precalculus

[sleek path]

Cant seem to do this at all

[patent beacon]

There's three forces on the mass. The tension in each rope, and gravity. These forces should sum to zero

If you're comfortable with component notation:
A[cos(30), sin(30)]

• B[cos(135), sin(135)]
• [0, -20g]
  = 0

Gives two equations and two unknowns

[sleek path]

are A and B the tensions?

[patent beacon]

Yes! Forgot to mention

[sleek path]

ok ok. How did you get to cos and sin

is that just the x and y components?>?

[patent beacon]

Yeah, that's the components of each of the vectors

[sleek path]

wouldnt it be B[cos(45), sin(45)] then? or why are you using 135 instead of 45

[patent beacon]

Because if you take the angle from the x-axis, then vectors always split in terms of [cos(), sin()]

You should probably draw the diagram and split them the way you like

[sleek path]

ohh the 135 is the exterior angle? got it

[patent beacon]

But that's the procedure. Split the vectors, add them componentwise to get two equations

[sleek path]

ok

thank you so much for helping

[patent beacon]

Np. Feel free to ask if you have anything else!

[hearty plinth]

Hey guys I need help for this

Find the std.eq of all circles having center at a focus of 21x^2-4y^2+84x-24y=36 and passing through the further vertex

And

Find the std.eqn of the hyperbola of which has focus and vertex that are the same as these of x^2-6x+8y=23 and whose conjugate axis is on the directrix of the same parabola

[viscid thistle]

Just complete the square on the first one?

[finite jolt]

can someone help me with 11? please

<@&286206848099549185>

[sullen shoal]

#❓how-to-get-help

[stark violet]

<@&286206848099549185>

[elfin night]

hmm

Gotta post a q m9

[thick raptor]

asks question in #help-4

pings helpers in #precalculus

[sullen shoal]

Fucking hell

[half ridge]

anyone here?

I have an urgent question

[fallen cloud]

ask it

[half ridge]

use absolute value to describe the interval [2,8]
I know the answer is |x-5| < or = 3

[gritty blaze]

And ?

[cedar chasm]

when u find the domain of a rational function and like put numbers in to see if its positive or negative

do u use the simplifed version you get by cancelling out terms

or the original equation

[viscid thistle]

Forty-five percent of a radioactive substance decays in four years.

By what percent does the substance decay each year?

Round your final answer to two decimal places.

[mental trench]

Wouldnt you use a a decay formula? The whole A(1-r)^t

[viscid thistle]

yeah but im lost about the specific one year part per 4 years @mental trench

The amount of radioactive substance as a function of time in years is an exponential function of the form y=ab^t . Use this formula and the fact that after 4 years only 55% remains to find the growth factor b. The growth factor can then be used to find the annual percent growth rate, or growth factor, which will be negative as the substance is decaying.

[mental trench]

And youre given no principle amount?

[viscid thistle]

yep no initial value

it is only based off of the percentages and times

[mental trench]

Okay, i tried it and i got 0.86 which is 86 percent but that sounds extremely sus.

[viscid thistle]

it still displays as incorrect lol. very odd

[graceful egret]

<@&286206848099549185>

[viscid thistle]

@graceful egret

[graceful egret]

what

[viscid thistle]

[graceful egret]

i might have posted my question in the wrong section so here it is again

<@&286206848099549185>

[viscid thistle]

Where did the 8 come from at the bottom of this?

[bitter nova]

counting to 8

[viscid thistle]

Counting what though?

Thanks btw.

[bitter nova]

kinda like a clock

counting pi/6 angles from 0

tbh it's pretty strange

[viscid thistle]

Oh okay. Thanks a lot.

I don't know how I didn't get that but that makes sense thank you.

Does that only work in that quadrant? Because if I do 150 degrees I get 6pi/6 not 5pi/6

Ignore that.. figured it out.

[rocky bison]

240 = 8*30

So there's 8 intervals of 30

and 30º is pi/6 radians

[viscid thistle]

When determining the period of -sin((x/2)+(pi/4)+5 why does x/2 = 1/2?

[solar abyss]

=tex -sin(\frac{x}{2} + 5) + 5

[solar abyss]

the period is

=tex \frac{2pi}{b}

[solar abyss]

=tex \frac{2\pi}{b}

[solar abyss]

where b is the coefficient in front of the x in the sin

so for that function the period is

\tex = 4\pi

$$ 4\pi $$

[forest prawn]

Hi all, I’m having trouble with factoring out a function in order to find it’s zeros.

I’ve tried the rational root theorem with synthetic division, as well as factoring out via grouping

I’m wondering if I’m just missing something, or if I am doing something fundamentally wrong

Also, as far as I know regarding the rest of the question, (Please correct me if I am wrong) it is an odd function as f(-x) = -f(x), it’s end behavior is positive approaching Inf. and -Inf, it is to the sixth (sextic) degree with a leading coefficient of 1.

[patent beacon]

@forest prawn
It is not odd. Odd polynomials only have odd exponents. This is a neither.

End behavior is ∞, but since the degree is even, the "start" behavior is -∞.

You're correct about the degree and leading coefficients

[forest prawn]

Thank you for pointing that out, not sure how I forgot how to find an odd/neither/even function

Regarding the second point, that is due to the fact that, as the line comes in from the left, it is sloping downward causing it to be -Inf?

Lastly, could you give me a starting point for factoring out the polynomial so that I can get off the ground with that? I feel like I have exhausted every option that I know

I appreciate the help.

