precalculus

[scenic charm]

hmm ok

so i completed the square and got y= 1+-sqrt(1-x)

idk if that helps

[spice canopy]

yes thank you

oh but how did u complete the square

like the steps

i didn't know what to do at the beginning of completing the square

[scenic charm]

y^2+2y=x

turn y into a perfect square

-(y^2+y)=x

*forgot to put 2y

-(y^2+2y+1)=x+1

-(y+1)^2=x+1

(y+1)^2=1-x (since I need to isolate (y+1)^2)

y+1=+-sqrt(1-x)

y=-1+-sqrt(1-x)

I messed up the first one on accident sorry

[fickle moat]

what do do here

[hexed ermine]

Computes the amount of days after x amount of money at $26 per day

[fickle moat]

oh so you dont even have to solve

[hexed ermine]

No not reall

y

[fickle moat]

i always thought of

x = rent per day

i just wanted make sure

[hexed ermine]

x=days for f(x) and y=cost

inverse function flips it

so x=cost, y=days

[fickle moat]

i seee 😄

🙂

[hexed ermine]

x is your independent variable

👍

[fickle moat]

i m ready for that final

lol

i m good on trig

the answer should be B correct?

[serene heath]

yes

[fickle moat]

why friend is saying its A

but i think its be

-2<5

9< equal 5

[serene heath]

0<5

therefore f(0)=0-2=-2

[fickle moat]

yep

9-0 =9

[hexed ermine]

Do you know the half angle for sine?

$ \sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}} $

[obsidian monolith]

PJS:

[hexed ermine]

since its squared, the sqrt cancels

So.. $ \sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2} $

[obsidian monolith]

PJS:

[hexed ermine]

The 2's cancel

so you are left with all together, $ \frac{1-\cos\theta}{x^2} $

[obsidian monolith]

PJS:

[hexed ermine]

Well you can apply Lhopitals for that

Since its 0/0

It reduces to sin(x)/2x

And now you can apply it again

1/2*cos(x)

1/2*cos(0) -> 1/2

[rocky bison]

HOPITAL

[hexed ermine]

Frigg off

[scenic charm]

Find cos(11pi/12) exactly

help

[fallen cloud]

well

you know the half angle formula right?

and the sum angle formula?

[tawny stag]

Can someone help me with this problem

I want to learn how to do it

[scenic charm]

Which problem ? the whole thing?

[tawny stag]

for part a

the domain and the vertical asym is 3 and -3 right

[rocky bison]

YUe

GO STUDY DATA STRUCTURES

DO UR DJIKSTRA

[solemn tiger]

Part a asks for the domain. That is, it asks for all x-values such that the function is defined.

I'M TAKING A BREAK

[rocky bison]

NO

[solemn tiger]

As you should already know, dividing by 0 is bad.

So we check the denominator.

[rocky bison]

[solemn tiger]

When x^2 - 9 = 0, the function is undefined at that point.

From the relation (a - b)(a + b) = a^2 - b^2

We can factor x^2 - 9 to (x - 3)(x + 3)

Which means x will be undefined at 3 and -3.

[rocky bison]

x will be undefined? 👀

[solemn tiger]

:L

The output.

Because dividing by 0 makes the world end or something

[tawny stag]

domain is x cannot be -3 or 3

[rocky bison]

yes

Do you understand why

[tawny stag]

yes

and for part b

to find out if it inverse you do the horizontal yest right

[viscid thistle]

What is the order of translation sfor the function asin(b(x+c))+d

[meager topaz]

f(x) = sin(x) > sin(x+c) > sin(b(x+c)) > asin(b(x+c)) > asin(b(x+c)) +d.

[viscid thistle]

I think I mispoke, I meant order of transformation

@meager topaz

[meager topaz]

That's the order of transformation only I guess ? Route from f(x) = sinx to g(x) = asin(b(x+c)) + d ??

@viscid thistle

[viscid thistle]

I want to be sure I'm understanding your pervious comment. Are you saying that those are the transformations from left to right? So first we do the horizontal shift, then the horizontal stretch/compression etc..?

