help-27

[cobalt plinth]

Or are you right

Nvm thank you

.close

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[restive river]

Ok

[restive river]

Wait let me get the question out

So how would I solve this out

I have no idea where to start

[grizzled yew]

Plug it in

[restive river]

So do I plug 2 into the 2y?

Wiat what do I plug in for the x then

<@&286206848099549185>

Wait nvm I got it

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[rotund torrent]

Help how do I do these

[main gull]

Do you know how to factor in general?

[rotund torrent]

A bit

[main gull]

A bit meaning?

What do you know?

[rotund torrent]

I know up to difference of 2 squares?

Grouping in pairs

[main gull]

Do you know something like AC method?

[rotund torrent]

Nope

[main gull]

Well here you don't need something that intense since a = 1

But a hint is, find the factors of c that add up to b

[restive river]

product of a and c

adds up to b

[main gull]

Well here you don't need something that intense since a = 1

[rotund torrent]

I'm still confused

[main gull]

find the factors of c that add up to b

Take the first one

What is c?

[restive river]

could just set to 0 and use quadratic formula

[rotund torrent]

c is 12 right?

[main gull]

Don't think they know that yet cause it's factoring

[main gull]

Yes

What are the factors of c?

Aka 12

[rotund torrent]

1,2,3,4,6,12

[main gull]

Which factor pair adds up to b?

What is b in that problem?

[rotund torrent]

7 right?

[main gull]

Yes

[rotund torrent]

so 3 and 4

[main gull]

Yes

So the general form for the problems you are doing is (x + a)(x + b)

[restive river]

(3)(4)=(1)(12)

[main gull]

Where a and b are the factor pair you just found

[rotund torrent]

I see

[restive river]

3+4=7

[rotund torrent]

so (x + 3)(x + 4)

[main gull]

Yes

[rotund torrent]

Alr got it thanks

[main gull]

Now you just apply that concept to the rest

[rotund torrent]

Alr thanks

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[clear hedge]

can anyone explain this??

[supple knot]

Do you know log rules

[clear hedge]

nope wasnt there during the lesson but this is the only log related question i have to deal with for now

i dont understand it at all

[winter hound]

Make x as the subject

[clear hedge]

what do you mean by that?

[winter hound]

do you know what’s a subject in an equation

[clear hedge]

no...

[devout snow]

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[fiery finch]

hiii

[fiery finch]

not math but could someone help

pls

[hazy quail]

#old-network physics server

[fiery finch]

ty!

[hazy quail]

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[mortal jungle]

quick question, the answer would be no right

[mortal jungle]

[1 _ _ _ _
0 1 _ _ _
0 0 0 0 1]

that means that it'd be 0x + 0y ... = 1

(doesn't have to be one ik)

but that means its inconsistent right

oh i just thought about it for a second, yea i guess it would be inconsistent

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[soft sphinx]

The answer is (16√3 - 8π)cm^2

[real trench]

Uhhh is that a question?

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[restive river]

How do I graph constraints on desmos again?

[winter patrol]

{ condition }

no colon

-x{x<=1}

[restive river]

thx

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[void current]

tan + 1/tan = csc/cos

[void current]

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[violet wind]

the sum set up doesn't solve the problem

I think your friend treated it like putting $600 in a bank every month

and getting 6% interest compounded monthly from that

hmm

well putting $600 in early will prevent you from paying 600+interest later

So in that sense it's a little bit like putting money in a bank

but in that case the final "sum" would represent how much money it would cost if you paid all of it at the 4 year mark

ah nice

np

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[kindred sedge]

I'm stuck once more..

[devout snow]

Please don't occupy multiple help channels.

[kindred sedge]

I don't know what to do

[late oriole]

@kindred sedge

On the right,

sin^2 * 1/cos^2

That becomes tan^2

Can you solve it from there?

[kindred sedge]

May I ask how to make it a secant?

[late oriole]

Okay

So it's sec^2-1=tan^2

First of all, that is an identity (tan^2+1=sec^2)
But you may be wondering how to get that

So sec^2 = 1/cos^2 -1 = 1/cos^2 - cos^2/cos^2 (you can see that is still equal to 1)

Then, (1-cos^2)/cos^2 when you simplify
Then, there is another identity that sin^2+cos^2=1

Solve from there

[void sigil]

Do you have to prove that it's equal to sec² -1?

Or is it something else

[kindred sedge]

Hello if you don't mind me asking, is it sec^2=1/cos^2-1

The one you said

[kindred sedge]

Yes

[void sigil]

Oh

You can apply identity in sin²

And write it as 1- cos²

Multiply the sec² inside

Sec² times cos² is always 1

And you multiply it by 1 too

Did that solve it

[kindred sedge]

So is it like this

Wait if you don't mind me asking how did sec² times cos² is 1

Is it a fixed identity

[void sigil]

Because sec is just 1/cos

Just like how cosec is 1/sin

And cot 1/tan

[late oriole]

$sec^2x = \frac{1}{cos^2x}$

[woven radish]

Anti-Clownfish

[void sigil]

Yep there you go

[kindred sedge]

Ohhhh

So it's the same as

Csc²x times sin²x = 1?

