precalculus

[short pilot]

How?

[willow bear]

0.01 = 1/100

1/100 = 1/10^2

1/10^2 = 10^-2

[honest void]

help me guys

[short pilot]

Now i get it, thanks!

[rich lynx]

yes one day ill understand all this

[sharp mist]

Atta spirit

[low slate]

y=2^{x-2}+1 how would u write this in logarthmic form?

[viscid thistle]

by solving for x?

isolate the exponent using the log function

[low slate]

It’s for graphing

[viscid thistle]

3^2=9 in logarithmic form would be log_3 (9) = 2

[low slate]

But I would solve it when it’s y=

How would i*

[viscid thistle]

isolate the exponent

subtract by one
log_b (x) = y, x=b^y
subtract by 2
x is isolated

[low slate]

hmm

i dont think i understand

cuz the 2 would be the base

[viscid thistle]

do you understand the logarithmic function?

correct, 2 would be the base

[low slate]

but the y is throwing me off

and the +1

cuz other wise it would just be log_2 x-2

[red tree]

How is y throwing you off

[low slate]

nvm nvm we gud

i just read the instructions wrong

[red tree]

How was logarithm defined in your precalculus class?

Just seeing how it's done in your class

[short pilot]

Do u guys know why it is -5/2?

[short pilot]

What is wrong in my answer?

[uncut mulch]

wtf is that?

[short pilot]

the exponents are added right?

[uncut mulch]

the notation itself is nonsense

[short pilot]

Let me do it again

[uncut mulch]

and clearly show your steps

[short pilot]

Shit

Now it less crearly xd

Let me do it again pls

Now is crearly

[uncut mulch]

you're making the same notation crime

[short pilot]

What is ?

[uncut mulch]

$a^m \cdot a^n = a^{m+n}$

[obsidian monolith]

ℝamonov

[uncut mulch]

however there is another issue

with how you only got
$$3^\frac13 \cdot 3^\frac12$$
on the numerator

[obsidian monolith]

ℝamonov

[short pilot]

And that is 3⅚, right?

[uncut mulch]

note that the cube root covers the whole of $3\sqrt3$ and not just the 3

[obsidian monolith]

ℝamonov

[uncut mulch]

$\cbrt{3\sqrt{3}} \neq \cbrt{3}\cdot \sqrt{3}$

[obsidian monolith]

ℝamonov

[short pilot]

Aaaaa

[short pilot]

Its not this

[short pilot]

How can i express that multiplication?

[uncut mulch]

you could consider first expressing 3sqrt(3) with exponents

and/or intuitively identify the cube root of 3sqrt(3)

[short pilot]

Something like this right?

[uncut mulch]

yes

that works

[short pilot]

Aaaaaa

Man

Thank youuuu

U really help me

[uncut mulch]

and in case it wasn't clear before, that red $\red{a}$ doesn't belong there
$$\overset{,,, ,,,,m + n}{a , , \red{a}}$$

[obsidian monolith]

ℝamonov

[frigid cradle]

someone please explain how to solve these questions

[ionic drum]

Its been a while since I have done these but I feel the method for v38 is that you use ijk notation and then you can express the terms using trig. Do you know trig ?

[soft cargo]

for 39) forces of the two men need to be bigger than the normal force, to be able to lift (not 100% sure just a guess)

with normal force I mean F_n not sure if it's called normal force in English

[red tree]

What is a variable? What is a function? How do you use a variable in a sentence? How do you read a sentence containing a variable?

Take a shot at it

[weak needle]

This is basic

Look in any webpage or wikipedia for definitions

#prealg-and-algebra

[red tree]

Is that what you would say to someone struggling in reading math with variables?

[weak needle]

No, that's what I tell to someone that hasn't looked for answers in the most basic sites

[willow bear]

maybe you would do well to wait another year before going on discord

[weak needle]

@willow bear rly?

[red tree]

I'm not the one struggling, I am just proposing some basic questions to answer the other person's question

[willow bear]

discord TOS requires you to be 13+

[weak needle]

Oh, then it's a misunderstanding

[weak needle]

I hate to say this, but real life doesn't work like that

@misty obsidian Fire away dude

[red tree]

Although let's be real; if the internet was sufficient to learn how to read math with variables, we wouldn't have so many people struggling to do so, even many people among the top quartile of math students

and over 18, etc

[weak needle]

Mate, cool it with the spam

[weak needle]

Well, that's true. But the message read as if someone came here and asked "What is a variable?"

[red tree]

That's one use of variables

[stuck lark]

r u actually 12

[red tree]

Another one is as follows: for all integers x, there's an integer y such that x > y. True or false?
There exists an integer x such that for all integers y, x > y. True or false?

[stuck lark]

man

ur aware of discord terms of service?

its illegal for ppl under 13 to use discord. im sorry, u can rejoin in 6 months

[weak needle]

Dude....

It's for fucking real?

I knew Americans were sheltered but this is just plain ridiculous

[stuck lark]

https://discord.com/terms

[weak needle]

As long as nobody is messing with him or he isn't messing with anybody, he could have more guidance by being exposed to math in this server

[red tree]

action on the server and/or owner is scary

Imagine Discord shutting down a server of 10k people because of a single 12 year old allowed to roam

[weak needle]

So, the lesson for a child that comes to look for help here: Do not disclose age.

[ornate cairn]

Yes

[viscid thistle]

@low slate

[low slate]

could u simplify this to just log4 x^2-1?

[viscid thistle]

well you can use log properties to simplify it, but I don't know about log_4(x^2 - 1)

[low slate]

would this be most simplified

[viscid thistle]

yeah

well

I think so

it'd be $log_4 ((x)^{\frac{1}{7}} (x-1))$

[obsidian monolith]

kanga gang drug mule no. 2

[viscid thistle]

since you have to distribute that 1/7 into the square brackets

the second log's 7 would be "cancelled out" by the 1/7

[fiery creek]

Anyone know how to solve this problem

I’m confused on the diagram even or how to interpret the problem

[soft cargo]

Amplitude is obvious, period also,

phi is the phase swift

https://courses.lumenlearning.com/physics/chapter/16-3-simple-harmonic-motion-a-special-periodic-motion/

This might help, but what do they mean with range?

[outer briar]

Hi

I'm looking for a pre-calclus book with a solution manual online. But i couldn't get it

[viscid thistle]

PreCalculus by Axler has full worked out solutions to exercises included in the main text itself.

[outer briar]

Okay, does it has a lot of examples?

...... Do you guys know any book?

[shut folio]

(DS1+DS2) x 0,7 + TP x 0,3) x 2/3 + DS3 x 1/3

if the total of all this is 100, what is the value of DS1, DS2 and TP?

[willow bear]

are all those x's meant to be multiplication symbols?

also, you've got a missing ( somewhere but it's not clear where

(and as it stands, one equation is not enough to find the values of three unknowns...)

@shut folio do you have any context at all for this equation or did you just pull it out of thin air?

[shut folio]

yes the x represents the multiplication, it is to calculate my academic grades x) the DS1, DS2, TP represent the evaluations

and I don't understand anything about their stuff x)

[willow bear]

you don't understand anything?

