help-13

[crimson sedge]

fill in the blank

[gray oasis]

more than 1?

[crimson sedge]

yeah

there ya go

[gray oasis]

omg 😭

[crimson sedge]

so the intersection of both intervals is x >= 1

so for the other one

[
-x \ge 1, ; x < 0
]

[crimson sedge]

what is the interesection for this

oh actually

before that

lets simplify this

,align
-x &> 1 \
x &\overset ?{>} {???}

[crimson sedge]

what is ??? and is the inequality sign correct in the second step

[gray oasis]

??? = 1
? = yes

hehe

[crimson sedge]

wrong to both

[gray oasis]

oh no

whats the answer?

[crimson sedge]

ao like multiply both sides by -

what is 1 multiplied by -

[gray oasis]

is this for the sqrt(1-x)?

[crimson sedge]

no

we will get to that

we are working on sqrt(x^2 -1)

[gray oasis]

-1

[crimson sedge]

we are trying to find the domain for that

[crimson sedge]

ok cool

[gray oasis]

ah okay i see

[crimson sedge]

so in an inequality when u multiply both sides by a negative

what happens to the inequality

[gray oasis]

umm it will be flipped?

[crimson sedge]

yea

[crimson sedge]

this will be

,align
-x &> 1 \
x &\le -1

[crimson sedge]

is this fine

[gray oasis]

oh i will add equal too?

[crimson sedge]

yeah u must

[gray oasis]

okok

[crimson sedge]

so

now

$x \le -1, x < 0$

[crimson sedge]

what is the intersection of this

[gray oasis]

-1 to 0?

[crimson sedge]

no

i tell u i have lesser than 0 dollars on me, but my friend tells you that i actually owe some dude a dollar so i have lesser than -1 dollars

so i have lesser than __ dollars on me

fill in the blank

[gray oasis]

-1

[crimson sedge]

yeah

[gray oasis]

lesser than -1?

[crimson sedge]

yeah

so great!

we are almost done

so

we our x can be either

$x \ge 1$ or $x \le -1$

[crimson sedge]

thats what we got

thats our domain

[gray oasis]

for x^2 - 1?

[crimson sedge]

which you can represent in interval notation as $(-\infty, -1] \cup [1, \infty)$

[crimson sedge]

for sqrt(x^2 -1)

[gray oasis]

why is the bracket different

[crimson sedge]

also this is taking much longer than i expected and my energy levels are kind of depleting, so i will go once this question is done. we can continue this later if u wanna

[crimson sedge]

hm?

[gray oasis]

yes suree. i have to review for another subject too hehe sorry i think i drained u

[gray oasis]

parathesis and bracket

[crimson sedge]

nah its good i just havent eaten LOL

[crimson sedge]

( means it excludes, [ means it includes

so when i say (-infty, -1] i am saying that the domain includes -1

but we cant "include" infinity

so we dont put a [ there

[gray oasis]

Oh damn do u wanna go eat first. I'm not rushing with the answers here. It will serve as my review only.

[gray oasis]

okokok

Maybe we can contiue tommorow?

[crimson sedge]

its okay lets finish this one first

i will friend you

you can message me there if u want to continue

[crimson sedge]

sure

[gray oasis]

friend in discord?

reallly? we can do that?😱

[crimson sedge]

LOL yes

anyways i guess i will go now hit me up once u r free with ur other stuff

[gray oasis]

thank you so much for being patient

[crimson sedge]

no worries

[gray oasis]

I know I'm hard to teach especially with math LOL

should i close this?

[crimson sedge]

i have probably helped thousands of times by now in this server and honestly its just algebra is always the hardest to teach

like its the "easiest" if u r looking at math in terms of difficulty

but darn man teaching it is so so hard

because you need to get into the mind of someone who is not experienced with like doing math

which is hard

[crimson sedge]

but yeah

you should

[gray oasis]

Do u like teaching?

[crimson sedge]

yes

it helps me understand as well

[gray oasis]

r u an actual teacher?

college prof?

[crimson sedge]

no i am not that old 💀

im 20 im a student in college

[gray oasis]

DEYM 😭

I'm so embarassed

[crimson sedge]

lmao i can say the same about what i feel with some other people here. some people on this server are like 14 doing graduate level work

its crazy

but anyways we can continue this else where

lets close this now haha

[gray oasis]

I'll message u in the private one, is that alright?

[crimson sedge]

yeahh

sure

[gray oasis]

Thank you again!!!

[crimson sedge]

nww

[gray oasis]

how do i close it? HAHAHA

[crimson sedge]

.close

[cedar kiln]

Channel closed

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[crimson sedge]

there, did it for you lmaoo

[gray oasis]

.close

[cedar kiln]

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[fervent jackal]

I dont really understand this, can any1 give me a hint thanks

[fervent jackal]

!status

[cedar kiln]

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

[fervent jackal]

1

[lyric narwhal]

do you know of the euler's formula?

$e^{i\theta}=\cos\theta+i\sin\theta$?