@viscid thistle

@meager topaz

[meager topaz]

@viscid thistle yeah you can first shift the graph and after that work on stretching it

[oak tulip]

I need help

I need help with the evaluate (25 over 21)

The question says evaluate the summation notation

[oak tulip]

<@&286206848099549185>

[serene heath]

that is 25C21

[oak tulip]

Oh

Lop

[oak tulip]

So my question says find the sum of the infinite series

The problem is (3x-2)^9 and then it says find the 5th term

[serene heath]

google images

[buoyant ore]

Hi, i got this problem on a test and i was wondering what the right steps would be to solve for x, im not sure if i did it right

[hexed ermine]

4y=2e^x-3/e^x first

[solemn tiger]

Looks like quadratic formula? 🤔

[hexed ermine]

Yeah

[obsidian monolith]

Yue:

Yue:

Yue:

Yue:

[solemn tiger]

Then some quadratic formula shenanigans for the two solutions 🤔

[viscid thistle]

Can someone help me simplify (7sqrt(5) - 2sqrt(6)) * (4sqrt(5)-8) or
28sqrt(25) - 56sqrt(5) - 8sqrt(30) + 16sqrt(6))

[reef hearth]

,tex $$ (7\sqrt{5} - 2\sqrt{6}) * (4\sqrt{5}-8) $$ \nline $$28\sqrt{25} - 56\sqrt{5} - 8\sqrt{30} + 16\sqrt{6}) $$

[obsidian monolith]

Pastah:

[runic blade]

If I may ask another question.
How do I find the answers for 13 and 14?

Number 13 Requires the Compound Interest Formula, but what about 14?

[solemn tiger]

Your goal is to find the time for your principal amount to double.

That is, set your total amount to 2*principal

And solve from there using logs

[runic blade]

Is there a formula for the Logs portion?

[zinc granite]

@runic blade for the logarithms, there isn't really a formula for it. You have to use the log function on your calculator, and, in a way, logarithm is the inverse operation of exponents.

[eager thunder]

here are the log rules if it helps https://cdn.discordapp.com/attachments/451980214428499971/522269593041895427/7bb67e9d2a4bcff3dedfeb1e6b86982b.png

so you should have the equation 4000 = (2000)(1.05)^t

divide by 2000

so 2000 = 1.05^t

log both sides so: log2000 = tlog1.5

then divide by log1.5 so t = log2000/log1.5 (tap that into calculator)

thats for 14

[earnest finch]

What are the root sum and product laws again? I'm starting to study for my final.

[limber bone]

?

what

[royal gull]

@limber bone umm you told me youre dead how are you typing then?

[limber bone]

uh oh

[scenic charm]

5^2x-1=10^x+2

can someone help me plss

[keen root]

is this odd, even or neither

[solemn tiger]

What is the definition of an odd function?

What is the definition of an even function?

Does that function fit either of those parameters?

[viscid thistle]

Anyone have a pencil I can burrow?

[serene heath]

✏

here u go

[viscid thistle]

Ty

np

[meager topaz]

Smh

[viscid thistle]

How do we know step 1/4 is true?

<@&286206848099549185>

[serene fable]

See rules

Gotta wait 15 minutes before pinging

[solemn tiger]

The range of the sine function is [-1, 1]

A(t) will clearly reach its maximum when sine is at its minimum.

[serene fable]

But the function represents its altitude relative to the earth, so when A(t) is at the minimum, it'll have reached the lowest altitude, i.e. the point where it's closest to the earth, i.e. the perigee

@viscid thistle

Does that answer your question?

[viscid thistle]

Yes, thank you. It makes more sense now 😃 @serene fable

[leaden tinsel]

I am working on this problem:
When I stand 40 feet away from a vertical tree, the angle of elevation of the top of the tree is 𝟒𝟓° and the angle of depression of the bottom of the tree is 𝟏𝟎° . How tall is the tree, round your answer to the nearest foot?

What does it mean when it says angle of depression of the bottom of the tree?

This is what I have so far

[meager topaz]

Apply tan45 = perp /base

[leaden tinsel]

So the 𝟏𝟎° is therte just to throw me off?

I initially thought the same thing

[viscid thistle]

How can I determine an equation given two x roots

In step 3/4 how does t+23.2=23.2+92.8n?

,tex $$ (6 + sqrt{3}), (6 - sqrt{3})$$

[obsidian monolith]

cix:

[solemn tiger]

When the sine function is pi/2 or any coterminal angle to pi/2, the sine function reaches its maximum.

[leaden tinsel]

@viscid thistle What equation are you trying to get? slope?

[solemn tiger]

If that's a polynomial equation, think about how you'd find the zeroes given a factored equation.

Work backwards from there.

[gritty blaze]

Use \sqrt.

[viscid thistle]

Thanks I got it now. @solemn tiger

[solemn tiger]

👍

[viscid thistle]

How do we determine the range over which the principal value of an inverse trig function is defined?

[solemn tiger]

The y-values?

We know that the inverse of anything switches the domain and the range.

So we can check the domain of the original for the range of the inverse.

[viscid thistle]

Got it, so the inverse is only defined over the range where it remains a function right? So the domain of arccos must be between 0 and 180 degrees because outside of those parameters the function would fail the vertical line test?