[shut folio]

Yes, but I just managed to find the values ​​I think

[willow bear]

do i understand correctly that you are given zero explanation or info about how your academic grade is calculated from DS1, DS2, DS3 and TP?

[shut folio]

yes

[willow bear]

then where did your formula come from?

[outer briar]

Do you guys know any book?

[low slate]

2nd one would have 3 i believe

not sure tho

[teal pivot]

I saw I video online about finding i^173 but their explanation was not helpful so could someone explain it please

[tranquil haven]

I've been struggling with the epsilon delta definition for a week, will it do damage if I just skip over it for now?

I still cannot come up with proves for many problems

[spiral bolt]

So if we look at the powers of i (just by multiplying) we can find a simple pattern.
i^0=1 (obviously)
i^1=i
i^2=-1 (def of i)
i^3=-i
i^4=1
i^5=i
Here we can see that i^1=i^5=i^10...
So for any i^k=i where k is a multiple of 5
Then we can say that i^173=i^170*i^3
and then i^170=i
So we have i * i^3=i^4
i^4=1 from the table, so the answer is 1. Hope this helped!

[willow bear]

i^5=i^10

no

i^10 is not equal to i^5.

and i^k = i need not be true for k a multiple of 5. for example, i^20 = 1, not i as you claim.

@spiral bolt do not spread misinformation

the pattern is that i raised to multiples of 4 gives 1.

$i^{4n} = 1$

[obsidian monolith]

Kanga Gang Annihilator Ann

[spiral bolt]

wait sorry I'm messing up today

🤦 seems like I'm not doing very well today...

[teal pivot]

Thanks alot!!Helped 10 fold

I cant thank you enough

[spiral bolt]

Sorry I messed up a bit with that lol

Someone else posted a solution using the same method, I gave you the wrong pattern...

[teal pivot]

I saw dont worry (:

Still helpful Tho!!!!

[shy kettle]

how do i find the formula of an exponential function given two coordinate points?

[pale talon]

could someone help me PLEASE ASAP with this problem? <@&286206848099549185>

[willow bear]

@pale talon do you still need help with this or did the deadline happen already?

[modern peak]

know any place I can find review problems for pre-cal unit 5 stuff? (trig and inverses)

[pale talon]

i still need help

[pale talon]

on khan academy

[willow bear]

so, have you made any progress so far?

[pale talon]

yeah what i wrote in the box is all my progress so far

[willow bear]

oh, that was stuff you wrote

what's the second paragraph about?

it appears that you have already given your answer in the first, and your answer even has the nice property of being correct.

[pale talon]

yeah i'm not sure i just made an observation that i thought would be helpful to include

[pale talon]

yeah but i need help elaborating more on it

[willow bear]

do not reply-ping me so much.

it's annoying.

have you received feedback from your teacher that the explanation you've written is insufficient?

[pale talon]

yea, i need to write a bit more they said

[willow bear]

that's weird, because what you've written looks perfectly fine to me.

[pale talon]

idk how to explain more which is weird

i'm going to bed, so anything else u wanna add?

[willow bear]

no.

[red tree]

?

Congratulations

[tepid cloak]

can someone explain to me what concentric means and what does it have to do with the problem below thanks

how do I solve for that

[uncut mulch]

first complete the square and find the centre of the given circle

[tepid cloak]

the center is (2, -3)

[uncut mulch]

is not the correct centre

[tepid cloak]

oh

oh wait I solved a different problem sorry

center is (-2 , 1)

[uncut mulch]

yes

and then apply formula for distance from a point to a line

[viscid thistle]

my friend is being the hell

[tepid cloak]

169x^2 + 169y^2 + 676x - 338y - 1156 = 0??

[uncut mulch]

what's that supposed to be?

[tepid cloak]

the general equation im not sure

[uncut mulch]

,w distance from (-2,1) to 12x-5y-20=0

,w (49/13)^2

,w radius of 169x^2+169y^2+676x-338y-1156=0

looks ok

[tepid cloak]

ok last how do you determine if the point is on, inside, or outside the circle

[uncut mulch]

compare distance to the centre to the radius

[tepid cloak]

is this correct welp

I did point a

[uncut mulch]

yeh

[tepid cloak]

ok thankss

[azure karma]

Can someone help me understand what I'm doing wrong here

[vestal zenith]

you can't apply the log rule $log_{a}(x^{b})=b \cdot log_{a}$ to the square roots

[obsidian monolith]

seven

[azure karma]

how would I go about fixing this then

[vestal zenith]

you almost did it right, but try to do the problem again, but without using the log rule on the roots

[azure karma]

ahh I see.

so I can't simplify the square root part basically

[vestal zenith]

yeah

[tame canopy]

hello i need help with like everything could some hop in a call with me and help?

[normal bone]

Why cant we take the roots out tho? Dont they act like exponents too?

[willow bear]

yes you can. roots are just fractional exponents.

do not spread misinformation.

[normal bone]

Heyy

x² + 2x +1

(x+1)²

Thats why

Then we can take the 2 out

[vestal zenith]

Yeah, I know, but i’m just saying if they want the right answer, they shouldn’t apply the log rule to the roots.

[willow bear]

why not tho

[vestal zenith]

because it says that their answer is wrong, so I assume it must be $3ln(x)+ln(\sqrt{x+6})-ln(\sqrt[7]{x^2+2x+1})$ instead

[obsidian monolith]

seven

[tranquil haven]

How to deal with an epsilon delta proof, when the interval in which x has to be, is undefined for let's say 5

(x, 5) and (5, y) where x may be in

If c is in (5, y) should I count 5 as the left-endpoint?

$|\sqrt{x - 5} - 2| < \epsilon$

$- \epsilon < \sqrt{x - 5} - 2 < \epsilon$

$2 - \epsilon < \sqrt{x - 5} < \epsilon + 2$

$\epsilon^2 - 4\epsilon + 4 < x - 5 < \epsilon^2 + 4 \epsilon + 4,\ x \neq 5$

$\epsilon^2 - 4\epsilon + 9 < x < \epsilon^2 + 4 \epsilon + 9,\ x \neq 5$

[obsidian monolith]

sentinel

[tranquil haven]

For the limit $\lim_{x \to 9} \sqrt{x - 5} = 2$

[obsidian monolith]

sentinel

[tranquil haven]

So short together, when figuring out delta, does the interval in which delta may lay, need to be continuous?

[tranquil haven]

Thanks for the answers! I would never figure it out on my own, especially since the book says nothing about this

And especially because all the other explainations basically void this

Even teachers ignore me so that's nice

[broken pelican]

can anyone give me some mathematical induction examples? even 3 items will do, thanks!
if it's okay of course...

[tight compass]

Prove $\sum_{j=1}^{n}$j = $\frac{n(n+1)}{2}$.