[wraith dagger]

kheerii

[fervent jackal]

$e^{i\pi} + 1 :=: 0$

[wraith dagger]

googol

[fervent jackal]

yea why

I don't really understand the question

[lyric narwhal]

well

the question asks you to find all complex numbers z such that e^z is real

[fervent jackal]

yeah I don't understand what that means

[lyric narwhal]

i don't think the question can be any more straightforward

[fervent jackal]

my progress is

$e^{z} :=: e^{x}e^{iy}$

[wraith dagger]

googol

[fervent jackal]

and then I'm stuck

I don't know what to do next

[lyric narwhal]

can you expand that second term using the euler's formula?

$e^{iy}=???$

[wraith dagger]

kheerii

[fervent jackal]

$cos(y) + isin(y)$

[wraith dagger]

googol

[fervent jackal]

then??

make y=0?

or pi?

[lyric narwhal]

so you have $e^z=e^x(\cos(y)+i\sin(y))$

[wraith dagger]

kheerii

[lyric narwhal]

hmm, you are not wrong

but what are ALL the values?

[fervent jackal]

npi

[lyric narwhal]

correct

[fervent jackal]

that's all?

bruh

.close

[lyric narwhal]

since we want sin(y)=0

[cedar kiln]

Channel closed

Closed by @fervent jackal

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[lyric narwhal]

doorslam wow

okay then

[civic coral]

He’s spoiled

[fervent jackal]

doorslam is crazy

[lyric narwhal]

you realise people like me are going out of their way to help you

[fervent jackal]

bro thanks

I'm so sleepy rn cant think properly

[lyric narwhal]

all good

[cedar kiln]

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[lavish bobcat]

The figure below shows the graph of the function g, which is the transformation of the function f(x) (given below), where p is some polynomial function without zeros.
The asymptote of the function g is drawn in green. Write down the formula for g as a transformation of the function f, i.e. find the corresponding a, b, c, d for which g(x) =
af(bx + c) + d. Justify your choice of parameters well.

[lavish bobcat]

Can someone help please?

[cedar kiln]

@lavish bobcat Has your question been resolved?

[lavish bobcat]

<@&286206848099549185>

.close

[cedar kiln]

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[cedar kiln]

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[proper plaza]

Why is the limit of g(x) 2? Shouldn't it be zero?

[earnest socket]

approaching 0 from the right, not 1

[proper plaza]

Is the answer to that question 0? I'm not sure why it might be 2?

[proper plaza]

right

Okay

Looks like I'm super slow today haha

Thank you, I was looking at x = 1

.close

[cedar kiln]

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[lime gyro]

Hi! Im working on this cal 2 problem and know the answer is 128, i just dont know where i got the signs wrong, if anyone can help me it would be greatly appreciated

[distant jacinth]

You forgot the brackets here:

Ah, you already changed the sign. Never mind, it's ok.

[lime gyro]

Okay then where did it go wrong?

[distant jacinth]

I will look.

[lime gyro]

Okay 👍🏻

[distant jacinth]

What about this one:

$$\left[\frac{x^4}{4}\right]_{-4}^0 = \frac{0^4}{4} - \frac{(-4)^4}{4} = -\frac{256}{4}$$

[wraith dagger]

OneTrackPony

[lime gyro]

Okay thank you!

.close

[cedar kiln]

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[distant jacinth]

np

[cedar kiln]

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[rocky basin]

HOW DO I DO TJIS

PLEASE

IM DYING

[crimson sedge]

set up equations

calm down first of all

,rccw

[rocky basin]

I tried

To set up

Equations

Do u wana see what u got

I got

[crimson sedge]

yes

that would be helpful

[rocky basin]

Okk

Is it possible to do that

Turn just the log base 2 (x-p) = 0 into 2^0

[crimson sedge]

none of what you wrote, frankly, makes sense at all to me

[crimson sedge]

what operation are you performing here? and why equate it to 0 even?

[rocky basin]

Wait

Um

[rocky basin]

I just

Realized that

LOL

[crimson sedge]

[rocky basin]

I figured out WHAY mistake i did

But can I ask u smtjn

Abt Thai

This

[crimson sedge]

ok

go ahead

[rocky basin]

Over her

[crimson sedge]

ok

[rocky basin]

Can I turn the log into a exponential form

Even with the quadratic there

[crimson sedge]

no you cant

[rocky basin]

Like it could be 2^0 + the quadratic = x-p

No?

[crimson sedge]

no

[rocky basin]

Okk

Why

Even after subbing in a point?

[crimson sedge]

because what you are doing is like [
y= \m{\log_2}{x-p} \Iff 2^{y} =x-p
]
but it isn't $y$ in the equation anymore you are saying something completely new

[rocky basin]

Oh

So wait

If I sub in the point

Like (9,41)

Can I do that after

?

[crimson sedge]

how about, better idea instead of like what you are doing rn,

sub in the point

isolate the log on one side of the equation

THEN take 2^ of both sides

[rocky basin]

Huh

Wait

Let me look at it again I forgot

So like

Move the quadratic to the other side?