[patent beacon]

= pup roots of x^6 - 5x^5 - 5x^4 - 45x^3 - 108x

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@patent beacon

[patent beacon]

Not sure how you're supposed to find these

Sorry, I misstated, the "start behavior" is ∞. It's on the same side because the degree is even

[forest prawn]

Hmm, might have to bring this up to the teacher. He hasn’t officially handed this out, but I just happened to check the website and saw it was there so I figured I would take a crack at it.

Just to clear it up, End behavior as x approaches -Inf can be found by starting at 0 and going left correct?

[patent beacon]

Agreed, those aren't zeros you can find. Run it by him!

= pup graph y = x^6 - 5x^5 - 5x^4 - 45x^3 - 108x

[viscid thistle]

I don't understand this. There are no videos to help me.

[patent beacon]

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

[viscid thistle]

Oh, hold up.

How would I get the cos(a) though?

[vale pewter]

using the first identity you learn in trigonometry?

[north bough]

can someone help me with this proof?

=tex \frac{sin\left(x\right)}{1-cot\left(x\right)}+\frac{cos\left(x\right)}{1-tan\left(x\right)}=sin\left(x\right)+cos\left(x\right)

[north bough]

I can't solve by doing something like subtracting cos(x) from both sides

I already tried multiplying the part with sin(x) in the numerator by sin(x)/sin(x) and the part with cos(x) in the numerator by cos(x)/cos(x)

I have also tried multiplying them by the conjugate of the expression on the right hand of the equation

[viscid thistle]

about rectangular hyperbola
(proving y=1/x is a hyperbola)

the fourth equation, i have no idea where the square root of 2 came from

even how the y-x appeared there

[spring thunder]

$$\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2 = 2\frac{\left(\frac{x+y}{2}\right)^2}{2}+2\frac{\left(\frac{x-y}{2}\right)^2}{2}$$

[spring thunder]

$$\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2 = \sqrt{2}^2\frac{\left(\frac{x+y}{2}\right)^2}{2}+\sqrt{2}^2\frac{\left(\frac{x-y}{2}\right)^2}{2}$$

$$\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2 = \sqrt{2}^2\frac{\left(\frac{x+y}{2}\right)^2}{2}+\sqrt{2}^2\frac{\left(\frac{x-y}{2}\right)^2}{2}$$

[spring thunder]

:O shuts

(Fking internet)

$$\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2 = \frac{\left(\sqrt{2}\frac{x+y}{2}\right)^2}{2}+\frac{\left(\sqrt{2}\frac{x-y}{2}\right)^2}{2}$$

[spring thunder]

$$\left(\frac{x+y}{2}\right)^2 - \left(\frac{x-y}{2}\right)^2 = \frac{\left(\frac{x+y}{\sqrt{2}}\right)^2}{2}+\frac{\left(\frac{x-y}{\sqrt{2}}\right)^2}{2}$$

[tame plaza]

hi friends ! can I ask for help with understanding this...I'm not sure what you call it

[spring thunder]

And (a-b)^2=(b-a)^2 for all a,b anyway @exotic path

@tame plaza ya just ask

[tame plaza]

should I ask here or in the questions

;o?

@spring thunder

[spring thunder]

(Let it here : my internet is shit tho ATM)

[tame plaza]

ah, im not sure if you can see it

it's a file haha

[fickle moat]

can someone explain to me x^2 = +,- i

if your finding all zero's

[rocky bison]

What do you mean @fickle moat

=tex \pm i

[fickle moat]

yes

[rocky bison]

Can you give full context please?

[fickle moat]

ok one moment

[rocky bison]

Are you trying to find all of the imaginary roots of a quadratic?

[fickle moat]

let me take a picture

i get every answer EXCEPT x^2 = 0

the answer from book says +,- i

how????

[rocky bison]

I can't read your question

just post the original polynomial plz

[fickle moat]

2x^5 -11x^4 +12x^3 -3x^2 +10x +8

[rocky bison]

oof

large degree

Ok

[limber bone]

hehe xd

[rocky bison]

What did you do to try factor it

[fallen cloud]

omgr....

[limber bone]

gl bois

[fickle moat]

i used syntheic division

[fallen cloud]

synthetic, yep i learned it and then forgot it, never used it since

[fickle moat]

now after doing that i gotten 2x^3 +x^2 +2x+1

grouping

[fallen cloud]

well... lets multiply that out to check if thats correct

[rocky bison]

[fickle moat]

x^ (2x+1) (2x+1

so x = -1/2 and x^2

[rocky bison]

Well for a start your polynomial

[fickle moat]

but x^2 =0 HOW is it =,- i

[rocky bison]

has a constant on the outside

So there should be no x^2 isolated

Because that means all terms have a common factor

which they don't

[fickle moat]

i factored out x^2 from grouping

[rocky bison]

ok I'm lost

[rocky bison]

but in order to get pm i

it should be (x+1)^2

[fickle moat]

can you just explain to me how do you get X^2 = 0 is +.- i

[rocky bison]

You don't

x^2=0 doesn't have solutions pm i

[fallen cloud]

i think something happened with the synthetic division

[rocky bison]

It's x^2+1

[fickle moat]

the professor says x =0 is wrong answer

[fallen cloud]

i works

[rocky bison]

that has sqrt -i and +i

[fickle moat]

the answer is +,- i

[rocky bison]

It is

Clearly

But in order to get x=i, x=-i

You have to have x^2+1 as a factor

[fickle moat]

i get it when you have x^2 = -1 you get x = +,- i

[rocky bison]

yes

[fickle moat]

but how the heck x^2 =0 gets +,- i

[rocky bison]

It doesn't

x^2=0 is wrong

You've done something wrong to get to that

[fickle moat]

give me a sec

let me retake the picture

@rocky bison

[rocky bison]

I'm not familiar with synthetic division so I won't be helpful

I thought you were just doing standard division with some extra information

[fickle moat]

i got a 91 on test but i want to know where did that +,- i come from

so is there any helper could help out ?