[void sigil]

Yes

[kindred sedge]

Okay I'll try to solve this one

[late oriole]

$sec^2 -1 = \frac{1}{cos^2x} -1 = \frac{1}{cos^2x} - \frac{cos^2x}{cos^2x} = \frac{1-cos^2x}{cos^2x} = \frac{sin^2x}{cos^2x} = tan^2x$

That was what I typed above but I converted it into a cleaner format

[void sigil]

Seems like it mistyped it

It didn't subtract the 1 in the first equality but it did in the second one

[woven radish]

Anti-Clownfish

[kindred sedge]

This is hard

I'm sleep deprived I can't even think anymore

I'm stuck in this part

Sorry for the handwritinf

[void sigil]

You're on the right path

You got sec² ( 1 - cos²)

Which when simplified

Will give sec² - sec² cos²

Now since sec² cos² is 1

You get sec ² -1

[kindred sedge]

Ohhh

Okay I get it now

Oh god

Thank you

[void sigil]

It's no problem

[kindred sedge]

.close

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[steel escarp]

The points A = (-7,4,-3), B = (53,-101,42) and P lie on a straight line. The distance from A to P is four times greater than the distance from B to P. Determine the two possible (sketch!) positions of the point P.

[steel escarp]

I solved one of the positions which is 1/5(OA)+4/5(OB)

The other one is 4/5(OA)+1/5(OB) but how do I prove that?

See the position 2 part

I've failed to prove it, how can I fix it?

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@steel escarp Has your question been resolved?

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@steel escarp Has your question been resolved?

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@steel escarp Has your question been resolved?

[steel escarp]

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[primal tusk]

can someone please explain this to me

i have no idea what this wants or how to go about it

[lunar harbor]

Expand $r\sin(\theta+\phi)$ and match the coefficients of $\sin \theta$ and $\cos \theta$

[woven radish]

Civil Service Pigeon

[primal tusk]

what are the coefficients of sin and cos theta in this?

[lunar harbor]

Expand it out and you’ll see 🤷‍♂️

[primal tusk]

bruh

if i knew how to do that then i wouldnt be here asking

[lunar harbor]

You do know that $\sin(x+y)=\sin x \cos y +\cos x \sin y$ right?

[woven radish]

Civil Service Pigeon

[lunar harbor]

Also if you really wanted an answer quickly you would google it, this is a well known technique

[primal tusk]

i have googled it

im asking here to understand it not be told "just google it lol"

e-e

[lunar harbor]

Well apply this formula, it’s just that the variables are $\theta$ and $\phi$ rather than $x$ and $y$

[woven radish]

Civil Service Pigeon

[lunar harbor]

Comparing coefficients, as I said earlier, with the original expression yields a system

You can divide the equations to find $\tan \phi$, the rest is trivial.

[woven radish]

Civil Service Pigeon

[primal tusk]

what do i do with tan thi once i have it?

[lunar harbor]

Use that to solve for $\phi$…?

[woven radish]

Civil Service Pigeon

[primal tusk]

and i have no idea how to do rsin thi/ rcos thi

[lunar harbor]

Given the fact that it’s literally just substituting into a formula, idk what you want me to say.

I suppose I’ll leave this for someone else

[primal tusk]

,n,

[devout snow]

@primal tusk Has your question been resolved?

[primal tusk]

<@&286206848099549185>

[devout snow]

@primal tusk Has your question been resolved?

[urban spindle]

I'll try

[primal tusk]

thank you

[urban spindle]

so we try to solve a$\sin\theta$ + b$\cos\theta$=$r\sin(\theta+\phi)$

[woven radish]

Maths Enjoyer

[urban spindle]

as the other user said, we expand $r\sin(\theta+\phi)$

[woven radish]

Maths Enjoyer

[urban spindle]

I believe it would be useful if you expanded it (useful for you). so, try to use the aforementioned formula to transform that sin of a sum into 2 pieces.

and then tell me your result

[urban spindle]

apply this formula to $r\sin(\theta+\phi)$ and tell me the result 🙂

[woven radish]

Maths Enjoyer

[primal tusk]

which formula sorry?

can you just quote it

unless you mean

rsinθcosϕ+rcosθsinϕ

[urban spindle]

yes, that

[primal tusk]

ok, im not really sure what to do with that tho

im aware that cosϕ=rsinθ

i think?

[urban spindle]

so now this is equal to a$\sin\theta$ + b$\cos\theta$

[woven radish]

Maths Enjoyer

[urban spindle]

we notice both of these things have $\sin\theta$ and $\cos\theta$, and therefore I believe the idea is to equalise their coefficients

[woven radish]

Maths Enjoyer

[urban spindle]

we see, for example, in the place of the letter "a", we have r$\cos\phi$ here

[woven radish]

Maths Enjoyer

[urban spindle]

so a should be equal to r$\cos\phi$

[woven radish]

Maths Enjoyer

[urban spindle]

makes sense?

[primal tusk]

um... why should it be = to rcosϕ?