[crimson sedge]

yes

[rocky basin]

And then turn into dxponentinal form?

Exponential*

[crimson sedge]

you can just take the 2^ of both sides yes

[rocky basin]

Okay

Thanks

Wait

Can u show me

What u mean

I DONT gey ot

[crimson sedge]

ok sub in the point

what do u get as a value

[rocky basin]

Um

Log 2 ( 9-p) -81+36+q = - 41

?

And then

Log 2(9-p) -45 +q = -41

Then

[crimson sedge]

yeah

take the -45 and q to the right

[rocky basin]

Log 2(9-p) + q = 4

Ohh

Log 2(9-p) = 4-q

[crimson sedge]

yes

ta da

[rocky basin]

Then 2 ^ 4-q = 9-p

And then what ;-;

[crimson sedge]

hol up

[rocky basin]

Wat

[crimson sedge]

ok you reaaaaally need to add parentheses to what u r writing

but anyways

solve for p

[rocky basin]

Sorry haha

[crimson sedge]

in this

[rocky basin]

It’s hard on phone

Okay um

[crimson sedge]

lmao its ok

[rocky basin]

P = 9-2^(4-q)

[crimson sedge]

yuh

[rocky basin]

Then should I do it for the other one

[crimson sedge]

sounds legit

[rocky basin]

Like sub in

The other point

[crimson sedge]

yes

[rocky basin]

And then. Set equal to and solve

[crimson sedge]

simplify everything

yes

[rocky basin]

WOOHOO U SAVED ME!!

Thanks

Let me ss this so I rmeeber

[crimson sedge]

i can tell u the values of p,q later if u wanna verify ur answer

[rocky basin]

It’s okay dw

Thanks sm

[crimson sedge]

oki

[rocky basin]

.close

[cedar kiln]

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[cedar kiln]

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[burnt seal]

Hi quick question, whats another way to represent sqrt (x+1)

[burnt seal]

like how can i break it apart

i want the sqrt x to cancel out with another sqrt x in my denom

[buoyant latch]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[mighty shuttle]

sorry for that message, you're basically asking how to factorise $\sqrt{x+1}$?

[wraith dagger]

Why am. I here

[burnt seal]

yeah i want to make it 2 seperate things multiplying eachother

[mighty shuttle]

One way would be to write it as $\frac{x+1}{\sqrt{x+1}}$

[wraith dagger]

Why am. I here

[mighty shuttle]

If you have x+1 in the denominator in your OG problem, this should help

[burnt seal]

sorry, in my denom is the sqrt (x+1) im looking to cancel a sqrt(x) in my numerator.

[mighty shuttle]

to cancel a $\sqrt{x+1}$

[wraith dagger]

Why am. I here

[burnt seal]

yes

my problem is lim n--> inf of the absolute value of -sqrt(n)/sqrt(n+1)

or is there a better way to solve this limit?

im so dumb should i just use lhopital

[mighty shuttle]

so you want the limit of $\frac{\sqrt{x}}{\sqrt{x+1}}$ at infinity?

[wraith dagger]

Why am. I here

[burnt seal]

yeah

[buoyant latch]

@burnt seal can you just send the original question

[mighty shuttle]

I'd probably divide the numerator and denomenator by $\sqrt{x}$

[wraith dagger]

Why am. I here

[burnt seal]

im doing the ratio test to find the interval of convergence

i got to the point where my x is on the outside of my limit

and i managed to cancel out (-1)^n

[burnt seal]

so now all thats left is x outside the limit multiplied by this

thank u why am i here i figured it out with ur dividing by sqrt

[cedar kiln]

@burnt seal Has your question been resolved?

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[earnest marsh]

So would it become like: x(x+2) ≥ 80??

[crimson sedge]

Yep

[flint plinth]

👍

[earnest marsh]

okay thanks!

.close

[cedar kiln]

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[cedar kiln]

@dry fossil Has your question been resolved?

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[cedar kiln]

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[nocturne iris]

I followed the quadratic formula and got 6+sqrt40/2 and 6-sqrt40/2

[nocturne iris]

do these simplify any more?

[sharp pike]

$\sqrt{40}$ can be simplified

[wraith dagger]

Ari

[nocturne iris]

do you make it look like 10sqrt4?

or 2sqrt10

since the question asks for an exact answer decimals can't be used right?

and the final answer would be 3+sqrt10 and 3-1sqrt10

[sharp pike]

👍

you can just write sqrt10 don't put the 1 in front

[nocturne iris]

so in this example the answers would be 1+sqrt14 and 1-sqrt14?

[sharp pike]

ye

[nocturne iris]

(x-4)² = 9

would you start byrooting each side to get x-4 = 3

[sharp pike]

x-4 can also equal -3

be careful of that

[nocturne iris]

so this has 2 answers being 7 and 1?

also this problem is tricking me

I divided both sides by x to get x=49 but that wasn't right

[quasi hinge]

what if x is zero

[nocturne iris]

0² = 49*0

oh

[quasi hinge]

x^2 - 49x = 0

x • (x - 49) = 0

x = 0 and x = 49
@nocturne iris

[nocturne iris]

.close

[cedar kiln]

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[hollow elbow]

Anyone knows how to graph this?