[viscid thistle]

What are x restrictions if y=root(x-2)

[hexed ermine]

Well when we are talking about the reals, the radicand has to be positive

So x-2>=0

So the restriction for x is >=2

[viscid thistle]

I don’t think that’s right

My text book says x>=-2 or no restrictions

No you’re right

Hmmm

[limber bone]

do u know wwhy

[gusty kraken]

would this be correct

finding derivative of (7x^6- 4/x^3 + 6√ ̅x^4)

equaling (42x + 12/x^4 + 4/6√ ̅x^2)

my teacher is really confusing, so I may be way off with this one

1 a.m. and im still doing homework yikes haha

imma go to bed, please @ me if you can confirm my math. thanks! 😄

[limber bone]

use https://www.derivative-calculator.net/

[gusty kraken]

Oh wow, didn’t even know that existed haha

Thank you! Appreciate it

[limber bone]

np

[heavy sedge]

@gusty kraken that's also #calculus, not pre-calc. Not that it matters too much.

https://www.symbolab.com/ is another good one

[limber bone]

ye

[pseudo nexus]

Anyone help me with this precal paper

1-4 I’m not sure where to go from there

[hexed ermine]

Multiply by pi/180 to convert to radians

or 180/pi to convert to degrees

[pseudo nexus]

So to go from radians to degrees id rinse and repeat?

O wait I see

What about the rest I missed a lot in class

[hexed ermine]

if its in 0 to 90 its Q1

if its from 90 to 180 its q2

if its 180to270 its q3

and from 270 to 360 its q4

thats for degrees

[pseudo nexus]

Alright

Would you go backwards for negatives

[hexed ermine]

0 to pi/2 for q1. pi/2 to pi for q2. pi to 3pi/2 for q3 and 3pi/2 to 4pi(0) for q4

[pseudo nexus]

What about the degree-minute-second thing

[hexed ermine]

multiply the decimal portion by 60

the part thats not a decimal is the minute portion

multiply the decimal portion from the minute to find the second

[pseudo nexus]

Is that it?

[hexed ermine]

yeah

[pseudo nexus]

Thanks man

[hexed ermine]

to convert to decimal you divide the minute by 60 and divide the second by 3600 and add the two decimals

to the degree

[viscid thistle]

How do I create a quadratic from these solutions?

[limber bone]

what i know is that u can create a quadratic from its solution

s

but a system

idk 😦

[flat turret]

for the first just take any linear and any quadratic equation that has these two solutions

and for the second any two quadratic equations

[limber bone]

thats totally arbitiraly

" how do u spell that word "

[flat turret]

arbitrarily

learn your vocab

[viscid thistle]

[serene heath]

oof

[past veldt]

The linear function you take is determined by those points

[limber bone]

😦

[flat turret]

for the quadratic just take anything

[limber bone]

thats totally arbitrarily

tho

what type of math is this

delete this question

NOW

[viscid thistle]

arbitrary*

[limber bone]

jk

oh rly?

wtf

thats totally arbitary

what type of math is this

delete this question

NOW

[past veldt]

Your linear thing is y=linear function

you have two points

so that determines the line

I assume what they're meaning for a linear system is

y=linear function
y=quadratic function
solve

Think about it graphically if it helps:

You've got a straight line and a parabola and you can move them about as you please

you want them to intersect at some specific points

@viscid thistle

[viscid thistle]

So what are the steps to get quadratic?

[echo smelt]

Using reciprocal, quotient, or pythagorean properties, how would you prove that cos(x) * [sec(x) * cos(x) + sin(x)] = csc^2(x)

I'm very lost on this one

[fringe stream]

are you sure you have the right question?

[viscid thistle]

How come area is a vector but volume isn't

[crisp radish]

Area isn't a vector

[vestal dew]

I think they were referring to the area vector

[viscid thistle]

Solving a system of equations by elimination. I can’t seem to get the correct answer (-0.5,4) and (3,25)

[serene heath]

-4x2 =8?

[viscid thistle]

?

[serene heath]

your second equation

its y-4

but it turns into 2y+8?

[viscid thistle]

Oh

Didn’t see that let me try again

No it’s right?

I have to double it to have y on top = y on bottom

And flip the signs/subtract to remove the term

Then I get 2x^2-5x+3

But that yields incorrect solution so idk

[serene heath]

yes but it should be 2y-8

not 2y+8

[viscid thistle]

-3

[solemn tiger]

🤔

Complete the square on both sides

Or rather, isolate y first, then do that.

Set them equal to each other to get the two x-coordinate solutions

Plug them back in for the y-coordinate solutions

[viscid thistle]

@solemn tiger

[rose roost]

Completing the square

Can someone help me out please

[frozen needle]

Do +25 on both sides

and there you have your completed square

[rose roost]

That easy?