[urban spindle]

a$\sin\theta$ + b$\cos\theta$ = rsinθcosϕ+rcosθsinϕ

[woven radish]

Maths Enjoyer

[urban spindle]

I assume the question is...why is xsin(a) + ycos(b) = tsin(a) + vcos(b) necessarily implying that x=t and y=v?

[primal tusk]

thats the question there

so... kinda?

i just want rsin(theta+thi)

[urban spindle]

while this explanation might not be correct mathematically, your question has exactly one answer. and since the identification of coefficients of sin and cos would lead to 1 result...:)

[urban spindle]

it's fairly clear it's the answer. we identify the coefficients of $\sin\theta$ and $\cos\theta$ in here (the message I replied to)

[woven radish]

Maths Enjoyer

[primal tusk]

um. im sorry i dont know what "clear answer" it is

i feel so lost

is it litterly rsinθcosϕ+rcosθsinϕ?

[urban spindle]

a$\sin\theta$ + b$\cos\theta$ = rsinθcosϕ+rcosθsinϕ
in this, we try to solve a=rcosϕ and b=rsinϕ and express r and ϕ based on a and b

[woven radish]

Maths Enjoyer

[urban spindle]

in other words, we try to solve the system of the 2 equations: a=rcosϕ and b=rsinϕ . we consider a and b given, and want to find r and ϕ. 2 equations, 2 things to find, so it should work 🙂

[primal tusk]

wait wait wait, to be clear on something

we want tan^−1(b/a) right?

becouse that =ϕ

so it would be tan^-1(sinϕ/cosϕ)

[urban spindle]

tan^−1(b/a) = ϕ , yes

a mention: in our exercise, we know a and b

don't worry about putting the ϕ you found back in the system (I think it's more complicated). there's another method

square both equations and add them up. tell me what you get 🙂 (these equations: a=rcosϕ and b=rsinϕ )

[primal tusk]

how do i put ϕ into my calculator?

[urban spindle]

what do you mean

also, have you found r yet?

[primal tusk]

no?

[urban spindle]

have you tried raising both equations to 2nd power, like I said before 🙂

[primal tusk]

i... dont know how

[urban spindle]

square both equations and add them up. tell me what you get 🙂 (these equations: a=rcosϕ and b=rsinϕ )

so, if a=rcosϕ

a^2=?

[primal tusk]

is there like a special rule im not getting here?

[urban spindle]

if you square what's left of =, you square what's right of =

not really, you need to square the rcosϕ

[primal tusk]

as far as i can tell rcosϕ^2=....rcosϕ^2...

[urban spindle]

hint: we do this because we know cos^2 +sin^2 of the same angle = 1

[urban spindle]

but we have (rcosϕ)^2

since you have to square everything right from the "="

and (rcosϕ)^2 =/= rcosϕ^2

let's remember (ab)^2=

[primal tusk]

i might just entirely give up on this topic, it dosnt make any sense what so ever

[urban spindle]

hm

if 2 numbers are equal, their squares are equal too, right?

[primal tusk]

yes

[urban spindle]

so if a=rcosϕ

then a^2= r^2 (cosϕ)^2

correct?

the power goes to all factors

[primal tusk]

i guess?

we have a

which is sin(theta) right?

[urban spindle]

we just said a=rcosϕ

so it isn't sin(theta)

btw we cannot express a based on theta (or anything)

give me a sec

[urban spindle]

in your question at start, tell me. how much is a and b

at question 1.a

[primal tusk]

i have do idea dude

[urban spindle]

right now we struggle to find r and phi based on a and b from the system of 2 equations I mentioned earlier

we got phi pretty easily

[primal tusk]

wait we have phi?

when did we get phi

[urban spindle]

tan^−1(b/a)

[primal tusk]

ok

[urban spindle]

this?

[primal tusk]

and what, in this equasion

is a and b

please just write out the entire equasion for me

[urban spindle]

ok. so we know tan^−1(b/a)=ϕ

we got it from these 2 equations: a=rcosϕ and b=rsinϕ

you had a good idea to divide the 2 equations

now we try to find r

from the same 2 equations

[primal tusk]

wait hold up

[urban spindle]

you gave me this answer earlier

about phi

[primal tusk]

what is tan^-1(rsinϕ/rcosϕ)

i know it =ϕ

but is there a number to it?

[urban spindle]

that is the thing, it depends on a and b. so it's not fixed

[primal tusk]

ok

[urban spindle]

let's find r and then we'll see

ok?

so we have a=rcosϕ and b=rsinϕ

try to find r like you found ϕ

I said earlier, we need to use another trick, like the one you used for phi

and I said, let's square the 2 equations. a=rcosϕ and b=rsinϕ

and then add them (after raising both equations to 2nd power, sum them)

tell me if you manage

but first, try 🙂

[primal tusk]

ok so, i remember you saying that it was something like r^2cosϕ^2

for a

[urban spindle]

yes

when we square, we square both terms of product 🙂

very good

[primal tusk]

so a^2=r^2cosϕ^2

same for b but with sin instead

[urban spindle]

now let's sum them

so, the side with a and b are added on the left

= the sum of the terms on the right

a^2 + b^2= [you calculate]

from the 2 things you just told me

[primal tusk]

so a^2+b^2=r^2cosϕ^2+r^2sinϕ^2

?