[hollow elbow]

I used desmos, but I still dont get how to properly get it, especially if im on a test

??

[dull cedar]

Take the -2 as a common factor
Its point of symmetry will be (0 , -1/2)

A hyperbola

[cedar kiln]

@hollow elbow Has your question been resolved?

[hollow elbow]

Ok well

I have another issue now

How do i do f

Do i just flip it?

.close

[cedar kiln]

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[solar heron]

binomial expansion isnt my strong suit.

by the first three terms what is it asking for exactly?

[drifting marlin]

"In ascending powers of x" should be your clue as to what it's asking for

[solar heron]

so kinda like if

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

and the question asked for the second term in ascending powers of a

it would be a^2

?

[drifting marlin]

It would not

[solar heron]

I don't get it

[drifting marlin]

The powers of a would be a^0, a^1, a^2

[solar heron]

ohhhhh

[drifting marlin]

The second term would thus be that which features a^1 (or just a)

[solar heron]

genius

can i write the terms as factorials tho?

actually, sorry wrong question

is there a faster way to do binomial expansion without all the calculating factorial and ks hassle?

because this question had 3 marks and i cant spend this much time on it?

[drifting marlin]

Not really, though for the first 3 terms, at least 2 are trivial

[solar heron]

wdym?

atleast 2 are trivial?

[drifting marlin]

Do you know what nC0 is? It's the same for any n

$\binom n0$ if you prefer

[wraith dagger]

Steakanator

[solar heron]

i think its what we use as

( n k )

[drifting marlin]

Sure

But can you compute that?

[solar heron]

ig yea

[drifting marlin]

Then do so now

[solar heron]

you mean the calculator function

[drifting marlin]

I'm providing you a shortcut so you can avoid this issue

I don't think you need a calculator for this

How is (n k) defined?

[solar heron]

as

n factorial over n-k factorial

sorry

[drifting marlin]

You're missing a term

[solar heron]

n! / k! (n-k)!

[drifting marlin]

Much better

Now sub in k=0, what do you get?

[solar heron]

just n

[drifting marlin]

Not quite

[solar heron]

uh

so if n is 2

and k is 0

oh

youd get 1

so n/n

[drifting marlin]

You would get 1

(n 0) and (n n) are both 1 for all n

[solar heron]

and k refers to the term in order?

so k=1 is the 1st term?

[drifting marlin]

No

k=0 is the first term

[drifting marlin]

Recall

[solar heron]

ah yes cant make that mistakes thanks

Yes

ill try solving this

again

but i coudnt understand the markscheme

whats C?

[drifting marlin]

That's their notation for (n k)

[solar heron]

uh

wdym

[drifting marlin]

I mean what I said

[solar heron]

so for them

[drifting marlin]

5C1 and (5 1) are the same thing

[solar heron]

(n k) = C

[drifting marlin]

Not exactly

${}^nC_k = (n \ k)$

[wraith dagger]

Steakanator

[drifting marlin]

It's just a notation thing

They mean the same thing

[solar heron]

ok

so in this

why is there an "or"

there should be three terms

so 5C1 , 5C2 and 5C3

oh sorry for this again

5C0, 5C1 and 5C2

[drifting marlin]

That's a weird way of presenting it

I'm not sure what they're getting at

[solar heron]

I'm so confused by this

since the final answer is usally

ab

a product

but in this case if its a=2 b = -2/5

then should i write the minus as well?

llike a . -b?

[drifting marlin]

i think you need to recall the precise form of the binomial expansion

[solar heron]

what would be the answer

for

5!/2! 3!

1?

got it

finally

thanks a lot mate!

.close

[cedar kiln]

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[left socket]

Can someone help me with this?

[crimson obsidian]

What did you try?

[left socket]

Putting it in slope intercept formula

[crimson obsidian]

You mean y=mx+c?

[left socket]

y=mx+b

But yeah

That

[crimson obsidian]

You know y, x and m you can find b

[crimson obsidian]

What problem did you encounter

[left socket]

I got b=27/7

[crimson obsidian]

then again put it fresh in the equation y=mx+b with m's value as well, then you can manipulate the equation into what's being asked...

[slim quiver]

m = A/B

@left socket

[left socket]

Oh so then the correct answer would be A, right?

[crimson obsidian]

m=-3/7 is given though

Oh

[slim quiver]

https://sciencing.com/how-to-convert-slope-intercept-form-to-standard-form-13712257.html

b in your slope intercept form is NOT B in your standard form

b != B

case matters

[crimson obsidian]

mx+(-1)y+b=0, isn't this the form Ax+By+C=0?

Just manipulated y=mx+c

[slim quiver]

Not quite.. notice you have mx+(-1)y... and Ax + By...

I know you said "form" in general, but where is the -1 coming from?