Proof. Let P(n) be a statement of natural number that satisfy that above formula.The formula clear holds for n=1, now assume it holds for any natural number k. That is $\sum_{j=1}^{k}$ j = $\frac{k(k+1)}{2}$. To show it holds for k + 1 we have that is $\sum_{j=1}^{k+1}$ j = ($\sum_{j=1}^{k}$ j) + (k + 1) = $\frac{k(k+1)}{2}$ + (k+1) =
$\frac{(k+1)(k+2)}{2}$. Thus it formula for k+1, therefore by principle of induction it holds for all natural number n.

[obsidian monolith]

plegasus

[normal bone]

Ah nop

It was due to the (x+1)² expansion that they had to simplify and then take the 2 out

[tight compass]

Sorry It has grammar errors due to being on my phone.

[broken pelican]

okay thank youuu!

[raw oriole]

hey guys im currently in precalc but im struggling and am wondering if anyone in here could help me or how i could get help, i keep watching videos but cant get it down,

[red tree]

Videos mostly all teach the same way

[tight compass]

Even when I already know most of this stuff, I still find this site very helpful https://www.mathsisfun.com/algebra/index-2.html. Going through the log section part and there practice problems for it.

Also go through the exponent section too since most people lack of understanding for logs comes from a lack of understanding of exponents.

[copper vigil]

wow

[shut folio]

Don't understand why the integral of that :

have this result :

When we see that only x depends

[fleet yew]

it's a definite integral

so it will have the form F(L) - F(0)

@shut folio

there should not be an x once evaluated

[shut folio]

Here it's a bit the same thing, except that they add (-1)^n and ² unlike the first integral that I had to show yesterday

[copper vigil]

yeah ok and?

there should not be any terms in x

because x is a dummy variable

[muted owl]

best channel for that

[gloomy elbow]

Could skmehelp me with this

[gloomy elbow]

<@&286206848099549185>

do i just find the intersection

of the 3 lines

[gleaming owl]

I think yes

I mean that's what we need for solving those equations, intersection point

[mild swan]

You can also check this on a calculator

[graceful gulch]

Solve for the sum of all positive integers, from 5 to 1255, that can be divided by 5.

pls help

[mild swan]

Start by writing out the first few terms to get a feel for how the sum behaves. Given that each number you sum will be a multiple of 5, factor out 5 to get the sum of the first n positive integers

Then use the appropriate formula after you find n for your sum

[amber shuttle]

i have a quiz tomorrow on graphing trig functions

i have everything down except tangent and cotangent

does anyone have resources to hep learns

im really confused

[tepid cloak]

helpp

[mental steeple]

How do I find all the roots for x^5-27x^2=0

?

[upper quartz]

helpers in this server rarely help people, so i suggest you find another server

you can look at the questions above, nobody help them

[vivid void]

@mental steeple try #❓how-to-get-help

the available channels under there

[mental steeple]

Okay

[viscid thistle]

write $x^5-27x^2 = x^2(x^3-27)$. Now $x^2(x^3-27) = 0$ if and only if $x^2 = 0$ or $x^3-27 = 0$. What are the $x$'s that make either of these equalities true?

[obsidian monolith]

Botnuke

[mental steeple]

3?

[viscid thistle]

that's one

[mental steeple]

0

[viscid thistle]

that's good too

[mental steeple]

okay but what I do next

[viscid thistle]

are you solving this equation over the real numbers or the complex numbers?

[mental steeple]

This is what my teacher got for the answrr

[viscid thistle]

ok, complex numbers then

[mental steeple]

He got that without a graph

[viscid thistle]

you can factor x^3 - 27 with the knowledge that (x - 3) is a factor

[mental steeple]

Where is the x^3?

[viscid thistle]

divide x^3 - 27 by (x-3) or do something equivalent to that, then you can write x^3 - 27 as (x-3) times some 2nd degree polynomial

[mental steeple]

Ohh nvm

I see the x^3

[mental steeple]

Can you do that on a paper?

Or digitaly is fine

[viscid thistle]

hmmm there should be plenty of online videos or sites that gives you a step by step guide on how to divide polynomials

[mental steeple]

Ik how to do it I just don’t understand that specific question and the format

[viscid thistle]

you know how to divide x^3 - 27 by (x-3), but you don't know why it's relevant to the problem?

[mental steeple]

There’s no need to divide

Cause I’ve done many other problems and didn’t divide once

[viscid thistle]

yea you don't need to, I'm proposing one way to do the problem

[mental steeple]

Just multiplied

[mental steeple]

Just haven’t learned that way so it’s confusing

no hate

[viscid thistle]

ok what else do you know about complex numbers then? you know polar coordinates?

[mental steeple]

This is the way I do it can you show me how to do #11 that way

Oh

okay so what I tried was x3-27 = 0 add 27 both sides and then I get x^3=27

Is that correct?

[viscid thistle]

sure

[mental steeple]

So what do I do now lol

[viscid thistle]

not sure how that's going to help unless you want to think in polar coordinates

[mental steeple]

I don’t even know what that is

[viscid thistle]

also not sure how to factor x^5 - 27x^2 = 0 by grouping

unlucky

maybe you'll have to have to come up with another way to factor it

[mental steeple]

Can I use the quadratic formula

Or something

[viscid thistle]

well, yea, kinda

[mental steeple]

If I can’t factor than maybe that’s the way

[viscid thistle]

above, I suggested a way to turn into a problem where the quadratic formula can be used

[mental steeple]

Where?

[viscid thistle]

here

[mental steeple]

So basically factor x^3-27?

[viscid thistle]

yes

[mental steeple]

3x(x^2-9)

Wait no

3(x^3-9)

[viscid thistle]

no

[mental steeple]

What

[viscid thistle]

3(x^3-9) = 3x^3 - 27

[mental steeple]

Ohh

How do u factor that

[viscid thistle]

you know (x-3) is a factor

so you know x^3-27 is (x-3) times some 2nd degree polynomial

one way to determine that 2nd degree polynomial is with polynomial division

another is to write (x-3)(ax^2 + bx + c) = x^3 - 27, multiply the left side out, and determine the values of a, b, and c by comparing the coefficients of each side of the equation

division is quicker if you know how to do it

[mental steeple]

no I don’t

[viscid thistle]

the multiplication way is always safe if you forget/never learned how to divide polynomials

[mental steeple]

Okay let’s do it your way

[viscid thistle]

which way is that? division or multiplication?

[mental steeple]

Multiplication

[viscid thistle]

ok, multiply out (x-3)(ax^2 + bx + c)

[mental steeple]

How do Ik what to put for ax bx or c

[viscid thistle]

you'll know after, when you multiply that expression out and set it to equal to x^3 - 27

just leave them as letters for now

[mental steeple]

Okay

[viscid thistle]

two polynomials in x are equal to each other for all x if and only if their corresponding coefficients are equal

[mental steeple]

wdym

Can you show me how to do it I understand better like that

like an example

or something

[viscid thistle]

the way I wrote it was confusing but what I mean is (x-3)(ax^2 + bx + c) is a third degree polynomial, and so is x^3 - 27. If these expressions are to be equal for all x, then the coefficient of the x^3 term in (x-3)(ax^2 + bx + c) must be the same as coefficient of the x^3 term in x^3 - 27 (i.e. 1), coefficient of the x^2 term in (x-3)(ax^2 + bx + c) must be the same as coefficient of the x^2 term in x^3 - 27 (i.e. 0), etc. etc.

when you multiply (x-3)(ax^2 + bx + c) out, you will see that the coefficients are in terms of a, b, and c

[mental steeple]

Okay so where do I begin do I just mutpilply x*ax^2?