[urban spindle]

yes

we can do some magic on the right side

do you notice something there?

[primal tusk]

no?

[urban spindle]

maybe a common factor

[primal tusk]

no...?

[urban spindle]

r^2cosϕ^2 and r^2sinϕ^2 , don't they have something in common

a factor

[primal tusk]

,n,

im not seeing it at all

[urban spindle]

ok, another hint: ab+ac= a(b+c)

[primal tusk]

where is c coming from??

[urban spindle]

these a , b and c are supposed to show you a formula

the hint is a formula

r^2cosϕ^2 + r^2sinϕ^2 in this, we see r^2 common factor

saw it?

[primal tusk]

ok, why does that matter?

[urban spindle]

let's use the formula I mentioned earlier, ab+ac=a(b+c) on our equation

r^2cosϕ^2 and r^2sinϕ^2 this is our equation

[urban spindle]

when did I mislead you? haha

r^2cosϕ^2 and r^2sinϕ^2 let's use that formula to make r^2 ( something + something)

the things in the ( ) will be important

[primal tusk]

the issue that i see is that to find ϕ we need r. but to find r we need ϕ

[urban spindle]

the issue will disappear soon

[urban spindle]

we already found ϕ without needing r

earlier, that tan

remember?

[primal tusk]

but the formula has r in it???

[urban spindle]

tan^−1(b/a)=ϕ not really?

yes, if we replace b and a with the other results, sure. but a and b will be given in the problem

we'll see them, don't worry

ok, we found ϕ, let's find r please

r^2cosϕ^2 and r^2sinϕ^2 let's use that formula to make r^2 ( something + something) focus on this

[primal tusk]

ok ok...

[urban spindle]

ohh, another thing to consider. when we wrote cosϕ^2, you have to understand that it is cosϕcosϕ and not psi being squared. just to make things clear(er)

[primal tusk]

ye im aware cosphi is one function not 2

[urban spindle]

well, the angle (as a number) could have been squared too... that's what I was trying to say. I am glad there is no confusion 🙂

[primal tusk]

r^2(cosϕ^2+sinϕ^2)?

[urban spindle]

yesss

[primal tusk]

wait seriously?

[urban spindle]

and what is the sum of those things between ( )

[primal tusk]

hoh

uh

[urban spindle]

yes , but you forgot a ϕ there, at sin

[primal tusk]

oh

miss type mb

[urban spindle]

np, what is cos^2 + sin^2 of the same angle?

[primal tusk]

wait

[urban spindle]

it's the most famous relation in trigonometry

[primal tusk]

is that not 90degress?

or is it 180

[urban spindle]

hmm nope

it's...1 (not degree)

[primal tusk]

RIGHT RADIANS

and 1rad=180deg

[urban spindle]

so, that ugly parenthesis

we replace with

1

[primal tusk]

im gona write down cos^2+sin^2=1rad

[urban spindle]

not 1 rad, just 1

[primal tusk]

seems important for future stuff

oh

[urban spindle]

here's why

how much is sin30 for example

are you familiar with it?

np it's 1/2. sin and cos of an angle, are numbers. that's why their sum cannot be in degrees or radians

so that sum couldn't be anything but a number (without rad or degrees)

but this is a formula...cos^2+sin^2=1 (as long as it's cos and sin of the same thing)

[primal tusk]

so... r^2=1

[urban spindle]

nope

[primal tusk]

hmm

[urban spindle]

we had a^2+b^2=r^2(cosϕ^2+sinϕ^2)

hopefully we agreed the parenthesis is 1

so what are we left with?:)

[urban spindle]

we didn't have r^2=( the parenthesis)

we had a^2+b^2=r^2 (the parenthesis)

we are very close, almost done 🙂

having trouble? tell me

[primal tusk]

so... a^2+b^2=r^2

thus

[urban spindle]

yeeees?

[red tiger]

i ned help pls

[primal tusk]

dude.,..

[urban spindle]

wrong room, this one's taken already

lol

[red tiger]

oh sry

[primal tusk]

xd

[urban spindle]

I'd help but cannot participate in 2 sides at same time :))

scroll up and look for

math help (available)

pick one of those rooms

someone might come to help

oof so nattly

[urban spindle]

what do we find now

[primal tusk]

a^2=cosϕ^2
b^2=sinϕ^2
so cosϕ^2+sin^2=

r^2

[urban spindle]

we leave it in a and b, it's better

if we have a^2+b^2=r^2

who's r

[primal tusk]

would r just be... 1?

[urban spindle]

hmm not really

say a=2 b=3, r^2 would be 4+9=13

let's not replace a and b with those things

let's just find r from a and b

[primal tusk]

ok so a^2+b^2

hmm

[urban spindle]

a^2+b^2=r^2 => r= (something with a and b)

if we had 16=r^2 how do we find r

[primal tusk]

well we would root it right?