Your coefficient for y isn't necessarily negative

[crimson obsidian]

if we have y=mx+c form in general, can't we substract y both sides and get this form?

In this form (-1) is the coefficient of y

[slim quiver]

Ah, I see what you're doing... yeah, functionally the same

[crimson obsidian]

Ohh

[slim quiver]

Just depends on which side of the equals sign you're moving things to

[crimson obsidian]

Yeah

[slim quiver]

The site I linked (and my general habit) is to move everything to the left, so mx ends up negative, and b ends up negative

You just flipped it.. so yeah... it's the same

[crimson obsidian]

I see

[slim quiver]

Any of this making sense, @left socket ?

[left socket]

Yeah I'm sorta understanding it.

[cedar kiln]

@left socket Has your question been resolved?

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[cedar kiln]

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[flint lion]

for question 1c

how do you know what x coordinate to plug in

[humble karma]

You can just trace where the outputs of h go and see if they are possible inputs of f.

For instance, h sends -4 to -1, but f doesn't send -1 to anything.

However, h sends -2 to 4 and f sends 4 to -5, so in total -2 is sent to -5

[flint lion]

so just keep plugging the coordinates?

[humble karma]

Yeah, or you can just look at the inputs of f and try to find them in the outputs of h.

[cedar kiln]

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[void sand]

what is c

[pale lake]

any idea?

im stcuk tbh

up to that part

.close

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[cedar kiln]

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[grand forge]

how r u s o funny

[livid hound]

<@&268886789983436800>

[cedar kiln]

Channel closed

Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[livid hound]

ty

[grand forge]

was gonna have fun with him :(

[livid hound]

don't feed trolls

[cedar kiln]

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[pale lake]

Can you take constants out the cross product

[humble karma]

Yes

As in scalar mutliplication

[pale lake]

So say i wanna do the determinant cross product method

And one vector I can take a constant out

And the other vector I can’t

How would I do it

Does the constant have to be the same? For both vectors

[humble karma]

You can take it out of one and compute the cross product of what's left

(ra) x b = a x (rb) = r (axb)

[pale lake]

What about if there is two constants

so one for a and one for b

[humble karma]

(ra) x (sb) = rs (axb)

[pale lake]

Ok cool

And what about the norm

Of one vector

Can u take constants out of that as well?

[humble karma]

Usually, norms are such that ||cx|| = |c| ||x||.

[pale lake]

So say I had a vector, and the constant I took out was like 1/(a^2 + b^2)

Would the constant just stay the same?

[humble karma]

Yes, you could take is out of the vector.

[pale lake]

And does this work the same way for the inner product.

[humble karma]

Yes

[pale lake]

So the two constants u multiply and then u multiply that by the inner product sum

[humble karma]

Yes

[pale lake]

Yh cool that makes sense

Thank u a lot for ur help

👍

.close

[cedar kiln]

Channel closed

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[pale lake]

.reopen

[cedar kiln]

✅

[pale lake]

@humble karma sorry one last question

Say I’m doing the cross product

Does the ‘constant I take out’ does it matter what letter it is

So ie

Say I have a vector with respect to x

And a vector with respect to y

And I factor out something to do with x in the first vector

Can that still be classed as a constant?

[humble karma]

It doesn't matter.

[pale lake]

Ok cool

[humble karma]

It's still a constant

You could take a constant in y in the first and a constant in x in the second

[pale lake]

Yh cool that makes sense

Thanks again

.close

[cedar kiln]

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[restive stratus]

what is 5x^2 + 5x + 2 = 0?

[foggy merlin]

an equation of degree 2

or a quadratic if you want

[restive stratus]

is it

[sharp lotus]

yes

[restive stratus]

(x+1)5x + 2 = 0

then we divied the whole thing by 5x

[umbral blaze]

Isn't that a quadratic?

[sharp lotus]

yes, that's still the same equation rewritten slightly but it doesn't really help you solve it since it doesn't do anything to separate the variables

[restive stratus]

so x+1 +2/5x = 0

x+1 = 2/5x

i think im off track here

x+1 = (2/5)x

[humble karma]

This won't help you if you want to solve for x.

[sharp lotus]

but it's still good to play around with things yourself and see why not

[umbral blaze]

Okay I have a question too I am taking college algebra and this is a super basic question but I still would want to know in an inequality when graphing, I forgot when is the parenthesis a open circle, or a closed circle, or a bracket and a parenthesis? When can you tell when to use each!!

[restive stratus]

what

is parenthesis?

[junior cape]

this channel is already taken by someone else. please go to one of the channels in MATH HELP (AVAILABLE). those are...available

[umbral blaze]

Oh is it okay, no problem I appreciate you telling me that, thank you.

It was just out of curiosity, but I get it.

[junior cape]

are you familiar with the quadratic formula?

[restive stratus]

no

i just use the calculator

or chatgpt

[junior cape]

ah, ok. it's good to learn the quadratic formula, and why it works

[restive stratus]

i seen some on the internate

but it dont make sense

[junior cape]

it's okay, i can help walk you through it

see the a, b, and c in that?