@rocky bison

[rocky bison]

Probably

<@&286206848099549185>

[dapper oar]

you missed a + sign here

[fickle moat]

my hole point of question is x^2 = 0 it says +,- i

i get where i missed

[dapper oar]

wdym

[fickle moat]

ok so u got 2x^3+x^2

u factor x^2

u get 2x+1

i get you will get x = -1/2

but for x^2

how is it +,- i

[dapper oar]

how have you got 2x^3+x^2

[fickle moat]

you group

[dapper oar]

what do you group

[fickle moat]

oh my god

https://cdn.discordapp.com/attachments/363224154469826562/496789212578840577/unknown.png

do the grouping

insteed factoring that

insteed factoring out i grouped so i can factor out easier

(2x^3+x2) (2x+1)

now you can factor out x^2 for 2x^3+2

x^2 ( 2x+1)(2x+1)

now you solve that for all zero

u get x = -1/2

but MY QUESTION HERE is x^2 = +,- i

[dapper oar]

No
2x^3 + 1x^2 + 2x + 1 does not equal (2x^3+x^2) (2x+1)

[fickle moat]

ITS GROUPING

can you factor x^2?

for 2x^3 + x^2

or not

so what is it

2x+1

x = -1/2

but xlfjaskldfjweri gki

sorry frustration is rela

[dapper oar]

you keep saying ITS GROUPING, but this is where you're making the mistake

there is no factor of x^2

[fickle moat]

you try to factor https://cdn.discordapp.com/attachments/363224154469826562/496789212578840577/unknown.png

without grouping

[dapper oar]

I don't know what you mean by "grouping", but let me grab a pen and paper

[fickle moat]

ok so insteed using quadratic formula

i am spliting that each side

so i can find all acution zeros

[fickle moat]

2x^3+x^2+2x+1 same thing what i mean as group is ( 2x^3+x^2 ) ( 2x+1 ) l for 2x^3+x^2 what can you factor out?

[dapper oar]

Sorry for the bad lighting, but does this make sense?

specifically the second line

[fickle moat]

yes

3rd line is going to be

x^2 (2x+1) (2x+1)

[dapper oar]

No, you've gotten rid of the + sign again

[fickle moat]

the plus sign goes inside (

i did not miss the sign

ok one sec

let me retake pic

if you solve that you will get orginal equation x^3

i found all the x of real zeros

but X^2

[dapper oar]

you did miss out the sign

[fickle moat]

......

my question is X^2

x^2 =0 HOW do you get x +.- i

[dapper oar]

You don't, because you don't get x^2 = 0

[fickle moat]

please thats all i really want to know

[dapper oar]

Because if you don't miss out the sign, you get a different equation to solve

x^2 = 0 does not give you x = +/- i

[fickle moat]

so your saying x^2+1?

[dapper oar]

Yes

[fickle moat]

because me missing that + sign infront

x^2+1 =0

?

[dapper oar]

it's not a "sign" in the same way that you have positive and negative signs, it's an operation, it's addition

[fickle moat]

x^2 +1 = 0 => x^2 = -1 => x^2 root both side => x= root -1 ?

x= +.- i?

@dapper oar

[forest prawn]

@fickle moat

if you have (x^2+1)

you can set x^2 = -1

square root both sides and you get x = sqrt(-1)

which is equal to i

[fickle moat]

oh

ok now i see where my teacher got +.- i from

thank you

[forest prawn]

no problem

[neon vale]

http://prntscr.com/l1h1pl

please help

[patent beacon]

What are they doing to x³ - 9?

Or - what does h(x) do to x³ - 9?

[neon vale]

6th rooted lol idk

is it just 6sqrt x

[patent beacon]

Exactly. So this is a 6th root that's been "applied onto" x³ - 9. The outer function is - that yes

6th root of x

[neon vale]

wow thanks that was easier than i thought how embarrassing lol

[violet latch]

is anyone here at the moment?

I need help with a max/min kind of problem

quadratic equation

[viscid thistle]

Can someone tel me the domain and range of -3(x - 3)^2 + 9?

[hexed ermine]

What do you know about the domain and range for a standard quadratic

[forest prawn]

@viscid thistle If vertex form is y=m(x-h)^2 + k, your vertex is (h,k).

Your domain is = to all the values of x that your graph can possibly touch, with range being all values of y.

You can use the rest of your knowledge about graphing functions to determine the domain and range fairly simply as long as you know the vertex

[clear coral]

Does anyone know how to do 5 and 8

[hexed ermine]

Use log rule

[clear coral]

change of base?

but how

[hexed ermine]

loga^b = bloga

[clear coral]

i still don't get how to use it in the context of first problem tho

[hexed ermine]

log_9(5^log_5(27))

[clear coral]

yeah still no clue

the only exponent i see to bring down is a 3 from the 27 or 3^3

[hexed ermine]

Look at the 5

What do you know about 5^log(5)

[clear coral]

huh I'm reading it as (log base 9 of 5 ) * (log base 5 of 27)

I can't see where you're bringing down an exponent

[hexed ermine]

Yes

log(a)*log(b) = log(a^log(b))

[clear coral]

Is it the same thing as that integer coefficient thing I'm used to but just with a log instead?

so x = 3/2

[hexed ermine]

Yes

log_9(27)=1.5

[clear coral]

and 8?