Thanks man

[frozen needle]

You're welcome

[rose roost]

If í send a series of exercise

Can u help me out?

Like 8 exercises?

[frozen needle]

As a human being, I do need some sleep, and it's getting quite late here in France, so you'll have to wait for someone else to come help ( ^_^)

[rose roost]

Ok

Thanks man

[frozen needle]

You're welcome

[dusky parrot]

If you post them here someone will probably ehlp @rose roost

[dusky parrot]

1. if you recall, direct variation is given by the equation y = kx
   plug in the first point to find k
   18 = k(12)
   k = 18/12 = 3/2

y = (3/2)x | y=16
16 = (3/2)x
x= 32/3

2. to find f of g you plug gx into every place where x appears in f

f(x) = 3x^2 - 6
g(x) = 2x+5
f(g(x)) = 3((2x+5)^2)-6

sorry i dont know how to Latex

[rose roost]

Thanks

[bright dock]

am really struggling with proving trig identites

[viscid thistle]

Hello, mind if I ask for help?

This has been bothering me for days now.

[solemn tiger]

The price of the can can be directly calculated by finding the area of the top, bottom, and the side, and multiplying each component by the cost of the material.

To get it as a function of the radius, you may want to isolate h (height) in the volume formula, and substitute that into your cost equation.

[viscid thistle]

C=2Pi(r)

C=3PIr^2

How can I do b)?

[rose roost]

number 11 and 14 are the ones i am having problems

[hexed ermine]

$ f(x)=\sqrt{-3x^2+5} $ find $ f^{-1}(x)? $

[obsidian monolith]

PJS:

[hexed ermine]

Can't really see it

If so just switch the variables and solve for y

$ x=\sqrt{-3y^2+5} $

[obsidian monolith]

PJS:

[hexed ermine]

x^2-5=-3y^2

$ f^{-1}(x)=\left\sqrt{\frac{x^2-5}{-3}\right} $

[obsidian monolith]

PJS:
Compile Error! Click the reaction for details. (You may edit your message)

[rose roost]

thousand thanks man

i see that i have to use the laws of logarithm

but everytime i do it and compare my answer to the answer key i have it wrong

[hexed ermine]

$ \log_{4}(\sqrt[6]{3x^2+5}) $

Expand this out?

[obsidian monolith]

PJS:

[hexed ermine]

First of all bring out the 1/6

$ \frac{1}{6}\cdot\log_{4}(3x^2+5) $

[obsidian monolith]

PJS:

[hexed ermine]

There's nothing else you can really do

because 3x^2 and 5 are adding

[rose roost]

thanks

[hexed ermine]

move the 6sqrt(x) over

and square both sides

[rose roost]

do i end up having 2 answers?

[dusky parrot]

you should

[solemn tiger]

Yes, but think about what solutions are valid for a square root function.

[dusky parrot]

or none

or 1

it should be a quadratic

[rose roost]

am lost

let me see what i do

thanks for the help

[hexed ermine]

When you square both sides and test for valid x values, you have to negate negative values because you have the sqrt(x)

that would be an extraneous solution

[rose roost]

i dont get it

[solemn tiger]

When you solve the quadratic, you'll yield two solutions.

Looking at the original function, you have a square root.

Square roots are only valid for x in the set of [0, infinity)

So we ignore negative solutions.

[rose roost]

thanks i

is it right?

[hexed ermine]

Nope

you have to multiply by the conjugate

conjugate a+bi is a-bi

so rewrite 2i+5 as 5+2i

so the conjugate would be 5-2i

[indigo mulch]

uhhh I have a final tomorrow and I don't remember how to do #8 and I lost my notes lol

[serene heath]

try turn into the form $R\sin(x-\alpha)$

[obsidian monolith]

lemon catto:

[hasty wolf]

Well, $R\sin(x+\alpha)$ the way I've seen it

[obsidian monolith]

DMAshura:

[hasty wolf]

@indigo mulch : Do $R=\sqrt{a^2+b^2}$ and $\alpha = \arctan{\frac{b}{a}}$ look familiar?

[obsidian monolith]

DMAshura:

[viscid patrol]

Hello, could somebody help with this

Consider the demand function q = −3p + 150. Find the point elasticity when (a) p = 10.
(b) p = 40.
(c) Why is demand elastic when p is high and inelastic when p is low, in this example?

[viscid thistle]

I don’t understand quadratic-quadratic systems and I have a test tmo 😦

[limber bone]

equate them

what do u not understand

[viscid thistle]

Ignore q6...... 5a) i can’t seem to get the right answer. I substituted y=second equation

[limber bone]

which 1 do u want?

[viscid thistle]

I can draw them because vertex form but for exact points I need to use quadratic right?

5a

[limber bone]

1/3(x+2)^2 + 2?

