indeed, it's that obvious, but it's better if you factor again and justify why

[distant wren]

How would I factor that?

[odd jacinth]

by what would you want to factor to show that a divides it ?

I give you kn+k, with k, n integers

how would you show that k | kn+k ?

[distant wren]

By dividing the right side by K?

[odd jacinth]

by factoring by k

kn+k = k(n+1)

where k and n+1 are integers

[distant wren]

Ohhhh

[odd jacinth]

that comes back to your def of divisibility

a number n is divisible by m if you can write it like a product mq with q integer

[distant wren]

But how do I factor a^2 * x^2 + ay with the x there

[odd jacinth]

back to a | a^2 * x^2 + ay

how would you show a²x² + ay is divisible by a ?

[distant wren]

Give me a moment to reread your past messages lol

By factoring. But the x^2 doesn’t have an a to factor out of

[odd jacinth]

it's not by something in x that you would want to factor

you want to conclude it's divisible by a

[distant wren]

By multiplying? I’m sorry but I really don’t know

By the closer of integers under multiplication

[odd jacinth]

you want to factor by a

[distant wren]

Just want to say thank u for helping be btw

[odd jacinth]

a²x²+ay = a(ax²+y)

[distant wren]

Omg 😱

That makes so much sense

[odd jacinth]

do you see why a divides a²x²+ay ?

[distant wren]

Yes

[odd jacinth]

ax²+y is an integer

so a²x²+ay can be written as a * an integer

it's divisible by a

[distant wren]

I need to practice my algebra

Thank you. I know how to go from here. I screenshoted stuff to make notes

.solved

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[opaque burrow]

How can I find $m(\angle QMO)$?

[thorny flame]

Monkagoras

[opaque burrow]

without using trigonometry

[fluid belfry]

can we say that angle OMP and angle OPM are equal?

@opaque burrow

[karmic dove]

did you find <Q?

[opaque burrow]

No

Oh, sorry

yes you can

[fluid belfry]

why not, you have MO=OP

[opaque burrow]

because it is an isosceles triangle

[fluid belfry]

yup, now mark those 2 angles as 'x'

[opaque burrow]

ok

[fluid belfry]

no actually angle MOP is 90 degree, so those angles must be 45 deg, right?

[opaque burrow]

yes

[fluid belfry]

ok, now lets go to triangle MPQ

same logic here

what 2 angles would be equal here?

[opaque burrow]

Oh the angle of M and Q must be equal

[fluid belfry]

mhm, and you know the angle sum property i believe

[opaque burrow]

yes, got it, ty

[fluid belfry]

🍷

[opaque burrow]

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[blazing crescent]

if they're a multiple of each other

does that mean that 2(12i+5j) is just twice the size?

or does it translate

twice the size meaning the length x2

[fluid belfry]

,w \sqrt(24^2+10^2)

[thorny flame]

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[fluid belfry]

,w 2(\sqrt(12^2+5^2))

[blazing crescent]

hmm?

[thorny flame]

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[fluid belfry]

do you get your answer?

[blazing crescent]

ok yh

thanks

[fluid belfry]

np

[blazing crescent]

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[spice flicker]

Whats the trick to figuring out F(x)=a^x

[spice flicker]

I dont understand how to find 3

Oh wait the answer was for a different question

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[simple orchid]

Hello

[simple orchid]

How do I find the domain for this function?

I think it has a restriction being greater than 0

[acoustic pecan]

there are two restrictions i can see

[acoustic pecan]

whats restricted to >0

[simple orchid]

The squared root should be greater or equal to zero

And the ln function has to be greater than 0

[acoustic pecan]

are you referring to the arguments of the respective functions?

or the functions themselves

[simple orchid]

According to the multiplications of domain, which is A ∩ B, it should be greater than 0?

[simple orchid]

Together combined

[acoustic pecan]

what if x was 1.5

[simple orchid]

Ohhh

I have to factor the inside of the root first

I think?

[acoustic pecan]

you need the arguement of the sqrt to be >=0
and the arguement of the ln to be >0

[simple orchid]

Yes, that was what I was trying to say

[acoustic pecan]

aha, okay, start with the sqrt

[simple orchid]

Do I first factor the inside expression?