[hexed ermine]

Isolate 8^2x

Log both sides

[clear coral]

eh?

how

[hexed ermine]

I dont feel like explaining , I'm sorry, someone else can help you

[clear coral]

lol okay

I don't feel like doing this either

[hexed ermine]

With these types of problems I have to use paper bc it's hard keeping track of everything

[viscid thistle]

@clear coral whats the question?

[clear coral]

number 8

scroll up

[viscid thistle]

using ln/log?

[clear coral]

ehhhh ummm idk how to isolate the 8^(x)

[viscid thistle]

u can always ln it

[clear coral]

yes but howww

Idk how ln it

[clear coral]

I understand the power rule but there are no logs in this question

[viscid thistle]

in order to solve these questions, u needa ln everything

and solve from there

[clear coral]

could you try explaining it in the context of the problem

[viscid thistle]

would u like me to write it out?

[clear coral]

yeah

[viscid thistle]

@clear coral

im not that familiar with the rest... but does it help a little?

[clear coral]

yeah it does '

thanks much

[viscid thistle]

lol i would like to know how to solve it too 😛

do u mind posting ur results in the end?

[clear coral]

ooooooh okaaayyy I can tryyy

[viscid thistle]

lololol we are all in the same boat...

[hexed ermine]

You cant combine the second part

3^3*2^3x

That doesn't equal 6^9x

[clear coral]

I knew it

I just got there and got so confused

[viscid thistle]

@hexed ermine wait.... isnt that like an exponent rule?

oh no... nvm

sorry... 😦

[viscid thistle]

can someone message me if they are good with TVM solver or finance math (compounding stuff etc)

[drifting hollow]

The most you can simplify 3^3 * 2^3x would be either 27*2^3x or 27 * 8^x

[dense zealot]

smh

[viscid thistle]

hi guys

whats the multiplicative inverse of 2+9i

[patent beacon]

1 / (2 + 9i)

[viscid thistle]

k thx

[patent beacon]

Np here for ya

[atomic harness]

i am stuck on number 3

u know how to do that

[viscid thistle]

@viscid thistle normally you would want to take the imaginary part out of the denominator if you want to represent it in general complex number form.

U can do that by multiplying both nunerater and denominator by conjugate of the denominator

Similar to how u rationalise denominators when u have irrational numbers as denominators

[viscid thistle]

Yeah

[sterile portal]

$$8/(2+h) $$

[sterile portal]

Could I do 8/2 and get 4 and then have 8/h still or is that against the maffs rules

[civic plaza]

that is not valid

[sterile portal]

👍Thanks

I'm trying to find the slope of the tangent but I need an H variable in the answer for the slope of the secant

I tried getting rid of the fraction by just multiplying the denominator but I guess its not right

Ping me please & thank you :)

[severe verge]

rawrr

[sterile portal]

I need helps I've been stumped

[severe verge]

hmmmm

(f(2+h)-f(2))/(2+h-2)

(8/(2+h) - 8/2)/h

(8/(2+h) - 4(2+h)/(2+h))/h

(8-8-4h)/h

-4

seems legit

[sterile portal]

But then what would the slope of my tangent be

[severe verge]

-4

[sterile portal]

That was for the secant

[severe verge]

but all the secants are -4

and tangent is basically a secant but one of the good ones

[sterile portal]

And for the tangent u have a limit of H approaching 0

[severe verge]

lim_{h→0} of -4 = -4

[sterile portal]

Oop

Ur right lool

The answer says it's - 2 but I guess typo?

[severe verge]

ya prob

[sterile portal]

Ok thanks👍👍

[severe verge]

np

maybe I am just dumb

=wolf f(x)=8/x, f'(2)

[sterile portal]

My textbook got lots of error cause it's old and needs to be revised tbh lol

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@severe verge

[severe verge]

oof

where did we go wrong

what the heck

[sterile portal]

Idek

[severe verge]

I am bamboozled

[sterile portal]

Wooaahh woog bamboozled?!!?

[severe verge]

lol

ok now I will thonk

about where we could have possibly gone wrong

[thick raptor]

tl;dr you forgot the 2+h in the denominator

which divides the result by 2

[sterile portal]

Thanks guysss now I feel stupid cause it wasn't that complicated

[vapid oxide]

how do i convert this into polar? 4x^2+3y^2-2y-1=0

i brick myself with some nonsense fraction expression that isnt the same graph

[somber plaza]

It’s a circle graph imao

[vapid oxide]

yes i get that but i brick when i try to convert to polar

:/

can someone help me :?

[civic plaza]

use the substitutions: x=rcos(theta), y=rsin(theta)

[fallen cloud]

4xx + 3yy - 2y -1 = 0,

0 = rrcos(theta)cos(theta) -2rsin(theta)-1,

solve for r

r = (2sin(theta) +- 2)/(2cos(theta)cos(theta))

r = (sin(theta) +- 1) / (2cos^2(theta))

wait...

i think i did something wrong

yep, i forgot the + 3...

or something like that

[vapid oxide]

i always have a r left over on the other side for some reason D:

thats what i meant by brick

[desert jackal]

can someone explain compound interest y = ae ^bx to me

[patent beacon]

Have an example question?

[desert jackal]

you invest 100$ at 6% per annum compounded quarterly. how long will it take for balance to exceed 120$?

[patent beacon]

You want to find interest per compound. Banks are lazy af and they just decide that 6%/4 = 1.5% per quarter

[desert jackal]

would you use equation A=P(1+r/n)^nt

[patent beacon]

Yeah, you would!

[desert jackal]

ok

[patent beacon]

120 = 100(1 + 0.015)^(4t)

Make sense how I got there?