[viscid thistle]

yea, plain old distributive property

[mental steeple]

Okay

So the answer is ax^3?

For that

[viscid thistle]

what do you mean? x*ax^2 = ax^3, yes

[mental steeple]

Yes

Okay now what

Combine like terms?

[viscid thistle]

yea

[mental steeple]

Btw that c^2 is supposed to be an cx

My bad

How do I combine bx^2-3ax^2?

[viscid thistle]

bx^2-3ax^2 = (b-3a)x^2

[mental steeple]

Oh we do it like that

Did I do it right?

[viscid thistle]

yep looks right

[mental steeple]

Okay now what

[viscid thistle]

recall that we are trying to find the a, b, and c so that this equality holds (x-3)(ax^2 + bx + c) = x^3 - 27

[viscid thistle]

did you understand this?

[mental steeple]

Not really can you explain like in numbers lol not in text

If that’s okay

it’s fine if u can’t

[viscid thistle]

what's the coefficient of x^3 in (x-3)(ax^2 + bx + c)?

[mental steeple]

X^2?

[viscid thistle]

huh?

[mental steeple]

What’s the leading term

X?

[viscid thistle]

does the phrase 'coefficient of a term in a polynomial' have meaning to you?

[mental steeple]

a+ b

[viscid thistle]

don't think that's a reply to my question, nor is it a correct answer to anything else I've asked

[mental steeple]

The number that is being multiplied by a variable which is a and b

C is constant

[viscid thistle]

what

[mental steeple]

Wait so what’s the answer?

What’s the coefficient

[viscid thistle]

a is the coefficient of x^2 in the polynomial ax^2 + bx + c

(x-3)(ax^2 + bx + c) is a different polynomial, and I'm asking for the coefficient of the x^3 term

[mental steeple]

Oh

Well that would be a

[viscid thistle]

yes ok, good

now what's the coefficient of the x^3 term in x^3 - 27?

[mental steeple]

a?

[viscid thistle]

a is nowhere in sight

[mental steeple]

How do u see these

All I see is x and 27

[viscid thistle]

you don't see an x^3?

[mental steeple]

yes

That’s the coefficient

[viscid thistle]

huh?

[mental steeple]

Man idk

[viscid thistle]

what if I write it as 1*x^3

[mental steeple]

oh

Lmfao

Yes it’s 1

[viscid thistle]

ok great. Now the coefficient of x^3 in (x-3)(ax^2 + bx + c) must be equal to the coefficient of x^3 in x^3 - 27

so what can you say about a?

[mental steeple]

1x^2

[viscid thistle]

what is the meaning of that

[mental steeple]

X^2

[viscid thistle]

??

[mental steeple]

A is the coefficient

[viscid thistle]

??

[mental steeple]

What are u asking again

[viscid thistle]

I'm asking you to deduce something about the value of a from this mysterious fact I've posted

[mental steeple]

Yea it’s the coefficient

What else is there to say

[viscid thistle]

you can say a = 1...

a is the coefficient of x^3 in (x-3)(ax^2 + bx + c), which is equal to the coefficient of x^3 in x^3 - 27, which is 1

hence a = 1

analogous things apply to the coefficients of other terms of the polynomials

use them to determine the values of b and c

[mental steeple]

That’s what I said

[viscid thistle]

then you're done

[mental steeple]

Look

[viscid thistle]

where is "a = 1" written in any of your messages

[mental steeple]

It’s obvious I plugged in 1 for a

Anyways what’s next

[viscid thistle]

I explained above what's next

but no, nothing you said indicated that you deduced a = 1

[mental steeple]

Ok

[mental steeple]

I still don’t understand how to find b and c

[viscid thistle]

then you don't understand what happened above

basically the exact same thing

[mental steeple]

Oh so 1?

I plug in 1 for b and c

?

[viscid thistle]

What? No

What makes you want to do that

[mental steeple]

Idk what I’m doing tbh for b and c

[viscid thistle]

Explain to me how you deduced a = 1

[mental steeple]

From x^3-27

[viscid thistle]

That is not an explanation

[mental steeple]

I’m lost

[viscid thistle]

I explained it earlier

[mental steeple]

Yea I get that 1 is the coefficient

A=1

But the other stuff u said I’m lost

[fleet yew]

1x^3 + 0x^2 + 0x - 27

[mental steeple]

Oh

[mental steeple]

Wait so how do I get the answer tho

@fleet yew

[errant laurel]

hi do you need some help

[mental steeple]

Please

I’ve been stuck on this question for the past 2 hours

[errant laurel]

ok whats the question

[mental steeple]

X^5-27x^2=0

[errant laurel]

thats not bad

[mental steeple]

My teacher got this as the answer

[errant laurel]

are you confused on how to approach it

[mental steeple]

yea

[errant laurel]

ummm

[mental steeple]

On how to do it

[errant laurel]

ill see what we can do

[mental steeple]

okay

[errant laurel]

first step is to factor out x^2

x^2(x^3 -27) =0

[mental steeple]

okay

[errant laurel]

now its a difference of cubes

do you know how to set that up

[mental steeple]

Not really

[errant laurel]

ok

i can help

[mental steeple]

Bet

[errant laurel]

so we know that 27 is a perfect cube right

[mental steeple]

Yep

[errant laurel]

so a formula is

x^3-a^3 = (x-a)(x^2+ax+a^2)

so this will be

(x-3)(x^2 +3x +9)

so we get to

x^2(x-3)(X^2 +3X +9) =0

now we can solve a couple different parts

we can conclude that x=0 and x=3 are solutions just by looking at what we have

btw are you all cleared up rn

or do you want me to slow down

[mental steeple]

Yea slow down lol

[errant laurel]

oh sorry

if you want, we can go into a vc to go over this

[mental steeple]

okay bet let me get my headphones

[errant laurel]

ok ill be in off-topic-voice

[mental steeple]

What was the equation you used that you put into the quadratic formula can you write it here

@errant laurel

[errant laurel]

x^2 +3x +9

[mental steeple]

Ok this is what I have so far and didn’t get x^2 +3x +9

Ohh

nvm 🤦

Since 2 numbers can’t multiply to get 9 and add to get 3 we do quadratic formula right

@errant laurel

[errant laurel]

ye

exaclty

so use quadratic formula to get ur last 2 solutions

[mental steeple]

Ok so I got this

Squat root of -27 is -3 sqrt 3

Right

Thanks bro life savior lol

Hey can you help me with this one @errant laurel

I’m stuck on the long division

[errant laurel]

sure ig

but this ones a lil different

first lets solve by setting the equation to 0

[mental steeple]

It’s already set to 0

right?