[urban spindle]

yess

[primal tusk]

r would be 4

OH

SO

uh

not sure how to express that

[urban spindle]

so r = $\sqrt(a^2+b^2)$

[woven radish]

Maths Enjoyer

[urban spindle]

all of that under the radical

[primal tusk]

is that as simple as you can make it?

written down i mean

[urban spindle]

yep. so let's do a quick check up

we have r as expression of a and b

we have phi as expression of a and b

let's look again at our problem now

we solved asin(theta)+bcos(theta)=rsin(theta+phi)

if we know and a and b, we can transform any equation from left form to right form

by calculating r and phi with previous formula

makes sense this far?

[primal tusk]

yes

[urban spindle]

but now we want to transform sin(theta)+cos(theta)

that's what 1.a asks

so...who's a? who's b?

[primal tusk]

well a is sin(θ) and b is cos(θ)

right?

[urban spindle]

no, it's much simpler than that

we learned to transform for any asin(theta)+bcos(theta)

[primal tusk]

=1?

[urban spindle]

and now we wonder what's a and b in sin(theta)+cos(theta)

yess

a=b=1

they don't appear, they are 1

[primal tusk]

wait so, a and b both = 0.5?

[urban spindle]

nope

a=b=1

if and b both=0.5 we'd have

0,5sin(theta)+0,5cos(theta)

but we have sin(theta)+cos(theta)

which means the coefficients must be 1

you see?

[primal tusk]

eh...

[urban spindle]

only 1sin(theta)=sin(theta)

we cannot write 0,5sin(theta) as just sin(theta)

[primal tusk]

ok yea

[urban spindle]

ok

so in the question, a=b=1

let's see again phi and r expressed based on a and b

the formulas we discovered

r = $\sqrt(a^2+b^2)$ this is the r

[woven radish]

Maths Enjoyer

[urban spindle]

you'll have to find the phi in our conversation

remember, we found the formula for him

actually, you did

something with tan, a and b 🙂

it would have been good if you noted (some of) the things mentioned , so you remember the method and the results

[primal tusk]

so... tan^-1(1/1)

?

[urban spindle]

yepp

tan^-1 means we wonder

what angle has the tan = 1/1

[primal tusk]

idk

[urban spindle]

well it's sadly something that needs memorised

it's a 45 angle 🙂

[primal tusk]

ok so tan 1/1 = a 45 angle

[urban spindle]

tan^-1 of 1/1 is a 45 angle

tan 1/1 is some really ugly number

[primal tusk]

ok, i assume this is just a constant in maths

[urban spindle]

tan takes an angle and gives a number

tan^-1 is it's reverse; it takes in a number, and gives the angle

[primal tusk]

ahh

[urban spindle]

just so you understand the concepts

[primal tusk]

ok that makes sense

so,

[urban spindle]

who's phi then

[primal tusk]

phi is 45? or is phi soly in radians

[urban spindle]

very good observation 🙂 the exercise says phi is in radians

[primal tusk]

so 45x 2pi/180

1.57

[urban spindle]

the formula is mistaken

if you want to use such a transformation, we remember pi =180

so it would be 45 x pi/180

else it changes value

45 is pi/4

[primal tusk]

0.7854 then

[urban spindle]

ok. now you need the r

[primal tusk]

which was 1?

[urban spindle]

check the formula

[primal tusk]

um

[urban spindle]

yes? what was formula for r

based on a and b

I mentioned it a bit earlier

[primal tusk]

/(a^2+b^2)

[urban spindle]

sure

and if a=b=1

[primal tusk]

so sqr of 2

uh...

is that a number?

[urban spindle]

yep, it is

an infinite one that never repeats

[primal tusk]

1.41

[urban spindle]

non-periodic, we might say

well, that's the first 3 digits of it...

ok, these should be phi and r that you're looking for

does your test give you results automatically?

[primal tusk]

not sure how far im ment to go with it lol

god i wish

if i put in something wrong all i get is

"bad"

[urban spindle]

let's you send again?

[primal tusk]

yea

[urban spindle]

try the results

if it has a keyboard or something, and you can insert sqrt2 and pi/4

try those, I think

[primal tusk]

1.41(theta+0.79) is considered wrong

[urban spindle]

hm

[primal tusk]

sqrt2(theta+pi/4) is aswell

im going to say that its the websites problem tho

[urban spindle]

most likely

also it probably has a single way of writing marked as correct

and if you miss a ( or something, it gives wrong answer

either way, I hope now you know how to solve such a problem which matters more than finding the result.

[primal tusk]

so

before i close this

r=(a^2+b^2)
and thita= (invers tan(b/a)

[urban spindle]

r had a sqrt of the thing between ( )

r^2=a^2 + b^2 or r=($\sqrt(a^2+b^2)$)

[primal tusk]

(i ment to type / mb lol)

[woven radish]

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

[urban spindle]

yep, pretty much those are the results

but the method matters too 🙂

ok so , I got to go. been a long day. wish you luck.

[primal tusk]

ty so much

close.

.close

[devout snow]

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[strange radish]

Calculate the amount of interest earned by investing $13,500 for 6 months at a simple interest rate of 0.90% p.m. I know the what rate and principal are, but do i calculate for time?