[desert jackal]

yeah

thank you

[patent beacon]

Ok. Feel free to ask if you have anything else!

[vapid oxide]

can anyone help me convert this into polar? 4x^2+3y^2-2y-1=0

i keep ending up with an r on the other side that i cant get rid of :/

[viscid thistle]

why do people not use questions chat

understandable

[vapid oxide]

actually nvm i finally found out why i bricked

[thick summit]

I need hand simplifiying a really weird question:
x^-(3a+7b)-8(-a-b) / x^a-5b
Never dealt with a question before that had more then a single number as the exponent, so I'm rather confused.

and sorry if I should be asking this elsewhere, I tried using a question channel but my message wasn't sent for some reason.

[thick summit]

<@&286206848099549185>

[viscid thistle]

Uhm could you specify? @thick summit

[thick summit]

Like the equation or what I need done with it?

e.e

I just need to simply x to the power of -(3a+7b)-8(-a-b) divided by x to the power of a-5b.

[thick summit]

Anyone?

[hexed ermine]

Same base

And its dividing

What do you do with the exponents?

[thick summit]

Simplify somehow?

[bleak fractal]

could someone help me with a complex number question?

why do they say that (z+6)^8=81 is the same as z^8=81??

wouldnt the +6 offset the answer

[spring thunder]

we're translating the triangle, but there's no rescaling : the area will be the same @bleak fractal

[bleak fractal]

that makes sense, thx!

[spring thunder]

👌

[vapid oxide]

can someone show me the work for this :/
(convert to polar)

4x^2+3y^2-2y-1=0

[calm whale]

ah

[vapid oxide]

converting to polar my b

i always brick it and i dont get the right answer

[patent beacon]

4x² + 3y² - 2y - 1 = 0

x² + 3(x² + y²) - 2y - 1 = 0

r²cos²θ + 3r² - 2rsinθ - 1 = 0

[vapid oxide]

yea i get that but i brick afterwards

[patent beacon]

Well, I'm done. That's a polar form.

Solved for r?

[vapid oxide]

yes

r= ?

[patent beacon]

(cos²θ + 3)r² + (-2sinθ)r - 1 = 0
Any easier?

[vapid oxide]

:/

[calm whale]

ax^2+bx+c=0 ..

[patent beacon]

This is a quadratic in r. Know the quadratic formula?

[vapid oxide]

yes

OH

i get it now

i never thought about using quadratic formula LOL

[patent beacon]

Lel. Think you've got it from here?

[vapid oxide]

yea now i got it

thanks so much 😄

aighty poggers

finally got it

r=(2sintheta+4)/(2cos^2theta+6)

[fickle moat]

So the equation for this problem i have is
A = 1500e^(.02 x 10)
And i plugged into calculator it only showa same number

Can someone tell me where did mess it up?

[fallen cloud]

you dont have a variable

[calm thicket]

== 1500e^(0.210)

[granite stirrup]

1‚500×exp(2) = 11083.584148396

[fallen cloud]

it's a constant that you plugged in

unless 'e' is your variable

because 'e' is an actuall number, e = ~ 2.718281828

[fickle moat]

Oh

So e is for x

Is that what your saying or?

My english is pretty bad so..

[fallen cloud]

'e' should never be used as a variable

same with 'i'

[viscid thistle]

e and i represent numbers

[rare zephyr]

What was my mistake

<@&286206848099549185>

[viscid thistle]

=tex log_3(2x+5) - log_3(x) = log_3(\frac{2x+5}{x}) \neq log_3(\frac{x}{2x+5})

[viscid thistle]

And you forgot to flex.

[rare zephyr]

I'll try to keep forcing them muscles to flex itself

[earnest finch]

Yeah

[viscid thistle]

It's just a quiz.

[earnest finch]

Wow. Apparently it was a teacher's mistake. I actually got a 96% on that quiz. That's a bit better.

[rocky girder]

How do i do #3 ?

[viscid thistle]

look at the leading coefficient and the constant

[somber plaza]

@rocky girder look at rational root theorem

[earnest finch]

All the factors of the leading coefficient and the constant (last term)

[viscid thistle]

Yo guys I know the answers to this problem but I don't know how to get to them I'm looking for arc length, theta in radians and area of the section. This is a homework assignment.

[faint acorn]

Whats with the 7 and 12?

Oh I get it, its a clock

[rancid ether]

what does cos(arcsin(1/5)) equal to?

suppose to figure this out without calc

[primal yarrow]

Can you help me with a question?

It involves only high school Mathematics

@rancid ether

[rancid ether]

sure

[primal yarrow]

Sorry I am just really desperate

Let’s take this to dm then?

[rancid ether]

sure

[hexed ermine]

@rancid ether make a reference triangle with the given information

arcsin(1/5) indicates an angle value, we will call theta

with that, sin is opposite of hypotenuse

once you have that you can label the adjacent side of theta using the pythagorean theorem

once you do that you do adjacent over hypotenuse for that

[somber plaza]

@hexed ermine hes solved it already lol

i posted literally what u said earlier

[viscid thistle]

for a greatest integer function, ex. [2x-1], when do we know to stop graphing

[chrome grove]

Is this where I can ask about trigonometry?