[restive stratus]

yes

[junior cape]

ok. do you remember what a coefficient is?

[restive stratus]

the thing infront of x

[junior cape]

excellent

so the a, b, and c in that picture are the coefficients in your quadratic

[restive stratus]

you lose me there

[junior cape]

it's okay, i'm making a picture to help you

[restive stratus]

as i dont see any a b or c in equation

[junior cape]

they are color-matched. the red 5 is the a, the blue 5 is the b, and the green 2 is the c

ok?

[restive stratus]

how do you do that?

[junior cape]

you just have to remember those positions:

so in your problem, what is "a" equal to?

[restive stratus]

5

[junior cape]

good. and b and c are equal to?

[restive stratus]

52

5 2

[junior cape]

very good. so now you put those numbers in the quadratic formula that i posted

[restive stratus]

that's too complicated as i dont understand what he plus and minus do

[junior cape]

it's okay, i can help

[restive stratus]

(-5-+sqr(5^2 -4(10)))/10

[junior cape]

so the reason there is a plus/minus in it is because you may have multiple "roots", or, values of x that make the equation equal to 0

[restive stratus]

i dont think you can squre a negative number

[junior cape]

so you have to consider the plus and then the minus

[restive stratus]

what do you mean by root?

[junior cape]

a root is the value of x when the function crosses the x-axis

[restive stratus]

how do we know it's ture

[junior cape]

see how that function in the graph crosses the x-axis twice?

that means that there are two roots

[restive stratus]

yes

but this have nothing to do with a graph

[junior cape]

it does

i mean, i'm guessing that's what the problem is? you were given that function and you were tasked with finding the roots

[restive stratus]

you lose me there man

[junior cape]

what exactly are you trying to find?

[restive stratus]

x

[cedar kiln]

@restive stratus Has your question been resolved?

[junior cape]

yeah, that means that you're trying to find the value(s) of x that make it true

[cedar kiln]

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[short token]

If we flip 3 coins in the following order A,B,C. Where the success rate of A and C is 0.2 and the success rate of B is 0.5. What is the probability of getting at least two consecutive successes (1,1,0 ; 0,1,1 ; 1,1,1)

[short token]

is it just .2 * .5

[cedar kiln]

@short token Has your question been resolved?

[cedar kiln]

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[knotty acorn]

i am so confused this is an example given to me it says a rotation of 90 degreese counterclockwise and a translation of 2,7 but i am so lost how is this the answer

[wraith dagger]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[knotty acorn]

.reopen

<@&286206848099549185>

PLZ HELP

[rocky harness]

Which thing are you confused about?

[knotty acorn]

idk

the translation

i dont see hw it could be 2,7

[rocky harness]

What does a translation of 2,7 mean?

[knotty acorn]

it moves 2 over and 7 up

or down

[rocky harness]

How many units is it moving to the right by?

[knotty acorn]

2

[rocky harness]

How is it 2 dude

[knotty acorn]

idk

it says 2,7

so i thought x=2

y=7

[rocky harness]

But what is actually happening?

[knotty acorn]

idk

[rocky harness]

Is it only moving to the right by 2?

[knotty acorn]

yes i think

and 7 down or up

idk

are you just gonna keep making me look like an idiot or can you please help

[rocky harness]

Im trying, and that wasn't my intention

[knotty acorn]

oh ok mb

[rocky harness]

The translation should be more than just 2

[knotty acorn]

well thats what it says

[rocky harness]

Thats not whats happening though

[knotty acorn]

it says thats the answer

Describe a sequence of transformations that shows the congruence between triangle ABC
and triangle EGH

[rocky harness]

Ah I realise what's happening

We are rotating 90° about the origin

Where will A(-4,-3) end up if you rotate by 90° about the origin?

[knotty acorn]

5,4?

or counter

clockwise

[rocky harness]

Counterclockwise

[knotty acorn]

if so its -5,4?

[rocky harness]

A(-4,-3) ends up at (3,-4)

When you rotate by 90° counterclockwise

[knotty acorn]

oh i meant 3 not 5

[rocky harness]

Yeah

[knotty acorn]

so would i measure the distance from the two points

after the rotation

[rocky harness]

B(-4,-5) ends up at (5,-4) after rotating 90° counterclockwise about the origin

C(-1,-5) ends up at (5,-1) after the same rotation

[knotty acorn]

I UNDERSTAND

after you rotate you can just find the distance

so its 2,7 because you go 2 points to the right and 7 up AFTER the rotation

[rocky harness]

Yes

[knotty acorn]

my problem was the rotation part

ty

so much

[rocky harness]

They should've specified what point the rotation was about

[knotty acorn]

ye thanks tho

appreciate it

[rocky harness]

Alr

[knotty acorn]

.ckose

.close

[cedar kiln]

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[cedar kiln]

@sleek bronze Has your question been resolved?