[strange radish]

if I=PRT i already know what P&R are, but im a little confused at what i calculate for time. is it 6/12 or 6

[wooden wraith]

You said 0.90% p.m.

Does that stand for per month?

If so, you're counting months

So t=6

[strange radish]

yes it does

thank you

another question. if i convert an inerest rate from a decimal to percentage and they are saying round two decimal points how would I round this number. 0.115160428 obviously i convert this to a percentage but does that mean I round up after i convert it?

im unsure whether the correct answer is 11.51% or 11.52%

[devout snow]

@strange radish Has your question been resolved?

[devout snow]

@strange radish Has your question been resolved?

[wooden wraith]

round the 1 up to 2, since the next digit is 6

[devout snow]

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[bright gulch]

Do i have to calculate the magnitude of the vector for c?

what do those absolute value signs indicate?

[wicked turtle]

yea that's almost certainly the magnitude

[bright gulch]

magnitude or the actual abs value?

Got it

Thanks

.close

[devout snow]

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[jaunty dagger]

idk what to even do for this problem

[graceful cosmos]

Those curves all act as a boundary to some region. You want the area of the shape they make.

[jaunty dagger]

they really dont make any shape tho

[graceful cosmos]

Before anything, you might want a graph of the region. Even a crude one will do

[jaunty dagger]

x would jsut be a line

and x^3+2x is just a perabula

wait let me put it in desmos

idk if im missing something

[small raptor]

X=2

[jaunty dagger]

ohhh

so its legit

ah

maybe it was 0 to 2 then

[small raptor]

I’m thinking this part

Idk why it didn’t just say ‘the area of the curve between y=x and y=x^3 +2x between 0 and 2

[jaunty dagger]

yeah that makes

a lot mroe sense

lmaooo

jsut putting 0 and 2 as the bounds and intigrating

[upper schooner]

Idea is to find the point that they intersect?

[jaunty dagger]

f(x)-g(x)

they intersect at 0 tho

there is no y intersept added that makes no sense to figure out i think

BRO

IT WAS RIGHT

HOLY CRAP

WE GOT IT

TY SO MUCH

[small raptor]

Idk if you are at the point of integration why is there a need to test this

[upper schooner]

To be fair, there are a lot of questions that are like that I've seen like that, where they try and build up on knowledge you'd know rather than just giving you something straight up esp like in final exams it's a common flavour of question I think!

[small raptor]

True. And also to actually solve where they intersect you can’t divide by x coz u will have a complex number, so you just have to see it instead of solve it

[devout snow]

@jaunty dagger Has your question been resolved?

[devout snow]

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[distant cape]

If a fair die is rolled 6 times, what is the probability, rounded to the nearest thousandth, of getting at least 5 fives? This is veryu confusing and i dont get it

[rugged jay]

hold on my memory isnt being great

so i think its just 1/6^5

u dont care abt the last roll

and u dont care what order the 5s come in

so i believe its just

[distant cape]

i can do the first part but the part where you have to do the at least part is confusing

[rugged jay]

wait

no

k im sry

idk how to do this

[distant cape]

its binomial probility with like or as

its on delta math

.close

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[opaque terrace]

Hello! I have a confusing pre calc problem that I've been given no information on how to solve with little knowledge into pre calc!

[fresh horizon]

you can construct a hypothetical "triangle" with the given

triangle in quotes because we are allowing for negative length

[opaque terrace]

For me, I don't know how to solve for each function, like sin(t), cos(t), etc.

[fresh horizon]

just use the ratio-definitions on a triangle I guess

[opaque terrace]

I'll be dead honest, I have no clue what that is lmao

[opaque terrace]

ohh

And what about the hypothetical triangle?

[hidden raft]

trigonometry ratio

[devout snow]

@opaque terrace Has your question been resolved?

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[mild basin]

Is this a typo?

[hushed wraith]

no

[mild basin]

if t < -4, ok.. let's say -3 as an example

[hushed wraith]

for example if t = -5

t<-4

[hushed wraith]

no -3 is greater than -4

[mild basin]

oh you're right

[hushed wraith]

the numbers go like -5, -4, -3, ..., 0, 1, 2, 3

[mild basin]

[hushed wraith]

😄

[mild basin]

TY

[devout snow]

@mild basin Has your question been resolved?

[devout snow]

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[devout snow]

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[sullen nova]

can someone help me please? i have to solve #67 using the row echelon method

[sullen nova]

i got to the matrix you can see below and solved for x, but idk how to find y and z without using substitution

[warped relic]

You have 2 equations for 3 unknowns, which means you have infinite solutions

And the matrix you got is the best you will get

[sullen nova]

is it bc one variable will always stay unknown

or i mean 2

if you use matrices

[warped relic]

One will be based on a variable

[sullen nova]

yeah

[warped relic]

And the other 2 will be found through it

[sullen nova]

so the matrix is my answer + x=2?

that or writing the answer in terms of variables

[warped relic]

You write the answer in terms of a variable

[sullen nova]

ok yeah

so as an ordered pair it would be (2, z-3, y+3) ?

or dors it have to be the same variable

[warped relic]

Same variable

[sullen nova]

ok

so you would have to use substitution anyways right

[warped relic]

Nope

[sullen nova]

oh wait

yeah z is infinite solutions

[warped relic]

You can assume z = t

Yep

[sullen nova]

whats t

[warped relic]

t is a variable

[sullen nova]

just like a random variable ?