[errant laurel]

in this sort of scenario you have to guess and check factors

[errant laurel]

there are 2 things to solve for

f(x)=0

and f(x) = -4 +i

so lets do 0 firt

[mental steeple]

Here let me show you the I started off this is how my teacher showed us

[errant laurel]

ok

[mental steeple]

Ignore where it says x^2-8x+25 that’s a different problem

My long division is off

[errant laurel]

dude idek what ur teacher is saying

[mental steeple]

https://youtube.com/watch?v=88GEurUtzjU&feature=share

Basically this video explains the long division portion of what I’m doing

It’s short

[errant laurel]

ok

im really not the best in this topic

im sorry but i might not be able to help you as well as some others

maybe try pinging helpers

[mental steeple]

Ur good

Oh lmao I see what i did wrong

[sharp mist]

When do you learn how to divide polynomials, precalc or algebra?

[vocal jetty]

I remember learning that in Algebra II.

[mild swan]

Polynomial long division in either Algebra II or Precalculus

[viscid thistle]

aye

$20 + 6sin(Q) = 3$

[obsidian monolith]

codemonkey

[viscid thistle]

is impossilb e

right

because sin of -17/6 does not exist

since it does not lie on unit circle

[willow bear]

sin^-1(-17/6) does not exist, yes.

this equation has no solutions.

[viscid thistle]

so then should i email my professor

He does this a lot where he wants you to email him on the test that its not possible

like the whole question is to see if you can see its impossible, but doesnt put that as a possible answer

[willow bear]

yes, you should definitely email him

[viscid thistle]

Thank you as always you are my go to!

[mild stone]

isnt the factor a in the general quadratic equation pretty much the same as the factor m in the general linear equation?

[muted owl]

kind of

the A coefficient modifies the parabola's shape while the B and C coefficients just modify the locality

i recommend you geogebra/desmos for function graphing

[mild stone]

yeah i noticed that and my teacher said i was wrong for saying so

[west mortar]

Hey can someone help me with this one i am stumped hard

[copper vigil]

Οκ σιτρ

Whoops

[timid bloom]

For a) first answer what the domain of ln(u) is (u is a dummy variable). Then set u = x^2 - 9. (The +3 term in h(x) will not impact the domain). Solve for the new x value(s) that restrict the domain.

for b) I'm going to abuse notation since it's easier to type x and y than it is to type h inverse of x
if y = ln(x^2 - 9) +3,
find the inverse by setting x = ln(y^2 - 9) + 3, and solving for y

[west mortar]

ok thanks i appreciate it'

[timid bloom]

Np. Does that clear it up?

[west mortar]

Yes thank you

[manic stag]

Is precalc really harder than calc

My friend told me that half the stuff in precalc you dont use in actual calc and you're introduced to them anyway in calc

[red tree]

There's a very ancient saying: If you're struggling in calc, it's because of the algebra

[uncut mulch]

depending on how the course is structured, it helps with understanding calc

[haughty lance]

Really depends

Most of the stuff in precalc is useful if you do enough calculus

I can't think of anything that's not useful in calculus

[shut folio]

I have a physical problem with this expression
We want to determine the constants, B, D

$V(x) = B \sinh(\beta_{1} L) + D \sin(\beta_{2} L)$

[obsidian monolith]

le J

[shut folio]

We know the following boundary conditions:

$V(L) = 0$ and $V^{\prime \prime}(L) = 0$

[obsidian monolith]

le J

[shut folio]

So we calculate that with our starting expression

$V(L)=0$ so $B \sinh \left(\beta_{1} L\right)+D \sin \left(\beta_{2} L\right)=0$

[obsidian monolith]

le J

[shut folio]

$V^{\prime \prime}(L) = 0$ so $B \beta_{1}^{2} \sinh \left(\beta_{1} L\right)-D \beta_{2}^{2} \sin \left(\beta_{2} L\right)=0$

[obsidian monolith]

le J

[vivid void]

are you aware that this is calculus

[shut folio]

And there I am a little stuck, I do not know what to do in the correction, they say that:

[vivid void]

and definitely not precalculus

[shut folio]

It's a problem calculation, if you stop interrupting me maybe you'll know the final question

it's really disrespectful to interrupt right in the middle while I'm making this writing effort

[vivid void]

sure go on

but i already know that V’’ means the second derivative of V

[mossy forge]

serge lang basic mathematics is killing me lads

[vivid void]

<@&268886789983436800>

[stuck lark]

b&

[narrow cloak]

Can some one help

Tyb

[viscid thistle]

Are you allowed to use l'hopital?

[narrow cloak]

Nope

[uncut mulch]

consider factorising x^2 - 1
and doing some long division

[odd rivet]

Why?

[mossy forge]

It's incredibly dry to read lol

Also, I am not really that good at proofs, so when the "more complicated" proofs appear in the exercises, I freeze up.

[wicked solstice]

anyone care to explain how <b2 isnt 89.27

and how to get b2?

ignore the DNE, it was a hail mary attempt

[cloud oracle]

Hi

[red tree]

This question is way too hard because I have no idea what the answers are supposed to be or how to even start

[wicked solstice]

It's law of sines

sinB/b = sinA/a and so on

[cloud oracle]

I h8 the law of sines

[wicked solstice]

just not sure how to get the second angles or the second side. subtracitng 180 doesn't work and thats all I know lol

[red tree]

Damn does this teacher expect you to know what B1 and B2 stand for right off the bat

I would fail her class

[wicked solstice]

no, you just make the triangle with what you're given and go from there

its the ambiguous law of sines problem

[red tree]

With a triangle there’s ABC, I have no idea why there’s two different B’s

[hushed sphinx]

Its an SSA problem; those generally have two different solutions.

So, apparently, you're suppose to give one of the solutions in the left column, and the other solution in the right.

[red tree]

I was hoping there is an instruction or diagram or some text preceding this picture in the actual assignment

[hushed sphinx]

Might be too much to hope for it it's just one problem in a long sequence of triangles to solve.

[red tree]

The person asking for help didn't think giving the context was important either, indicating a gross misunderstanding there too

Like a misunderstanding of the nature of what a math problem is

[hushed sphinx]

Can't defend that. ¯_(ツ)_/¯

[shy kettle]

hey does anyone know how to solve this?

[vivid void]

they all have the same bases

[obsidian monolith]

quantum

[vivid void]

use these

@shy kettle

[shy kettle]

i did, but i ended up with some funky looking answers and i didnt know what to do from there so i just turned it in and hope they were actually right. they just looked weird towards the end so i must have done something wrong. thank you!

[vivid void]

no problem

[coral smelt]

Can you compare complex numbers?

I thought that their modulus could be used to do it

But Wolfram says that it is 'not well defined'

[willow bear]

there is no order relation on C that plays nice with its arithmetic the same way the order relation on R does.

[coral smelt]

interesting

[jaunty marsh]

could anyone help me solve a logs problem?

[willow bear]

maybe, but you will have to post said problem first

@jaunty marsh

[jaunty marsh]

I didn’t know if I was allowed to post a question here or not + I’m asking if anyone is willing to help in the first place

but here Solve the equation 15^4a = 81^a+2 for a. Express your answer in terms of ln 3 and ln 5.