[frozen needle]

yes

[chrome grove]

Ok I need help understanding these two questions.

https://i.imgur.com/PzmmwkO.png

so I was able to get this so far for the first one

amp = 5, Period=2π/3

How would I go about graphing? I know that the wave is contained 5 and -5

https://i.imgur.com/1JBXja6.png

[fickle moat]

simplify the expression how do you do that

#25

can someone explain to me? or Do i have to write it in y = b^x? formet

<@&286206848099549185>

[sullen shoal]

1

[limpid plover]

1

[fickle moat]

how tho

[sullen shoal]

8^1=8

[elfin night]

👀

[limpid plover]

Let $$log_8 8=x$$

[fickle moat]

what about 27

[sullen shoal]

lol it's \log

[limpid plover]

$$8^x=8$$

[fickle moat]

hm....

[limpid plover]

I am 9000% sure that I put a slash there

[fickle moat]

what about #27

the answer says 0

but how though

[sullen shoal]

Well, fuck you!

[limpid plover]

$$log_n 1=0$$

[sullen shoal]

n^0=1

[limpid plover]

That's true for all n>0

[sullen shoal]

^

[fickle moat]

i get the concept but

7.5^1?

[sullen shoal]

7.5

[fickle moat]

what am i missing

[limpid plover]

And probably all n but log isn't defined for all n so eh

Let $$log_{7.5}1=x$$

[sullen shoal]

==65465467^0

[granite stirrup]

1

[limpid plover]

$${7.5}^x=1$$

[limpid plover]

What's x

[fickle moat]

1?

[sullen shoal]

==7.5^1

[granite stirrup]

15/2 = 7.5

[fickle moat]

wait

🤦

okok

now i see

;;;;;

@limpid plover thank you

[sullen shoal]

[chrome grove]

oof

[fickle moat]

How do you simplify the expression

[thick raptor]

#❓how-to-get-help wait a minimum of 15 minutes after posting to ping helpers.

[fickle moat]

oh...woop sorry

[earnest finch]

5log5 cancel out so you're literally left with 3 @fickle moat

[sullen shoal]

$$5^{\log_5(3)}$$ in log form is $$\log_5(x)=\log_5(3)$$

[viscid thistle]

what is an equation that will move the vertex of the function y=x^2 to the point (-3,1)

nvm i got it

[sullen shoal]

xd

[limber bone]

y = (x+3)^2 +1

[chrome grove]

https://i.imgur.com/NB3yqK7.png

can someone break that down or link me something that covers this?

[odd lichen]

It's mainly identities and rearranging.
Identities found through symmetries in the unit circle. Like sin(-x)=-sinx

Oh and similar and common triangles.

@chrome grove

[viscid thistle]

hi hi

can i please get some assistance

this is my work

i've gotten to the graph part but im confused on finding the intervals for this problem

can i please get some help 😃

the first point of the graph should start at (-5,0)

im only graphing for one period

so when trying to find the last point of the graph (where it ends), i did
-5+(-20/7)

since the period is (-20/7)

my last point should've been -55/7

BUT, when i checked desmos it was wrong

i didn't get (-55/7,0) like i should have

@patent beacon i know, it's all sorts of fucked up

lmao

[hexed ermine]

Indeed -55/7 is a zero

And it looks from what I graphed it's a zero

Perhaps yours is a bit messed up

[viscid thistle]

but why is it giving a weird fucked up number on desmos?

a scientific notation sort of number?

[hexed ermine]

Well that's a very small number close to 0

Idk why it didnt give you zero though

[viscid thistle]

yeah, that what im wondering as well

also

check this out

if i had done -5+(20/7)
i would've gotten a zero

-55/7 is also less than -5

so how does ur last interval is less than ur first

like this question is weird af

the period is negative

BUT

if it was positive

i would've gotten the end point for the graph

it doesn't make sense

[hexed ermine]

-5+(20/7)?

[viscid thistle]

-20/7 is your period right?

is that what u got?

cuz that's what i got

BUT

if the negative wasn't there

i would've gotten a flat 0

and the graph would've been right

but since the period=(-20/7), we are both getting -55/7

if -5 is our first point, then how is -55/7 gonna be our last point

[hexed ermine]

-5-(20/7)?

[viscid thistle]

yes

that's what i did in order to find the last interval

but

it gives us a number less than -5

which shouldn't be right

we might need more peeps on this problem

lmao

im so confused with the intervals

[hexed ermine]

I'm not sure what you're asking

Why did you circle that value?

[viscid thistle]

can we talk about this in a vc, i feel like i can explain myself better

never mind, there is no vc's on this server

ok

which value are you referring to in which i circled?

[hexed ermine]

The circle after the -2

-5 is a zero so that means a zero will also be at +-20/7

[viscid thistle]

in the desmos graph?

that's the "last" interval

[hexed ermine]

So -5+(20/7) and -5-(20/7) which gi es you -15/7 and -55/7

[viscid thistle]

yes

-15/7 should be the last interval

BUT

since we are subtracting both numbers, we get -55/7

which is not a flat 0 for the y

[hexed ermine]

-55/7 =x gives you a 0 for y

[viscid thistle]

no it doesnt

[hexed ermine]

I know your shits fucked up

Trust me, it's a zero

[viscid thistle]

can i see ur graph

[hexed ermine]

==-55/7

[granite stirrup]

-55/7 = -7.85714285714286

[viscid thistle]

now check this out

the last interval should be greater than -5

not less than -5

you see what's wrong tho right?

[hexed ermine]

Your period is just a magnitude, if you add 20/7 to -5 you get a value greater than -5

[viscid thistle]

yes

there should be no point less than -5

if my first point is gonna be -5

all points succeding should be >-5

[hexed ermine]

Why do you say that?