[acoustic pecan]

it would probably help yeah

[simple orchid]

$\ln\left(\sqrt{\left(x-1\right)\left(x-2\right)}-x\right)$

[thorny flame]

LE SSERAFIM

[simple orchid]

x ≠ 1 and 2

[acoustic pecan]

x can be 1 or 2 with no issue

[simple orchid]

Oh

[acoustic pecan]

remember, you need the stuff inside the sqrt to be >=0

so you need to exclude whichever x values make it <0

[simple orchid]

So you mean, x should be greater than 1 and 2?

[acoustic pecan]

youre saying just x>=2?

[simple orchid]

Yes

[acoustic pecan]

no

[simple orchid]

I'm confused..

Hint please

[acoustic pecan]

imagine a parabola that dips below the axis and has roots at 1 and 2, where is it negative?

parabola is a U shape in this case, going up

[simple orchid]

x ∈ (1, 2)

[acoustic pecan]

yeah, so that set cant be in our domain

so so far we have x<=1 or x>=2

[simple orchid]

Okay

[acoustic pecan]

now you need to look at the ln argument

[simple orchid]

ln greater than 0

[acoustic pecan]

the arguement is greater than 0

ln can be negative

what inequality could you write out in this case?

[fluid belfry]

can't*

[acoustic pecan]

it can

,calc ln(0.2)

[thorny flame]

The following error occured while calculating:
Error: Undefined function ln

[simple orchid]

Uhhh

[acoustic pecan]

,w graph ln(x)

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by aldrnari_

[acoustic pecan]

anyway

[acoustic pecan]

back to this

[simple orchid]

What do you mean by argument greater than 0?

[acoustic pecan]

i mean the stuff inside the ln is greater than 0

the arguement is the input of a function

[fluid belfry]

my teacher told me it cant be (sorry to interupt)

[acoustic pecan]

they likely meant the value you put into ln cant be negative

ie the x in ln(x) must be st x>0

[simple orchid]

I guess it can have negative values

[acoustic pecan]

im not referring to the ln itself being negative, im talking about the arguement of it, the stuff in the brackets

[simple orchid]

Then x > 0?

[acoustic pecan]

no

this isnt ln(x)

[acoustic pecan]

all the stuff in that bracket must be>0

$\sqrt{(x-1)(x-2)}-x>0$

[thorny flame]

AℤØ

[acoustic pecan]

is our restriction

[simple orchid]

Ohhh

Yeah

I see what you mean

[acoustic pecan]

any ideas what you can do with that now?

[simple orchid]

I'm not sure..

The graph shows it's less than 1

[acoustic pecan]

we'll get to that

you need to rearrange though

[simple orchid]

What do we do after?

[acoustic pecan]

move the x to the other side

[simple orchid]

$\sqrt{\left(x-1\right)\left(x-2\right)}<x$

Like this?

[thorny flame]

LE SSERAFIM

[acoustic pecan]

your signs the wrong way around

what do you think you should do after that

[simple orchid]

Square both sides?

[acoustic pecan]

sounds good to me

[simple orchid]

$\left(x-1\right)\left(x-2\right)>x^{2}$

[thorny flame]

LE SSERAFIM

[acoustic pecan]

expand the lhs

[simple orchid]

$-3x+2>0$

[thorny flame]

LE SSERAFIM

[acoustic pecan]

yup

what does that tell you

[simple orchid]

x<\frac{2}{3}

Right?

[acoustic pecan]

so we had (x<1 OR x>2) AND (x<2/3)

so what is our domain for both to hold

[simple orchid]

x < 2/3

[acoustic pecan]

thats what we have

domain found

[simple orchid]

Ohhhh

That was cool

Thank you for the help

[acoustic pecan]

np

[simple orchid]

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[slate sparrow]

HELP ME

[slate sparrow]

can u guys just asnser this

question for me

?

[acoustic pecan]

nope

[fluid belfry]

no

[slate sparrow]

i cant do irt

[woeful drift]

Nada

[acoustic pecan]

we can help you do it yourself

but we wont do it for you

[slate sparrow]

ok

[acoustic pecan]

have you tried anything?