[willow bear]

okay, so what is troubling you here?

[timber bear]

what does the C and CC mean?

[uncut mulch]

clockwise and counterclockwise

[willow seal]

c = clowise, and cc = counteraclockwise

[obtuse dagger]

Hello. Can anyone tell me what steps occured that resulted in the answer in the picture, please? Thank you!

[full pagoda]

which step specifically

[obtuse dagger]

Well, I'm trying to figure out how ln(x+1/x-1/x^2-1) resulted in -ln((x-1)^2)

[full pagoda]

what step specifically are you lost on

[obtuse dagger]

Everything you see in the picture; I don't know how to clear the denominatorin order to solve the problem.

[full pagoda]

x^2-1 is a difference of two squares

they just simplified the fraction within a fraction

then used the log rule shown

[obtuse dagger]

but then wouldn't the answer just be 1?

[full pagoda]

?

how so

[obtuse dagger]

Idk. Since (x-1)(x+1) is in the denomintor, wouldn't (x+1/x-1) cancel since it is in te "numerator"?

[full pagoda]

$(x-1)(x+1)\neq\frac{x+1}{x-1}$

[obsidian monolith]

a disappointing son

[obtuse dagger]

Right

[full pagoda]

do you know how to simplify a fraction within a fraction?

[obtuse dagger]

No, I don't think so

[full pagoda]

multiply your fractions by the reciprocal over the reciprocal

$\frac{\frac{x+1}{x-1}}{(x+1)(x-1)}\cdot\frac{\frac{x-1}{x+1}}{\frac{x-1}{x+1}}$

[obsidian monolith]

a disappointing son

[full pagoda]

you're multiplying by 1/1 here, so it doesn't change the value of the function

so this makes the numerator simplify to 1 (multiplying something by its reciprocal), you just need to simplify the denominator

[willow bear]

it must also be said: x+1/x-1/x^2-1 is a horribly ambiguous and under-parenthesized expression

[full pagoda]

indeed it is

[obtuse dagger]

Hmm, ok. I probably could of thought of that lol. Thanks.

Btw, why is it that - from what I've seen - you have to multiply by the reciprocal of the denominator in order to clear the denominator of a rational expression? In other words, why do people usually multiply by the reciprocal of the denominator and not the reciprocal of the numerator, like they did here?

[full pagoda]

just depends on where the fraction is

if there's a fraction in the denominator (which is more common and probably why you normally see that instead of the numerator), you multiply by the reciprocal of the denominator

you just want to reduce that fraction down to 1 in order to simplify the entire thing

but yeah, entirely depends on where the fraction actually is

[obtuse dagger]

Hmm, I see.

Thanks

[full pagoda]

👍

[viscid thistle]

how do you find the direction of a vector when x or y is equal to zero?

[granite verge]

isn't that just 0 vector?

[vivid void]

no

that just makes one of the components 0

[granite verge]

oh sorry i saw it as x and y

[jaunty marsh]

idk how to start it lol

[willow bear]

that was an 11 hour delay

anyway

do you know how to solve simple exponential equations such as 5^x = 11

[jaunty marsh]

i went to sleep

[jaunty marsh]

tbh i dont really remember im just looking at some practice problems for a final i have coming up

i think its x=log5(11) but idk how to get there

[willow bear]

yes

there is no "getting there" in that particular example i sent

it's just the definition of log

everything else amounts to applying exponent laws and ordinary algebra to reduce the equation to a form resembling what i wrote

[jaunty marsh]

yeah idr what that is oops

like if i got a larger problem like the one i sent idk what steps i would take to get the answer

[willow bear]

if you don't remember what the definition of a logarithm is then you need to review it posthaste

like, it's something that you absolutely and undeniably must know

otherwise all you're doing is just symbolpushing with zero rhyme or reason

[jaunty marsh]

ok could you maybe tell me what it is so you could help me .__.

[red tree]

Fresh from the oven (wikipedia)

\begin{Definition}$\log_b (x)\coloneqq$ the unique real number $y$ such that $b^y=x$\end{Definition}

[obsidian monolith]

Icy001

[coral smelt]

Questions about limits go in this channel, right?

How do you calculate an iterating limit? Structure is ...f(f(f(f(x)))))... starting from infinity till it converges

Like sign(a)/(a+1) converges to 1/φ for all real a not equal to 0 or -1 (I think)

[sharp mist]

For self studying, what books are recommended?

[cedar crystal]

@sharp mist I assume since you posted in a pre-university thread that you want to study this kind of level of math then perhaps this would be of help ? https://www.cimat.mx/ciencia_para_jovenes/bachillerato/libros/algebra_gelfand.pdf

it was a book someone linked to me a while back

[sharp mist]

Sorry, I intended it for precalc books

Hence the category

Whoops

[outer briar]

Will you please share me important formulas which are mostly used in intigration?

Anyone please?*

[merry ledge]

here is a free resource: https://activecalculus.org/prelude/book-1.html

[viscid thistle]

hey how do i find the coordinates of the angles? is there a formula or something to find it?

[timid bloom]

What do you mean by "coordinates of the angles"? There are many formulae that can solve for angles. You could use law of sines or law of cosines. The angle between two vectors is related by the dot product between them (equivalent to the law of cosines). You're going to have to be more specific.

[viscid thistle]

.

like this one

for example the given angle is 35°, how can i find its coordinates

[willow bear]

you mean cos(35°) and sin(35°)?

generally you will not be expected to find cos and sin for non-"nice" angles without a calculator.

[viscid thistle]

i have calculator

what should i put in the calcu?

[willow bear]

okay then i just told you the names people use for what you called the 'coordinates'

[viscid thistle]

to find x and y

[willow bear]

cos and sin

[viscid thistle]

how to solve it

[willow bear]

do you see COS and SIN buttons on your calculator?

[viscid thistle]

yep

[willow bear]

okay then those are what you should use...

as i tried to tell you

[viscid thistle]

whats next

what should i put

[willow bear]

your angle???

i just hope you're not being thick on purpose rn

[viscid thistle]

sin(35) = -0.43
cos(35) = -0.90

so how about these?

[willow bear]

your calculator is in radian mode

[viscid thistle]

is that right

[willow bear]

you needed degrees

switch your calculator into degree mode

[viscid thistle]

how....

man im sorry our teacher didnt teach this lol

[willow bear]

if you show me what your calculator looks like i might know how, but otherwise you'll have to read the instruction manual for your model of calculator

it's different for different ones

[viscid thistle]

so here is the model

the given is 35°

[willow bear]

brb

press SHIFT, then MENU/SETUP in that order. your calculator should display a bunch of options

do this and tell me what your calculator shows afterward

[viscid thistle]

1 : input/output
2 : angle unit
3 : number format
4 : engineer symbol

1 : fraction result
2 : complex
3 : statistics
4 : spreadsheet

1 : equation/func
2 : table
3 : decimal mark
4 : digit separator

1 : multiline font
2 : qr code
3 : contrast

[willow bear]

select angle unit

[viscid thistle]

okay then here's next
1 degree
2 radian
3 gradian

[willow bear]

well you want degree mode

so press 1

[viscid thistle]

whats next

type the cos and sin?