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[halcyon haven]

i need help understanding law of sine and cosine

[slim quiver]

Hi there

[halcyon haven]

hello

[plucky owl]

https://www.khanacademy.org/math/precalculus/x9e81a4f98389efdf:trig/x9e81a4f98389efdf:solving-general-triangles/a/laws-of-sines-and-cosines-review

[slim quiver]

So, if you're just getting started in trig, you can ignore the pi values on this chart for the moment

What's really important are the (x, y) coordinates in parentheses on the outside of the circle

[plucky owl]

That's not law of sines/cosines

[slim quiver]

But this is the Unit Circle... a circle with radius = 1 centered on the origin, (0, 0)

[slim quiver]

Different people learn different ways... what worked for me wasn't just learning "opposite/hypotenuse = a number", but rather understanding where those numbers come from

[clear umbra]

law of sines and cosines is the one you use to solve sides and angles for non right triangles

[raven shard]

no they're right

this is not what law of sines is

[slim quiver]

And where they come from are the actual physical distances from the x and y axes to the point of intersection with the unit circle

sigh Yes... sin a / A = sin b / B

[plucky owl]

That's not law of sines/cosine still

[slim quiver]

c^2 = a^2 + b^2 - 2ab cos c

[clear umbra]

so what do you need help with it

[halcyon haven]

oh yeah that

[slim quiver]

Then what are they?

[halcyon haven]

so in class

[halcyon haven]

there was these formulas

it changes for sine

[plucky owl]

As mentioned to solve a triangle

[plucky owl]

Check the link I sent, it might help

[plucky owl]

.

[raven shard]

law of sines and law of cosines are geometric equations that relate the sides and angles of a triangle with each other

LoS: sinA / a = sinB / b = sinC / c
LoC: c^2 = a^2 + b^2 - 2ab cosC (as you said)

i am slightly confused by your question... so you know the equation for law of cosines but don't know what it's called

[halcyon haven]

how do i know which of the three equations to use

with the info given

[plucky owl]

Depends on the angle you are looking for

[slim quiver]

Depends on what you're looking to find... do you need the angles B and C? law of sines...

Side a? law of cosines

[halcyon haven]

i need to complete it

i have angle A and side b and c i need angle B and C and side A

[plucky owl]

You should use law of cosine first to find side A

[slim quiver]

You'll want to start with the law of cosines, then, since you'll need the length of side a to plug into the relationships for the law of sines

[plucky owl]

Then you can use law of sines for the angles

And you can find either angle B or C using law of sines, but you don't need to apply it twice to find the two angles

You can just use law of sines once and to find the other angle, sum of interior angles of a triangle add to 180

[raven shard]

(minor clarification

you can use only law of sines to solve the entire triangle
you can use only law of cosines to solve the entire triangle
or you can also use a mix of both)

[slim quiver]

You need one pair of a known side and the opposite angle

So far, he has neither for any of the sets

[plucky owl]

you can also use a mix of both

[halcyon haven]

so to find side a i use the a^2=b^2+c^2-2bcCosA formula

[slim quiver]

Yep

[halcyon haven]

oh oh okay thank youuu

[slim quiver]

And they're right, you could use law of cosines throughout the whole problem, but that involves manipulating the equation to arrive at a cosine value, and then applying inverse cosine to find the missing angle... a lot of extra work

[halcyon haven]

is it okay that i dont close the channel yet im trying to solve it rn but i get stuck alot

[plucky owl]

Yes

You can solve the problem and send your work when you are done, to get it checked

[halcyon haven]

i found side a i think

im stuck on how to switch to law of sines now

[plucky owl]

Do you know what law of sines is?

Because you sent an image of area of a triangle, not law of sines

[slim quiver]

Law of sines is just a comparison of ratios... sin A / a = sin B (or C) / b (or c)

[slim quiver]

Now that you have the length of a, you can set up a ratio with angle A

[clear umbra]

,rccw

[plucky owl]

This is not the best way to format it

[halcyon haven]

ok that makes more sense

except

i dont how to word my question

[plucky owl]

Where are you confused ?

[slim quiver]

You're going to end up with a small decimal value.. that's expected

[plucky owl]

You have all the sides now and angle A

[plucky owl]

As the angles? No

[slim quiver]

You'll have to use Shift+sin (usually) on your calculator to convert that small decimal number to an angle

[halcyon haven]

so its the sin of the angles but how do i get it to be just the angle

[slim quiver]

As the sin ratio... yes.

[plucky owl]

Inverse sin

[slim quiver]

Look at your calculator... you should see a small "sin^-1" printed above your sin button

This is inverse sin

[plucky owl]

Depending on your calculator, it's could be shown as asin, arcsin, sin^-1

[slim quiver]

It takes that small decimal value and converts it to an angle

And yeah, it goes by a few different names

[halcyon haven]

at what step do i do that

[slim quiver]

After you get rid of those fractions... you'll end up with a small value like 0.5835 = sin C

[halcyon haven]

inverse sin 12?...