[slate sparrow]

yeah

friends

[acoustic pecan]

have you actually tried to do the question yourself

[slate sparrow]

ngl its over

.close

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[acoustic pecan]

[slate sparrow]

me fr

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[chrome hedge]

Currently a grade 9 someone help me with this question discussed in class 😭😭😭 This is the question

The roots of the equation 3x^2-2kx+k+4=0 are alpha and beta. If alpha^3+beta^3=16/9. Find the possible values of k

[chrome hedge]

I tried to solve it but can’t

[fluid belfry]

thats a pretty good question for 9th graders

[chrome hedge]

It is about quadratic equations

But my small tiny brain can’t solve it😢

!status

[topaz sinew]

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

[chrome hedge]

2

<@&286206848099549185> can someone tell me how I should solve it or solve it for me.. I really tried hard but can’t get the answer

I don’t get how to link the alpha beta lesson to quadratic equations

[sly axle]

Use (a+b) ^3 formulae

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

[chrome hedge]

Can you check my answer

[sly axle]

One min

I am solving I will send pic

[chrome hedge]

Thanks you so much

[sly axle]

solve for k now by solving the equation

check my calculations once but the procedure is correct

[chrome hedge]

Is it solvable without that calcu since I didnt learn yet

Thank you so much

You saved my day

[sly axle]

your welcome

And also check if I have done some errors like addition subtraction but this is the procedure

Solve the last equation

Lemme sum up
First step is writing down what is addition and multiplication of zeroes of quadratic equation second step using the (a+b) ^3 and substituting values

Pic taking time to load

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[mint perch]

What exactly is the form for the particular solution?

The solution for the left side is Acos(2x)+Bsin(2x)

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[cinder owl]

how do u do this

[cinder owl]

im so confused

[topaz sinew]

@cinder owl Has your question been resolved?

[cinder owl]

<@&286206848099549185>

[odd jacinth]

what's H ?

[cinder owl]

parity-check matrix of the (7,4) Hamming code

[topaz sinew]

@cinder owl Has your question been resolved?

[odd jacinth]

I don't know anything about this topic so I went to check wikipedia and quickly read, so I may say something dumb:
I think that if C1 and C2 are both codewords, it's supposed to mean H(C1) = 0 and H(C2) = 0
But H is linear so H(C1-C2) = 0
However, if their distance is 1, as the hint says, C1 = C2+ei
so H(C1-C2) = 0 = H(ei)
Which, from my meager understanding of the wikipedia page, should be false

since ei shouldn't pass the check

@cinder owl

[topaz sinew]

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[humble badge]

ive started it

i need to make a chart

[topaz sinew]

@humble badge Has your question been resolved?

[humble badge]

<@&286206848099549185>

[topaz sinew]

@humble badge Has your question been resolved?

[humble badge]

<@&286206848099549185>

im going to go eat food

its not like anybody is coming anyways 😭

[humble badge]

<@&286206848099549185>

[heavy knoll]

yes

[heavy knoll]

Whats the question?

[humble badge]

hi

so it wants me to make a chart

[heavy knoll]

Do you need to find vf, vi?

[humble badge]

of what we know

[heavy knoll]

ok

[humble badge]

and what we dont

[heavy knoll]

Yes

[humble badge]

so what i got so far is

[heavy knoll]

Ok so, do you what vf, vi, and a means?

[humble badge]

final velocity

initial

[heavy knoll]

correct

[humble badge]

accel

i know that

accel for vertical is -9.8]

[heavy knoll]

k

yes

[humble badge]

horizontal is 0

[heavy knoll]

Ok what does the question want from you

[humble badge]

its just asking me what i know and waht i dont know

so theres a chart

VI=?

a=?

like that

[heavy knoll]

Ohhh ok

lets see now

hold on

Ok, so if the rock is gonna be pushed from the start, whats it starting velocity gonna be?

If it is on standby, until it gets pushed

[humble badge]

0

for vertical

[heavy knoll]

Yes, the Vi will be 0

Ok now lets see vf

[humble badge]

ok

for horizontal or vertical?

[heavy knoll]

Both

I don't think we know vf, but we do know the acceleration

[humble badge]

do i need a formula for that

accel for vertical is 9.8

then 0 for horizontal

is delta x for vertical 45?