[willow bear]

well now your calculator is in degree mode so yes calculate cos(35) as you did previously...

[viscid thistle]

so i got...
cos(35) = 0.82
sin(35) = 0.57

[willow bear]

okay this now looks reasonable

[viscid thistle]

is that the final answer?

[willow bear]

i dont know if you only wanted cos(35) and sin(35) directly or if you wanted to do something else with these values

in the former case yes this is the final answer

[viscid thistle]

i need the coordinates, x and y of that given angle

[viscid thistle]

like this one for example the 60°, the coordinates are 1/2, sq root of 3/2

[willow bear]

okay so as i told you before that's exactly cos and sin

[viscid thistle]

oh

so thats it?

[willow bear]

yes...

[viscid thistle]

aight thanks for helping!!

[obtuse dagger]

Does anyone mind explaining how to solve #65? I got it in quadratic form, but the middle term (u^x+1) stumped me.

I made u = 4^x

[willow bear]

you mean 4^(x+1)?

not (4^x)^(x+1) surely, despite that being what you typed there

anyway, consider: $4^{x+1} = 4^x \cdot 4^1$

[obsidian monolith]

Kanga Gang Annihilator Ann

[obtuse dagger]

Yes, sorry. I meant 4^x+1. So, are you saying the new equation should be u^2 + u + 1 (with u = 4^x)?

[willow bear]

no, i am not saying that at all

not only that, but u^2 + u + 1 is not even an equation.

it is an expression

the equals sign just mysteriously disappeared somewhere

[obtuse dagger]

= 0

Sorry, I forgot the = 0 part

[willow bear]

are you sure that's the equation you get?

how did that constant term end up as +1?

[obtuse dagger]

Well, you said 4^x+1 = 4^x * 4^1. So, I u^x+1 became u + 4. Therefore, the equation would be u^2 + u + 4 -3 = 0, which is u^2 + u + 1

Did I miss something?

[willow bear]

you missed a lot of things

So, I u^x+1 became u + 4

again it's 4^(x+1) and not u^(x+1)

and no, it doesn't become u + 4

4^x+1 = 4^x * 4^1

but no, that multiplication has to magically transform into an addition suddenly

it's u * 4, not u + 4

[obtuse dagger]

lol wow. Ok, so, if I am substituting u for 4^x, then the equation I get should be u^2 +4u - 3 = 0. Is that right?

[willow bear]

yes now you are correct

[obtuse dagger]

Ok. Hmmm, so, if I had an expression that was, say, 4^x+2. That would also be equal to 4^x * 4^2, am I understanding this concept correctly?

[willow bear]

yes

this is just basic exponent laws

it should not be news to you at all

[obtuse dagger]

I mean, I haven't really worked a whole lot with exponents until now

[red tree]

4^(x+2)*

[obtuse dagger]

Thank you for your help, though

[willow bear]

[obtuse dagger]

@willow bear, I got -2 (+-) sqrt of 7 = 4^x. Do you mind briefly explaining the next steps to finding the value for x?

[willow bear]

has the topic of logarithms come up in your class at all?

[obtuse dagger]

Yes, however, I was absent for the last month of class so I have been learning logarithmic equations on my own and with the help of my school's tutors

I have a slight understanding of what to do, but I am not really sure

[willow bear]

well what i would do is first observe that -2 - sqrt(7) is obviously negative and so the equation 4^x = -2 - sqrt(7) has no solution

however the equation 4^x = -2 + sqrt(7) does have a solution

and at the risk of stating the obvious or being tautological, it's x = log_4(sqrt(7) - 2).

[obtuse dagger]

Ok, so x is the power that 4 is being raised to to get the sqrt of 7 -2 (does it matter if the -2 is before or after the sqrt of 7?); so, you can change 4^x into logarithmic form without it "offbalancing" the other side, is that right?

In other words, you don't have to take the log of both sides to solve for x

[willow bear]

....

that's a very wonky way of wording it but i don't know how to fix it

[obtuse dagger]

Yeah, I didn't have the best math teachers in high school, unfortunately. So, I am not well-versed in mathematical language/terminology

^well kind of

[kind arrow]

it doesnt matter if the "-2" is before or after the sqrt of 7

also when you change an exponential form into log form, you're taking the log of both sides

so in this case, you would be taking the log base 4 of 4^x, which is x

and then you have log_4(sqrt7 - 2) on the other side

[obtuse dagger]

Would anyone mind explaining to me how log (7x+6) = 1 + log (x-1)?, please?

I have tried a couple of ways to solve this and I keep getting improper solutions

[kind arrow]

assuming you're using log base 10, you can substitute the 1 for log(10) and then continue from there

[obtuse dagger]

bleh

wth

How am I supposed to know or "see" that?

[kind arrow]

practice

also you should have learned that log of its base is 1

since x^1 = x

[obtuse dagger]

This almost feels like sorcery lol

log of its base -- you mean the log of the argument that is equal to the base. Ok, yeah, I did. I am just not used to substituting log_x x for 1 and vice-versa; most of the problems I have been working with have not included a lot of x + or - log terms

[obtuse dagger]

Does anyone care to help? This is beyond me 😐

[hushed sphinx]

Take the exponential of both sides.

[obtuse dagger]

I am stuck at e^4 = x^2 +2x 😫

[hushed sphinx]

That's a quadratic equation. e^4 is just some number.

[obtuse dagger]

how do you add it - or, in this case, substract it - into the quadratic? e^4 is approximately 55 (that's a rough approximation, by my standards)

[hushed sphinx]

Just leave it as it is. It's as good a number as 42 or pi already.

You have x²+2x-e^4 = 0, so when you plug into the quadratic formula, plug -e^4 for c.

[obtuse dagger]

Oh, ok. Yeah, I was gonna say, you can't factor -e^4 lol

Does this come naturally to you? Like, seeing solutions and whatnot?

[hushed sphinx]

Yes, but I have several decades of practice to build on. Don't worry, it will come.

[obtuse dagger]

Ha. That's -- well, that's not really a relief; I am trying to figure out how to excel in math now - while I'm in college lol.

Thanks for your help though

I suppose continued practice will help

[hushed sphinx]

That's it. Only way that works, really.

[obtuse dagger]

If you don't mind me asking, what did you do when you got stuck? Like, back when you first started learning math -- and when your teachers were not available lol

^ I try to solve a problem in different ways but sometimes that doesn't work, even after trying to solve a problem(s) for a long time.

[hushed sphinx]

Swear loudly. Take a walk. Decide angrily to give up on learning anything forever and ever. Sleep on it. Design gorgeous model railway layouts on the graph paper instead of making any meaningful progress. Revisit it the next day or next week.

[obtuse dagger]

Lmbo! Hmmm, I don't give up, though 😔

[hushed sphinx]

Sometimes it is the best short-term tactic. Your subconscious needs time to work it over.

[obtuse dagger]

I tried using the quadratic formula, by the way. I only made a little progress; I am stuck at (-2 +- sqrt_4 - 4*-e^4)/2. Am I missing something here?