[viscid thistle]

becomes $1/3x^2 + 1 + 1/3x + 2/3 + 2$

[obsidian monolith]

» cix «:

[viscid thistle]

@limber bone

1.3x

[limber bone]

simplify both of them out

equate them

u will find them cancel out

[viscid thistle]

Hey become one linear?

[limber bone]

just see

and check

https://media.discordapp.net/attachments/363224154469826562/524308529725702144/image0.jpg?width=494&height=658

[viscid thistle]

Aight

[limber bone]

expand both of them

then equate them

for example for the first 1

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

1/3(x^2+4x+2)+2

distrubute 1/3 on

it

and add 2

and do the same for the second

and equate them

see what u get for x

the x u will get is the x coordinate of the intersection

to find y just plug in the x for any of them

doesnt matter since the x will be qeual to both

to find the y coordinate

got it?

[viscid thistle]

The x^2 cancel out, meaning there will only be one solution right?

[limber bone]

just see

not only the x^2

just see

do it

pelase

please*

[viscid thistle]

Aight

[limber bone]

gl <#

<3*

[viscid patrol]

Consider the demand function q = −3p + 150. Find the point elasticity when (a) p = 10.
(b) p = 40.

[viscid thistle]

How would I graph this

[solemn tiger]

Are those parabolas?

Use vertex form to find everything except the horizontal stretch, then plug in a test point to find that stretch.

[viscid thistle]

I found a by dividing q by (x-q)^2

And then I did

EquationA=equationB

To

EquationC = 0

Then used quadratic, but I’m getting incorrect solution

[viscid patrol]

Consider the supply function q = 0.25p−10, where p and q take only positive values. Find the point elasticity (a) when p = 44

[dense zealot]

What's point elasticity

So p is 44 and q is 1

And then take the partial derivative of q

With respect to p so

It is .25

And then p/q that's that is

Times

Is

44 * .25 = 11

[viscid thistle]

What is the meaning of maximum and minimum point here?
The length of the segment is: |b-a| or |a-b|, and the midpoint: (a+b)/2.
Why we add them in case of finding maximum point? What we are trying to find?

[serene heath]

maximum as in the maximum of the 2 numbers

so max{4,3}=4

[viscid thistle]

I know what max is @serene heath

Just it doesn't make any sense to me.

[serene heath]

(a+b)/2 is the mid point between the 2 numbers if you plot them on the number line and |b-a|/2 is half the distance between the 2 numbers

so adding them togethor brings you to the larger number

and subtracting them birngs you to the smaller one

[viscid thistle]

Where we are going to use these formulae? Can you give an example? @serene heath

[muted mason]

SOme use it for physics, some use for astronomy

[dense fable]

hello is someone available to help? (:

[solemn tiger]

[dense fable]

can someone help me find h(x) = f/g
f(x)=7-x, g(x) = 2x+1

i hope i worded that right lol

[hexed ermine]

Just divide 7-x by 2x+1

[dense fable]

the graph would look something like this, right?

[hexed ermine]

Mhm

[dense fable]

h(x) would be h(x)=7-x/2(x+1) , x ≠ -1 ?

[hexed ermine]

Well not 2(x+1)

Rather 2x+1

And x cant be -1/2

[dense fable]

ooh okay, thanks a lot!

[hexed ermine]

Np

[echo smelt]

Can someone verify my proof work

Cos(x) * (sec(x) + cos(x) * csc^2(x)) = csc^2(x) Cos(x) * (sec(x) + cos(x) * csc^2(x)) Cos(x) * (1/cos(x)) + cos(x) * ((1/sin(x)^2)) Cos(x) * (1/cos(x)) + ((1/sin(x) * cos(x)) Cos(x) * (1/cos(x) + (cos(x)/sin^2(x)) Cos(x) * (sin^2(x)/sin^2(x) * cos(x)) + (cos^2(x)/sin^2(x) * cos(x)) Cos(x) * (sin^2(x) + cos^2(x)/sin^2(x) * cos(x)) ((sin^2(x)+cos^2(x)) * cos(x)) / (sin^2(x) * cos(x)) (Sin^2(x) + cos^2(x)) / sin^2(x) (1/sin^2(x)) (csc^2(x)/1) ∴ Cos(x) * (sec(x) + cos(x) * csc^2(x)) = csc^2(x)

[civic plaza]

latex plz

[hexed ermine]

^

[echo smelt]

what does that mean @civic plaza

[hexed ermine]

Oh shit

Nick Babick

@echo smelt What's up man

[echo smelt]

Not much, hbu

[hexed ermine]

I'm good, I have mentioned how I do know you right? Haha

[dense seal]

@echo smelt youre right >.>

Just that you used a very dumb way to get to csc²(x)

[viscid thistle]

can you come up with a formal proof that ax is perpinducular to -x/a

[iron bobcat]

@viscid thistle how much information do you have?