[warped relic]

Yeah

[sullen nova]

uh ok

how does that help me?

[jovial mauve]

You can write the other variables in terms of t

[sullen nova]

oh

um

let me see lmao

this is the best i could figure out😭

thats what it says in the back of the book so

[warped relic]

Hmm

[jovial mauve]

,w rref {{1,-1,1,5},{3,2,-2,0}}

[woven radish]

Results provided by WolframAlpha

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

Requested by Hazysleet#7757

[warped relic]

x =2
y = t -3
z = t

Works out

[jovial mauve]

(2,t-3,t) for t ∈ R is the solution set

[warped relic]

Yep that's the full description of the solution

[devout snow]

@sullen nova Has your question been resolved?

[sullen nova]

wait why

sorry im back

had to eat dinner

[devout snow]

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[warped relic]

.reopen

[devout snow]

✅

[warped relic]

Lol

[sullen nova]

was i correct or not💀

[warped relic]

Yep

You're all set

[sullen nova]

great

thank you sm for your help

i appreciate it

[warped relic]

[sullen nova]

have a good rest of your day !

[warped relic]

Thank you! You too!

[sullen nova]

thanks :))

.close

[devout snow]

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[worn briar]

In number 23, the book says the answer is 20. I got x=10. (X is a segment length, so it cant be 0.) Is this an issue with my algebra?

[worn briar]

If this helps, here is the question.

I also had the same issue on number 25, but I think if understand my issue in problem 23 I can apply it to the other one.

[red sierra]

You got until 4x^2-80x=0 right

[worn briar]

Whats wrong with my quadratic equation?

Let me get my calculator out again

[red sierra]

How did you learn to solve quads?

Sum and product or quad formula?

Or factor?

[worn briar]

Never heard of sum and product but i use quadratic formula because I tried for days to understand factoring but it was always super inconsistent w/me

[red sierra]

Can I see ur work on solving 4x^2-80x=0?

I'll point out what you got wrong

[worn briar]

Ok. Ill redo it, I probably should have done that earlier

[red sierra]

Oh wait

[worn briar]

I write with shortcuts and fill in the blanks in my head normally

[red sierra]

(80+80)/8 isn't 10

[worn briar]

oml

😭

Ok well I had a similar issue on number 25

let me make sure it isnt the same

[red sierra]

Can you send the question for 25?

[worn briar]

Ok, the book got 2.5

I see

i forgot to write x^2 so i just did bad algebra

[red sierra]

Yeah

You forgot 42x^2

[worn briar]

Thanks man, that shouldnt have been an issue lol

.close

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[worn briar]

I have to find x. Where do I go from here?

[devout snow]

@worn briar Has your question been resolved?

[devout snow]

@worn briar Has your question been resolved?

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[devout snow]

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[unreal moat]

Two numbers are chosen at random from the interval −20 ≤ 𝑥 ≤ 10. Probability of the product of the two numbers
is a positive value
a. 1/9
b. 1/3
c. 4/9
d. 5/9
e. 2/3

[devout snow]

Please don't occupy multiple help channels.

[unreal moat]

@dry robin

<@&286206848099549185>

@royal thorn

pls help

[untold lance]

We will help.

[unreal moat]

how?

pls

[untold lance]

Before that, please don't occupy multiple channels. We get the desperation.

[untold lance]

How do you choose numbers such that the product is positive?

[unreal moat]

idk

pick a number betwen the interval

[untold lance]

....calm down first.

tell me what is (-1) X (-2)?

[unreal moat]

2

[untold lance]

right.

[unreal moat]

then?

[untold lance]

and (-1) x (2)?

[unreal moat]

still 2

[untold lance]

sorry

[unreal moat]

its probality

[untold lance]

it's -2 isn;t it?

please listen.

[unreal moat]

yeaps

its must be negatif times negatif and poositice times positive

[untold lance]

So that means - we need to choose two numbers with same sign whenever we try to product.

[untold lance]

very good.

So, the number of ways to choose positive product

[unreal moat]

i do (20 C 2 + 10 C 2)/ 31 C 2

but it doesnt gave me the answer

i got 5/9 but in the wrong way

[untold lance]

wait, it doesn't ahve to be only integers I think.

and if that is the case, I might not be able to help, as I haven't learnt much about Probability yet.

Interval means a set of real numbers so

[unreal moat]

i mean you only can pick integers

its not the real number

0,1, 2/4 cannot be picked

so it must be integers

@dry robin

[regal torrent]

Bro

[unreal moat]

i do used bro

but i cant find

[regal torrent]

What r total cases

I mean sample space

[unreal moat]

idk

[regal torrent]

31^2?