"Your subconscious needs time to work it over." - good point!

[hushed sphinx]

That's the solution, ready to punch into a calculator if you want actual digits. (Well, almost done; you need to discard the negative solution as spurious, because ln(x) wouldn't compute ).

You could tidy it up a bit by simplifying to $\sqrt{1+e^4}-1$, but that's basically as nice as the exact solution can be written.

(I haven't checked the details of the algebra, assuming you've got that down).

[obsidian monolith]

Troposphere

[obtuse dagger]

@hushed sphinx how can what I wrote be simplified to what you wrote (specifically, the radicand)?

[hushed sphinx]

sqrt(4a)/2 = sqrt(a)

[obtuse dagger]

I don't get it

[hushed sphinx]

4 - 4*(-e^4) = 4 + 4e^4 = 4(1+e^4)

[obtuse dagger]

Well, today I learned that you can factor something that is under a radical. Thank you @hushed sphinx.

[hushed sphinx]

You can factor/rewrite anywhere, as long as the subexpression you rewrite still contributes the same value to the rest of the computation.

[obtuse dagger]

Hmmm, you learn something new everyday. Thank you (I did not know that applied to a radical).

[hushed sphinx]

The point is, it applies everywhere. When we write 4 - 4(-e^4) = 4(1+e^4) what we mean is that computing the two expressions will give the same value. Then, necessarily when we take the square root of (or do anything else with) those same values, we'll still get the same results in the next step of evaluating the expression. Therefore the condition for sqrt(4 - 4(-e^4)) = sqrt(4(1+e^4)) to hold is met. This is not specific to square roots.

When you do the same thing on the same numbers, you also get the same number out in each case.

[obtuse dagger]

Right, I just didn't know that rule applied to sqrts. Thanks @hushed sphinx!

[robust star]

is there another way to find the zeros other than the rational root theorem?

[willow bear]

idts

[full pagoda]

you're an idt :(

[willow bear]

???

no, that's an abbrev for "i don't think so"

[full pagoda]

i was jokin lol

[willow bear]

the joke did not land.

[full pagoda]

[jolly raven]

landed for me

[full pagoda]

<3

[obtuse dagger]

Hey, guys. So, Idk what I am doing wrong but I got a solution of x = 4. However, that does not check; idk what else to do 😦

[willow bear]

show work?

maybe you made an easily detectable mistake

[obtuse dagger]

So, I simplified and I got (x+1)(2x-3) = 3. Then, after multiplying the binomials together and subtracting the 3, I got 2x^2 -x -6 = 0

That's how I got x = 4

[willow bear]

let's see

(x+1)(2x-3) = 3 is correct

,w expand (x+1)(2x-3)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by Ann#0413

[willow bear]

okay

so then you solved the equation 2x^2 - x - 6 = 0...

how exactly did you get x=4?

[obtuse dagger]

factors of -12 whose sum are -1: -4 and 3 (x-4) and (x+3)

Solving for x, you get x = 4 or x = -3. However, -3 is not valid

[willow bear]

hold up

(x-4)(x+3) expands to x^2 - x - 12

[obtuse dagger]

omg

I just realized my mistake

lol

I forgot about the 2 in 2x^2 lol

It factors to (2x+3)(x-2), not (x-4)(x+3) lol

Gn!

[tepid cloak]

welp

[vivid void]

this just means that it’s in the form of x^2+(y-b)^2 = r^2

considering that it passes through (0,0), this shouldn’t be too hard

[willow bear]

@tepid cloak is there anything in particular that you're having trouble with here?

[tepid cloak]

I get it now thanks

[fresh marsh]

Is it wrong to define a Combination as just a subset of something?

And thus counting all combinations is just counting subsets of some size?

[sick steppe]

the number of subsets, of a set of cardinality n, of cardinality k is in fact just nCk

[blissful wave]

hello can someone help me with this problem?

Find the equation of the circle that goes through (—1, 1) and the two points of the intersection of the circles x^2+y^2−8x+2y+8=0 and x^2+y^2−2x+4y+1=0.

[willow bear]

have you made any progress so far?

@blissful wave

[blissful wave]

not yet

We're having exam and I forgot how to do it lol

[willow bear]

you're having an exam? right now?

[blissful wave]

yeah. Finished already. I got the formula I had to solve the other questions before i recalled how to do it lmao.

[inland birch]

Hi forgot how to do this part, when it says ABC triangle which side is the right angle one? like is B the one with the right angle? or how do you tell

[cedar crystal]

I don't see how its a right triangle though if the angle of AB is 30+50 = 80 ?

[inland birch]

This is what I mean which is one is the set up

[inland birch]

the angles are not 30 and 50, those would be the lenght of the sides

[cedar crystal]

hmm no idea but I know the hyp length is the sum of AB

[inland birch]

i know the answer would be 50, cos = adjacent/ hypotenuse. so cos 30/50, 50 has to be the hypotenuse

but havent done math in a while and i cant remember how to set up lol

[cedar crystal]

the right angle is always = 90

so the angle between AB and BC has to be 90 for it to be a right triangle

if thats what you are asking

[inland birch]

how did you do that? so C is the corner with the 90 degrees?

[cedar crystal]

maybe this helps?

ABC there doesn't really matter as its just a way of wording the vertex points

[inland birch]

ok that was original question, how would you set it up, so i guess all three are valid? i just swear i remember an old teacher talking about this, and i cant find anything online

but if thats the case thank you for the help

[cedar crystal]

yeah they are all right triangles as long as the angle there is 90

the other examples you have there would be the same as the one I painted

CA + AB should be 90 and AC + CB = 90

[inland birch]

cool thanks

[cedar crystal]

@inland birch you could have any letter you like really but ABC is commonly used because its what Pythagoras used

so it becomes an "unwritten" rule that everyone understands

[red tree]

Nope, this doesn’t imply that the leg and hypotenuse are 30 and 50. Just that they are in a 30 to 50 ratio

[inland birch]

sure, but its implied in the question

[red tree]

Is there more to the question?

Saying a number is 30/50 is the same as saying that the number is 0.6

There’s nothing more to it

[inland birch]

nope that's it, what is length of hypotenuse, nothing more, goes on to list options

[red tree]

Show options

[inland birch]

eh its a udemy course i would have to restart the test to get that question again

[red tree]

I hope one of them was “cannot be determined”

[inland birch]

you can determine it though lol, 50 was right

[red tree]

50 was right? Stay away from that course then, the writers have no idea what they’re doing

[inland birch]

sure

[red tree]

You don’t believe me?

[inland birch]

ill respectfully disagree, but any thoughts on my original question?

[red tree]

The right angle can be either A or B

[inland birch]

taking away angles, lenghts or anything, when given triangle ABC how to set up

[red tree]

It cannot be C because that would contradict its cosine value

[inland birch]

ah that makes sense

cool thank you

[silk thunder]

Hello is there someone to help me ?

With logarithms

[red tree]

Do you know the definition of logarithm or the definition of inverse function?

[silk thunder]

-\sqrt{log(x+1)}

Domain of this function