[viscid thistle]

It always intersects the circle at certain points where the arc length between them is pi/2

I guess I'd rewrite the circle in parametric terms

Just enter it in desmos and see for your self. But visual aid is not a proof

[iron bobcat]

i've done that in geogebra, and it's not always perpendicular

[viscid thistle]

Why did you raise it to a power?

You multiply it by x not raise it by x

[iron bobcat]

oh.

do you have knowledge of asymptotes?

[viscid thistle]

Si

10th grade stuff

what about them?

[iron bobcat]

it seems like ax is the asymptote of -1/ax

[viscid thistle]

I meant -x/a

not -1/ax

[iron bobcat]

hmm

[viscid thistle]

[iron bobcat]

ah, okay

this I suppose is easier because this time they're actually perpendicular

lol

[viscid thistle]

yeah. But can you prove it?

[iron bobcat]

yes

this is what I came up with

do you know how to prove two vectors are perpendicular?

I mean, what is the definition of perpendicularity with two vectors?

[viscid thistle]

I know but how do you prove that?

[iron bobcat]

so what is it?

[viscid thistle]

It is asking the same question

Circular reasoning

[iron bobcat]

hmm: if I understand correctly, you must prove f(x) = ax and g(x) = x/a are perpendicular, right?

[viscid thistle]

si

[iron bobcat]

so my question is: what is the definition of perpendicular on two vectors

[viscid thistle]

g(x)= -x/a

[iron bobcat]

your definition

[viscid thistle]

[a,b] is perpindicular to [1/a,-b]

Same question

[iron bobcat]

no, what I mean is: take two vectors A and B; they are perpendicular if A.B = 0 right?

( . is dot product)

[viscid thistle]

If that is true, then either a or b is 0

???

[iron bobcat]

nope, are you familiar with vectors?

[viscid thistle]

si

[iron bobcat]

ok: so, for any two vectors, if a . b = 0 then they're perpendicular

notice that . (dot product) is not x (multiplication)

[viscid thistle]

oh

ok

[iron bobcat]

do you see?

so my strategy was to transform f and g into two vectors in R2 (which they are in a way, because f(x) and g(x) are both lines)

and showed that their director vectors were perpendicular to each other

do you know the definition of dot product?

@viscid thistle: anyway:

if we take v = (x,f(x)
w = (x,g(x)

then v.w = (x,ax) . (x,-1/ax) = xx + ax(-x/a) = x^2 - x^2 = 0

thus v and w are perpendicular

this is the best I can come up with, maybe someone else has a better resolution that doesn't involve vectors

[spring thunder]

one non vector way would be just doing some pythagoras : take two points at the same x-coord (say 1) and check that OB²+OC² = BC²

less general than going about it vectorially, but works in your case

[viscid thistle]

Show that f(A)+abs(g(A))= ((f(A)^2+A^2)+abs(g(A)^2+A^2)^1/2 and you are done

@spring thunder I like your approach better

Just plug in a and -1/a for the 2 functions then complete the square

Bada bing bada boom and your done

Mission accomplished. +Respect

[spring thunder]

but vectars are cool also

[viscid thistle]

sshhhh

Don't hurt my brain.

Also why do you use geogebra?

desmos is the shit

[spring thunder]

cause i just happen to have it dl on my pc

:/

[swift quail]

Doing a supposed to be group project on my own can someone help

Supposed to determine how long a fence is supposed to be for a baseball field

My idea is to make smaller triangles to accurately determine this

I made a very rough sketch if I could pm it to someone that could help

not sure if im doing it correctly

[viscid thistle]

Awe man

Anyone got tips on this?

Question d

[elfin night]

Hmm

Proving it?

[viscid thistle]

Yup yup

Double angle thing

No clue how to do it

[vale pewter]

,rotate

[viscid thistle]

Oh

[slender river]

probs expand it

[viscid thistle]

OwO

[elfin night]

$ \frac{1-\sin\left(2x\right)}{\cos\left(2x\right)} \ \frac{1-2\sin\left(x\right)\cos\left(x\right)}{\cos\left(x\right)^2-\sin\left(x\right)^2} \ \frac{1-2\sin\left(x\right)\cos\left(x\right)}{1-2\sin\left(x\right)^2} \ \frac{\sec\left(x\right)^2-2\tan\left(x\right)}{\sec\left(x\right)^2-2\tan\left(x\right)^2} \ \frac{1+\tan\left(x\right)^2-2\tan\left(x\right)}{1-\tan\left(x\right)^2}=\frac{\left(1-t\right)^2}{\left(1-t\right)\left(1+t\right)}=\frac{1-t}{1+t} $

[obsidian monolith]

Colen:

[elfin night]

Feels overcomplicated but works

[viscid thistle]

Good stuff thanks man

Got a test tomorrow

[elfin night]

gl

[hollow plank]

can someone check my work for number 5 please

[honest path]

@hollow plank it is correct if the coefficient next to the x^2 is 8

[viscid thistle]

hey

if you have the line y=-3x

[honest path]

Yes?