[unreal moat]

Two numbers are chosen at random from the interval −20 ≤ 𝑥 ≤ 10. Probability of the product of the two numbers
is a positive value
a. 1/9
b. 1/3
c. 4/9
d. 5/9
e. 2/3

[untold lance]

bruh

[unreal moat]

theres is 31 numbers betwen the intervals

[untold lance]

is repetition allowed?

[regal torrent]

Ye

[unreal moat]

i guess its allowed

[regal torrent]

31^2

[untold lance]

that's why it's wrong

[unreal moat]

hmm

[untold lance]

it's

[red sierra]

It should be P, not C if repetition is allowed

[regal torrent]

If it it's integres

[unreal moat]

idk but somehow i got d as the answers

[red sierra]

Ohh hold on

It should be (20^2+10^2)/31^2

[untold lance]

((20x20) + (10x10))/(31x31)

[red sierra]

Since the two numbers can be the same

[untold lance]

yeah it's permutation with repetition

[regal torrent]

Xactly

[untold lance]

500/961 ig

[regal torrent]

Ye

[unreal moat]

i do with all over 30^2 and got 5/9

[red sierra]

Huh but that's not in the answer

[unreal moat]

@dry robin

[untold lance]

bruh...

[untold lance]

bro

stop

[unreal moat]

@oak widget

@elfin lark

[untold lance]

what the fuck

[red sierra]

Stop pinging

[regal torrent]

Dont

[untold lance]

stop

[regal torrent]

You'll get banned

[red sierra]

Somebody ban him or smth

[regal torrent]

No no

Dont

Ban

But dont repeat

[untold lance]

oh.

they're excluding 0 from the sample space

this question is ridiculuous

[red sierra]

Huh what

[untold lance]

reminded me again why I hated mcqs

[untold lance]

30^2 = 900 and 500/900 = 5/9, d

[red sierra]

0 is an integer bruhh

[untold lance]

we all know

the question just sucks

the answer should be 500/961

[red sierra]

Everyone knows except this question ig

[regal torrent]

They shdve said msthg like

[regal torrent]

Given that the number aint 0

Conditional probability op

[unreal moat]

okay

i found it

[regal torrent]

Kool

[untold lance]

aid

[unreal moat]

in order to get positif theres 2 way

• • • and - * -
      so to get both negatif is (20 C 1 * 19 C 2)/31^2
      do the same thing to get positif, so in th end you got 470/961

sp 5/9

[untold lance]

wat

[devout snow]

@unreal moat Has your question been resolved?

Channel closed

Closed by @unreal moat

Use .reopen if this was a mistake.

[unreal moat]

yeah

[devout snow]

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[restive river]

Hi i'm doing some chemistry stuff but I face a problem algebraic problem I have this equation: $$-104000 = -8.314 \cdot 298 \cdot \ln{K_{eq}}$$ the final result is: $$K_{eq} = e^{4.19765802} = 66.53033572$$ but unfortunately my book telling me that this is wrong

[woven radish]

Hin

[restive river]

(wrong in calculating the final result)

The calculation seems pretty much fine to me
(Ignore, I misread)

not a sceintific mistake

,w exp{10400/(8.314×298)}

[winter patrol]

where.s 4.19... coming from

,calc -104000/(-8.314*298)

[woven radish]

Result:

41.976580297162

[restive river]

I see

[winter patrol]

seems your value was off by a factor of 10

[restive river]

my book telling me that the answer is 1.7 * 10^18

[winter patrol]

,calc e^(-104000/(-8.314*298))

[woven radish]

Result:

1.6990149182697e+18

[winter patrol]

yeh, sounds about right

41**.976580297162 = ln(k)
you should've had that instead of
4.**1976580297162 = ln(k)

[restive river]

Ahh I see

thanks for the help

[restive river]

yeah I missed the final 0 in 104000 thanks

.close

[devout snow]

Channel closed

Closed by @hoary shuttle

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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.

[slim rivet]

Would simplifying surds consist of, root9 = 3(root3) ?

[sick fulcrum]

root9 is sqrt(9) right?

if so that is wrong

sqrt(9) = 3

[slim rivet]

oh right, that one can just simplify

your right

although what if its a number like, root42? does that become 7root6?

how does it work

or instead, 6(root7)?

[hushed wraith]

try to rwite the number as product of a perfect square and another number

and use the fact that $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$

[woven radish]

SilverSoldier

[slim rivet]

okat so it would become, (root7)(root6)?

[hushed wraith]

42 cannt be witten as the product of a perfect square and another number

[hushed wraith]

that is correct

$\sqrt{42}=\sqrt{6}\times\sqrt{7}$

[woven radish]

SilverSoldier

[slim rivet]

but then does, root6 become, (root3)(root2)

[hushed wraith]

yes

[slim rivet]

when does a number become a coefficient of a root?

[hushed wraith]

say u had sqrt(50)

$\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}$

[woven radish]

SilverSoldier

[hushed wraith]

and $\sqrt{25}=5$

[woven radish]

SilverSoldier

[hushed wraith]

so $\sqrt{50}=5\times\sqrt{2}=5\sqrt{2}$

[slim rivet]

oh i see