[viscid thistle]

because

the graph starts at (-5,0)

if im finding the period from there, then all the x values succeding should be >-5

so how is it that the last interval is <-5

that doesn't make sense

the graph doesn't go backwards

there's an order to these graphs. the last interval will never be less than the first interval. it doesn't make sense

the last interval should've been -55/7. i got that BUT, it's not giving a flat 0. it's giving me some fucked up scientific notation

also, -55/7 doesn't make sense because it's less than -5

[hexed ermine]

So you are saying that for a regular sin(x) graph, -2pi isnt valid

[viscid thistle]

im saying the graph shouldn't go backwards. that's what im confused about. why is it saying that the last interval is -55/7 when it gives me a scientific notation and it's less than the first interval

-2pi is valid

but since the amplitude and horizontal stretch are both negative, it reflects just as a regular sin graph should

they both cancel out

returning to it's natural form as if the negatives were not present

[hexed ermine]

The period is -20/7 so since you are saying -5 is the starting point you should go back to find the number after that since the period is negative

[viscid thistle]

im saying that the last interval shouldn't be -55/7

[hexed ermine]

Why not though?

[viscid thistle]

because it's not giving us a flat 0 as it should

desmos isn't fucking up, it's something with this equation

also, -55/7 is less than -5

which doesn't make sense

goow, what do u think

[spring thunder]

it's not giving a 0 cause floating arithmetic isn't accurate

[viscid thistle]

i don't get it

what's floating arithmetic

[spring thunder]

(And screw my internet)

[viscid thistle]

lmao

so ur saying that -55/7 is right?

[spring thunder]

The calculations with floating numbres

Yes

It's just that computers can't store infinite digits

So there's a small error

[viscid thistle]

Holy fuck...

Ur right

but im so confused

this is insane

[spring thunder]

trust your mind, not those computer scumbags

[viscid thistle]

but wait

ahh fuck this

im just gonna go off of -55/7

this question has fucked with my head too much

thanks for the help y'all

i appreciate it (:

[hexed ermine]

👌

[spring thunder]

🍻

[odd knoll]

I don’t know what phi symbol means

for number 131

it looks like theta but its wonky

[bitter nova]

it's the other 3d angle

[odd knoll]

huh

ive never seen that symbol before ;-;

[bitter nova]

yea it's just another greek letter for another angle

[odd knoll]

oh

so its just like

[bitter nova]

p h i

[odd knoll]

another variable

[bitter nova]

yea

[odd knoll]

wow

okay

then

welp that makes sense now thx xD

[grizzled solstice]

Guys I need help with my homework 🙏🙏

What is the polar form of this

[gritty blaze]

$$z=\sqrt{6}e^{i\arctan\left(\frac{\sqrt{2}}{2}\right)}$$.

[grizzled solstice]

Ahh thanks 🙏🙏

[odd lichen]

π/4 +2nπ >>> arctan(sqrt(2)/2)

[somber plaza]

@grizzled solstice think about a Pythagorean triangle.

The R in the polar form is the hypotenuse

So the angle must be the theta

[grizzled solstice]

I see . Thanks you very much 🙏😁 @somber plaza

[somber plaza]

But of course you may have to change the angle depending on which quadrant the point lies on in an Argand diagram .@grizzled solstice

[grizzled solstice]

I got it ! Thx again @somber plaza

[subtle narwhal]

how do i find the difference quotient for this

im so lost

[spring thunder]

still need ahelp ? @subtle narwhal

[vapid skiff]

Where can I post trig questions?

[spring thunder]

here is pretty relevant

[vapid skiff]

Not to sure how to do this

[vapid skiff]

This is what I've got so far

[somber plaza]

@vapid skiff graph sin(x + 2π /3 ) first

then shrink that graph by a third vertically

[subtle narwhal]

@spring thunder yes

[spring thunder]

ok so you know what a difference quotient is at least?

[subtle narwhal]

yeah

f(x+h)-f(x) all over h

[spring thunder]

well here the variable seems to be t, but w/e

$$\frac{f(t+h)-f(t)}{h}$$

[spring thunder]

so f(t+h) blocks you i believe....

[subtle narwhal]

yes

and square root too

[spring thunder]

if i took a function like g : x -> 2x+3, what's g(t+h)?

or you just have no idea

[spring thunder]

i mean, i didn't tell you to use symbolab

[subtle narwhal]

i didnt use symbolab

i just use their bad

bar

so its easier

to write

[spring thunder]

ah k lel

[subtle narwhal]

i didnt actually calcualte it

[spring thunder]

i was like wtf

[subtle narwhal]

thats stupid whats the point of coming here then lol

[spring thunder]

so you did 2(x+h) + 3 right (ie 'replaced' that x with x+h)

so for $$h : x \mapsto \frac{1}{4x}$$

[spring thunder]

waht's h(t+h)?

(and screw my notations but just do it)

[subtle narwhal]

1/4t+4h

all over h ( i dont know how to write it over h lol )

but how do i simplify it further?

[patent beacon]

Get a common denominator

[spring thunder]

put them at the same denominator and rationalize basically

[vapid skiff]

@somber plaza how would I do that?

[solid reef]

#48?

[ruby hill]

help D:

[gleaming vine]

?

[ruby hill]

i'm not good at writing functions from looking at a graph of sinusoidal function

[gleaming vine]

https://www.khanacademy.org/math/algebra2/trig-functions/constructing-sinusoids-alg2/v/trig-function-equation

maybe this?

idk i just read the title

[ruby hill]

ahhh.

[north bough]

how do I write this as an algebraic expression not using trig functions?

=tex sin\left(cos^{-1}\left(x\right)+sin^{-1}\left(x\right)\right)

[gritty blaze]

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$.

[north bough]

I'm aware of the addition identity for sine, but how would you represent the inverse cosine and sine functions?