[viscid thistle]

and you're given that the terminal side is in Q1

Q2****

then youd just draw a triangle with lengths 3, 1, and sqrt(10), right?

(if you're trying to solve for sec(theta))

[honest path]

So the line y = -3x is a terminal side of an angle?

just to be clear

if that is the case, then that would be correct

[viscid thistle]

yes

[honest path]

and then you use inverse cos to find theta

[viscid thistle]

would you have to do that?

cant you just like

use the triangle

h/a

= -sqrt(10)

[honest path]

Is the question asking for the reference angle or the length of the hypotenuse?

[hollow plank]

@honest path how did you get 8?

[honest path]

oh crap my mind is killing me you are right

its 9

[viscid thistle]

the question is asking for sec(x)

[honest path]

ok so yes you need to find the hypotneuse which is sqrt 10 and then use inverse secant to find x

[viscid thistle]

you're not solving for x though

lol

[honest path]

ok

[viscid thistle]

wait so

how do you account for the part of the angle in Q1

[honest path]

I thought of another problem

[viscid thistle]

if you only use the triangle in q2 doesnt that overlook the rest of the triangle

[honest path]

well the angle is supplementary

[viscid thistle]

waiiit

so you use the reference angle

[honest path]

yes

[viscid thistle]

which is the rest of the 180 degrees above the x-axis

and then you find it on there?

[honest path]

yep

[viscid thistle]

so you would find sec(x) on the reference triangle?

[hollow plank]

(thank you for checking my answer 😃 ) @honest path

[honest path]

Wait wait wait

give me a moment to think

Yes, but it is completely unnecessary to use secant

it isnt a function on the calculator and I recommend you use sin or cos

so you can do sin ^ -1 and cos ^ -1

to find the reference angle

then 180 - reference angle to get theta

unless it is in radians...

[viscid thistle]

i dont think you get what i meant lol

secant is not unnecessary

the entire question is solving for sec(x)

but ty

[pseudo nexus]

Anyone help me on this final??

[viscid thistle]

Are you allowed to cheat?

[pseudo nexus]

🤔well not exactly

But I missed a whole month

[civic plaza]

no no one can help

[pseudo nexus]

🤧damn

[civic plaza]

in fact if you read #❓how-to-get-help you'll see that you can be banned for asking such a question

[pseudo nexus]

Not sure what your implying

[civic plaza]

I'm not implying anything

6. Do not ask for help on tests. Any violation of this will lead to appropriate action being taken at mod discretion.

[pseudo nexus]

Well explain

Damn well then this is a wast

Go ahead remove me

[viscid thistle]

You can remove yourself from this server if you choose. If you insist on being banned, someone will probably do it, but I recommend you choose to leave if that's your choice @pseudo nexus

[pseudo nexus]

How do I leave

[viscid thistle]

Right click on the server name, and click the red "Leave server" link. Also, feel free to PM me

[pseudo nexus]

Thanks mate sorry for breaking the rules and sure thing

[viscid thistle]

Can someone help me with a proof

Not really a proof but rewriting something in terms of something else

[frozen needle]

[viscid thistle]

Ok

[serene heath]

watchu say bout my chickn strips??

[viscid thistle]

Fuk yo chicken strips

Ik u have to set up like this

208= (x+5)(2x+1)(x+2)-(x)(3)(x+5)

But after u foil what do u do after

solve it

So descriptive

Idk how to solve it

[daring crest]

Discord would not let me write long messages. Hope you can see everything.

@viscid thistle

[viscid thistle]

Yes!

Thanks so much

[daring crest]

yw

[viscid thistle]

How are y’all so good at math

[wraith escarp]

genetic luck or a lotta work

[clever inlet]

^

but generally, almost usually the second

[wraith escarp]

I got the first one

[clever inlet]

lucky

but i assume you still have to put some of the work in

[wraith escarp]

nope

I literally have never studied for math

[dense zealot]

More so environmental factors

[vale pewter]

=pup solve 208=(x+5)(2x+1)(x+2) - x(3)(x+5)

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@vale pewter

[dense zealot]

boi

[earnest finch]

Wtf

A+ on my precalc final

yeet

[solar abyss]

y=e^e^t

[limber bone]

show it to me

now

!