[slim quiver]

= (the length of the side you need) / (the angle opposite the side you need)

Set it equal to another ratio

Then solve algebraically

sin (some angle) will equal 9.2/sin 50

[plucky owl]

You should have an equation

[slim quiver]

Think of it this way... 1/2 = 3/x

Solve for x

[plucky owl]

Something like $\frac{a}{\sin A} = \frac{b}{\sin B}$

[wraith dagger]

CaptainNova22

[plucky owl]

What you wrote is not an equation

[slim quiver]

There you go!

[plucky owl]

Also the other way you can write law of sines is this, so the sines are in the numerator

[slim quiver]

Which is why I stated the other way to begin with...

Now he has to extract a trig function from a denominator

applause

[plucky owl]

First, "he" is a "she" look at the roles

[slim quiver]

Just take the inverse of both fractions, and you'll be fine

[plucky owl]

Second, it's not that hard, you can just flip the fractions

[halcyon haven]

so i do sin^-1(9.2/sin50=12/sinB)?????

[plucky owl]

No

[halcyon haven]

im sorry

[slim quiver]

You don't need to apologize... it's a learning process. You're fine

[plucky owl]

To make it easier, you can write it in this form

So sin(50)/9.2 = sinB/12

It's an equation where you are solving for the unknown

So you need to isolate for it

What should the first step be?

[halcyon haven]

um common denominators..? i think

[plucky owl]

No

[halcyon haven]

this unit is so hard

[cedar kiln]

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.

[plucky owl]

https://www.mathsisfun.com/algebra/trig-sine-law.html

[slim quiver]

Sorry, but... it's about to get a little harder =/

Remember that unit circle I tried to start with?

Well... you're going to get the same sin value for two possible angles

You've probably noticed by now, your answer is a little different

[halcyon haven]

oooohhh

[slim quiver]

Technically this is a negative equivalent... so there are 4 possible places you'll see the same number

two positive, two negative

These methods will always return acute angles.. so if you have an obtuse triangle, you're going to have to be careful

[halcyon haven]

wait huh

[slim quiver]

When you take the inverse sin of a number, like .7071, it will always give you 45°

But grab your calculator and find the sin of 135°... what is it?

Then find the sin of 225° and 315°

The number you keep seeing pop up is the y-value of those angles, where the line intersects the circle

So if you've got a triangle with one angle that's 135°, you're going to have to be careful when applying the sin and cos rules

[halcyon haven]

oh ok, i dont think my teacher is gonna do that to us yet

we just started

[slim quiver]

Ok, good!

So! What angle did you get for B?

[halcyon haven]

working on it still

[slim quiver]

Ok, no problem

brb

[halcyon haven]

sorry i took so long

[slim quiver]

No worries

[halcyon haven]

also ok take ur time

[slim quiver]

I'm back, btw

[halcyon haven]

oh hello

[slim quiver]

So! Still stuck?

[halcyon haven]

i got a different answer from when you did it

[slim quiver]

Yup... expected as much

[halcyon haven]

so i think i did something wrong

[slim quiver]

Nope, you probably didn't

What'd you get?

[cedar kiln]

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

[halcyon haven]

ignore the top

[slim quiver]

87.7? That's one possible correct answer!

My answer was different because I solved for C first...

You solved for B

Now then, you have two angles, and every triangle has how many degrees in its corners?

[halcyon haven]

180

[slim quiver]

Cool.. so take 180 and subtract the two angles you have, and there's your third

If you solved for C first, you'd get a different answer for B

Which is why our answers were different

eventually your teacher is going to hash all that out for you, so don't stress about it for now

But yeah, you got it

[plucky owl]

No she solved for B

It literally says B

[slim quiver]

That's literally what I said.

[plucky owl]

You work here has B...

94 is not 87

[slim quiver]

I said she solved for B first... I solved for C first... you'll get two different answers depending on which angle you solved for first

[plucky owl]

You should use the full decimal for better accuracy

[slim quiver]

It's not a rounding issue.. it's an issue with the fact that the laws of sin and cos always return acute angles

[plucky owl]

Your work has sin B

You got 94.4 for angle B

[halcyon haven]

i rounded it bc of this

[plucky owl]

Round at the end

[halcyon haven]

oooh okay

[slim quiver]

It's an issue with me including the answer I got from B after solving for C and subtracting the other two angles..

[plucky owl]

Don't round in the middle stpes

[slim quiver]

I'm literally telling you exactly what I did, and you still want to argue... roger that, bud.

[plucky owl]

,w asind(sind(50)*12/9.2)

[wraith dagger]

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Requested by captainnova22

[plucky owl]

,w asind(sind(50)*12/9.22018)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by captainnova22

[slim quiver]

,w asind(sind(50)*7/9.2)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ryanund

[slim quiver]

180 - 50 - 35.7 = .....

[halcyon haven]

huh

[slim quiver]

It's ok.. you got one good set of answers, so be proud of yourself!

Everything else here is Nova wanting to argue... sorry about that

[halcyon haven]

ummm

so

do i need to go back and not round

cuz i rounded at the start

wait i think i do

brb

so i dont need 94?

[cedar kiln]

@halcyon haven Has your question been resolved?

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[sudden mirage]

I dont understand this