[heavy knoll]

it is moving at 7 m/s

[humble badge]

yea

[heavy knoll]

7.5

[humble badge]

that means horizontal initial

[heavy knoll]

Yes

[humble badge]

does it also mean horizontal final velocity

idk if its constant or not

[heavy knoll]

Well as you can see it says the caveman is 100 m above, so it is vertical

[humble badge]

but the end result is at the other cave

which is at 55 above

[heavy knoll]

Ok

So it is moving at a costanst speed

[humble badge]

how do you know

[heavy knoll]

But, it the accelartion started from 0, so that was not constant

does that help you

[humble badge]

ohh

that makes sense

[heavy knoll]

yes, also one more thing

i am not a helper i was capping the entire time, but i guess what i said was true though

[humble badge]

oh lol

[heavy knoll]

so there you have it

[humble badge]

ty still

[heavy knoll]

i am doing physics rn so thats my current knowledge

[humble badge]

lol ive been waiting for like 2 hrs

[humble badge]

fr?

[heavy knoll]

Yes

[humble badge]

what grade

[heavy knoll]

10th

[humble badge]

nice

[heavy knoll]

what grade is this work?

[humble badge]

well

im in 9th

but there are seniors and people higher in my class

so idrk

[heavy knoll]

Oh understandable

This seems advanced

[humble badge]

its AP

[heavy knoll]

Goodluck however

[humble badge]

wait

[heavy knoll]

Oh ok

[humble badge]

theres 2 more 😭

[heavy knoll]

Send a photo

[humble badge]

would the delta X be ?

k

[heavy knoll]

I might be able to help, i did a bit of IB physics

Have you sent it yet

[humble badge]

sending

[heavy knoll]

ok

[humble badge]

i put ? for delta x horizontal

and im guessing ? for Vf vertical?

[heavy knoll]

What does delta x mean in this problem

[humble badge]

like

the distance

so for height

it was 45

cause the caveman was 100 m above

[heavy knoll]

Oh ok hold on there is a formula for height

[humble badge]

and the end position was 55

[humble badge]

wanna see the options

[heavy knoll]

Yes

I'm trying to figure it out..

[humble badge]

the options are

0

100m

?

55m

45m

btw "?" is a response

[heavy knoll]

just wait give me a moment

[humble badge]

ok

[heavy knoll]

$h=
2g
(v
f
2
​
−v
i
2
​
)
​$

[thorny flame]

ynwsk

[heavy knoll]

This is the formula

[humble badge]

is g acceloration

[heavy knoll]

wait

$v^2f-v^2i/2g$

[thorny flame]

ynwsk

[heavy knoll]

its not divide, it is over 2g

and g is acceleration

[humble badge]

whats the difference

[heavy knoll]

in what?

[humble badge]

wdym

[heavy knoll]

Like look how it says /2g right, the dash is over 2g

so ya my fault ur right no difference

its divided by 2g

[humble badge]

ok

so then

[heavy knoll]

Ok plug it all in

[humble badge]

i dont get it

[heavy knoll]

dont get what

[humble badge]

a= 0

you cant divide something by 0

wouldnt it be undefined?

[heavy knoll]

wait just wait

[humble badge]

ok

[heavy knoll]

you are trying to find horizontal or vertical?

[humble badge]

horizontal

[heavy knoll]

Ok so if you do 7.5 squared minus 7.5 squared divided by 2 times 0 you get 56.25

I think that is the right answer

[humble badge]

theres 5 options to pick from

[humble badge]

isnt it divided by 0

cause 2*0

[heavy knoll]

Oh yes, my apologies

[humble badge]

but then

[heavy knoll]

Since you are dividing by 0 in the denominator, this implies that the object is at a constant height and does not change its position. In other words, the height is constant when acceleration is 0, and the initial and final velocities are the same.

[humble badge]

you cant do -7.5/0

[heavy knoll]

so the answer would be "?" i guess

[humble badge]

ok

how abt the Vf for vertical

would that too be "?"

[heavy knoll]

ok wait a moment

vf​=√vi2​+2aΔx​

$vf​=√vi2​+2aΔx​$

[thorny flame]

ynwsk

[heavy knoll]

it is the square root of all of the by the way

[humble badge]

ok

i was just abt to ask that lol

[heavy knoll]

so now plug in the value

[humble badge]

ok

2*(9.8)(45)

square that

[heavy knoll]

so what do you get

[humble badge]

29.6984848098

[heavy knoll]

Is that a option?

in your choices

It should be 29.7

so you did it correctly

@humble badge