[civic plaza]

let a=cos^-1(x) and let b=sin^-1 (x)

[rocky bison]

exactly the same way

[civic plaza]

that should be easier to see

[north bough]

why can we arbitrarily do that?

[civic plaza]

what do you mean

we aren't doing anything

in the end a is still cos^-1(x) and b is still sin^-1 (x)

[north bough]

ok

can you represent the values in terms of ratios of x?

like sin B in this case would be x

[gritty blaze]

$$
\begin{aligned}
\sin(\arccos(x)+\arcsin(x)) &= \sin(\arccos(x))\cos(\arcsin(x))+\cos(\arccos(x))\sin(\arcsin(x)) \
&=\sqrt{1-\cos^2(\arccos(x))}\sqrt{1-\sin^2(\arcsin(x))} + x^2 \
&=1-x^2+x^2 \
&=1
\end{aligned}
$$.

[desert jackal]

@sour wolf

[sour wolf]

yes?

[quartz bronze]

I know where the mistakes were made but not how to show the correction algebraically.

[fickle moat]

I have a question for number 14 and 15, 18
For number 13
Log 3 35 then i know its log 3 5 + log 3 7
But for number 14 15 17¿? How do you solve it

[spring thunder]

@fickle moat yep so,

$$\log(ab) = \log(a)+\log(b)$$ right

[fickle moat]

yes

[spring thunder]

that's what you used for the first q

[fickle moat]

itf its fraction you -

ok so the 14 log 3 5/7 is log 3 5 - log3 7?

[spring thunder]

yea, do you get why we substract?

[fickle moat]

i wrote in equation here

but

lost bit

@spring thunder is it this correct?

something like this?

[spring thunder]

yea those are correct

[fickle moat]

so if then

log 3 5/7

log 3 5 - log 3 7?

[spring thunder]

yea i didn't say that was false

[fickle moat]

than for 18 log 3 45/49

its subtraction

but can i like divide by 9 and 7

log 3 5/7

log 3 5 - log 3 7

i am not sure even i am doing this correct

[spring thunder]

just go with the easier things first

(and yeah log_3(45/49) isn't log_3(5/7))

[fickle moat]

so the answer for 18 is log 3 5 -log 3 7

your just following the rules of properties and same time solving it

i just have so many doubting myself...

[spring thunder]

no?

$$\log_3(\frac{45}{49}) = \log_3{45} - \log_3{49}$$

[spring thunder]

see

let's just start from this

[fickle moat]

oh you dont solve it

ok

[spring thunder]

they just want to write that big log_3(45/49) with some log_3(5) and log_3(7), that's all they want

and i didn't say this had only one step

[fickle moat]

so in the end you can write this in any form of -

45 -49 and 5 -7

so for number 15. log 3 7/25 = log 3 7 -log 3 25

[spring thunder]

and 25 is 5^2

[fickle moat]

log 3 7 -log 3 5^2?

[spring thunder]

yup and then what can you do?

[fickle moat]

do you have to solve for ^2?

i dont understand by mean what can i do afterward

[spring thunder]

what do you even mean by solve?

it's not an equation

can you use a rule to get something*log_3(5) instead of that log_3(25)

that's what i mean

[fickle moat]

now next question i hve is log 3 175

[spring thunder]

we're not finished with that one

[fickle moat]

oh

um

[spring thunder]

$$\log(a^b) = b\log(a)$$ hmmm

[fickle moat]

log 3 7 - 2log 3 5?

[spring thunder]

yas^^

[fickle moat]

so for 16

log 3 175

i will have to change that to

log b x^power?

[spring thunder]

yeah that's the idea

[fickle moat]

and do the samething

as 15

175..hm...

[spring thunder]

it will be a little bit more complicated than just log_b(x^some power) tho

[fickle moat]

so

for 175..

[spring thunder]

do you know what prime factor decomposition is?

[fickle moat]

no

i might either forgotten or

[spring thunder]

each integer > 1 can be uniquely decomposed as a product of prime factors

and a way to find it to start by the smallest prime number existing (ie 2)

[fickle moat]

so its like finding constand and LC

constant"

constand of 10 is 1 2 5 10

etc

CONSTANT""""

175 / 5 = 35 / 5 =7

5 x 5 x 7?

[spring thunder]

yas

ie 5^2 * 7

[fickle moat]

so than log 3 5^ log 3 7

2 log 3 5 log 3 7

?

[spring thunder]

you forgot the + between the two but yeah

[fickle moat]

OOPs

okok so i got the right idea

i finished the homework but i wanted to get better at it for my exam next week

[spring thunder]

👌 sure

[fickle moat]

so far i studied 2.7 3.1 3.2 solving least 90 problems

including word problems

so i can understand the full chapter

thank you!

for this you need to use subtract

to do that do you have to x power to cube to cancel the root along with 3?

[spring thunder]

how would you write cubert(x) as a fractional exponent?

that's the trick @fickle moat

[fickle moat]

i solved it i think

one moment

Is this correct step?

[spring thunder]

yup that's the right step

now 1/3 hmmm......

$$\log(b \cdot \frac1{b}) = \log{1} = \mathbf{0 = \log{b}+\log{\frac{1}{b}}}$$

[spring thunder]

ie log(1/b) = -log(b)

or you can just rewrite 1/3 as a power of 3

[tardy olive]

find all solutions to 3x - 2^x- 1 =0

[thick raptor]

ez @tardy olive

The two real solutions are trivially 1 and 3

and you can show with basic calculus there are no other real solutions

The general solutions can be given by the Lambert W function.

[tardy olive]

uh how do we solve this without calculus

@thick raptor