[earnest finch]

@limber bone

[limber bone]

SHOW ME THE EXAM

NOWSW

[earnest finch]

what

The actual test

aaaa

[full token]

Congratulations

Also I kinda wanna see the test too

[bright dock]

send us the tes

t

we haven't taken it yet mate

[earnest finch]

In my school you don't get your final exams back physically but you get to temporarily see them after winter break, so I unfortunately can't send it

Also, I went on vacation the day after I took that exam.

But I am sure every one of you will do as good as me, 1st semester precalc is usually merely an extension of Algebra 2 and not too difficult

[calm thicket]

^

[dense zealot]

Precalculus is easy

U don't have to study

[viscid thistle]

I follow all of the steps in 1/4 easily but step 2/4 is confusing me a little
I took the inverse cos of both sides of "cos(9x)=2/7" to get "9x= arccos(2/7)" which I simplified to get "x=arccos(2/7)/9"
But this method gets me 12.5 degrees which is far from the correct answer, so clearly I'm making a some big mistakes here

<@&286206848099549185>

[thick raptor]

== acosd(2/7)/9

[granite stirrup]

20×acos(2/7)/π = 8.15538337788664

[thick raptor]

== acosd(2/7)

[granite stirrup]

180×acos(2/7)/π = 73.3984504009798

[thick raptor]

@viscid thistle you probably plugged it in wrong?

or maybe you're not in degrees?

[viscid thistle]

I divided 113 degrees by 9 to get 12.5

@thick raptor

[thick raptor]

did you use ?

== acosd(2)/7

[granite stirrup]

180×acos(2)/(7×π) = 10.7794470414596×i

[thick raptor]

well ig that's not right either hm

well in any case I have no idea what's wrong with your calculators, but your math is correct

[viscid thistle]

Hmm. Well Idk where the 9 multiplying the x went in step 2/4

Thats the main source of confusion for me

[thick raptor]

they simply did 9x = arccos(2/7) and computed arccos(2/7)

they'll probably divide by 9 on the next step

[viscid thistle]

ohhh

Should have looked farther ahead lol

[thick raptor]

[serene heath]

lol

[quaint elm]

Anyone know any fun things you can plot

[serene heath]

=pup batman symbol

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@serene heath

[viscid thistle]

Merry christmas everyone!

[earnest finch]

yeetmas

[quaint elm]

I'm currently messing with r = 1 - cos(a · theta)

Quite fun

[lucid oracle]

Do y^2=x^2(sinx^2+y^2)

And make sure you zoom out to y=15 and y=-20 for best results

[uncut sinew]

precalculus = trigonometry?

[fallen cloud]

the are in the same year but trig is slightly more advanced

[white ferry]

Best book for trig?

[blazing raven]

er @white ferry check out amazon.com if your country has it. Up to 100s of reviews for each book.

also #resources is supposed to have things of that type.

[odd yarrow]

Assuming one has learned Algebra II / Trigonometry, how long would it take to learn Precalculus

And how useful is it in calculsu

Calculus*

BC

[limber bone]

no need

just learn functions

and go learn calc

[patent beacon]

I've been hearing more and more that pre-calc isn't crazy important to learn calc. I don't know how education failed us so hard

[limber bone]

fuck

sorry

[prisma scroll]

There is obviously a reason why it is called "precalculus"

[raven birch]

what does pre calculus even consist of

[ornate cairn]

prepromorphisms

[raven birch]

pretty sure I learned it but I never called it that

[patent beacon]

I mean, how much of pre-calc is committed to finding the roots of higher order polynomials, and how useful is that in calc?

[raven birch]

oh that's pre calc

[patent beacon]

That, trig identities

Exponentials and logarithms

Those are useful for calc you have to know those

[limber bone]

ye u take these in algebra 2

aswell

?

[raven birch]

in high school I was taught a button of just highly specefic algebra, which you were supposed to memorize not understand, since you didn't have the chance to trial and error in a 2 hour exam with a million questions

I was home taught pre calc

and so far it only comes in handy every so often

[limber bone]

what are u studying now

[raven birch]

calculus

[viscid thistle]

In step 2/4 when applying the sine identity to solve for the other solution to the inverse sine functions we apply the sine identity and then divide by 20. Why do we solve the problem in this order? Why not divide by twenty and then apply the identity?

[serene heath]

youre trying to isolate x

so by applying the sin inverse you get 20x on its own

and then you divide by 20 to get just x

[viscid thistle]

ok this makes sense so far

so then we have x= arcsin(5/14)/20

We apply the identity to find the other value of sine and then divide

thats intuitive 😃 Thanks! @serene heath

[serene heath]

np

[buoyant dome]

Yeah guys ngl

I saw some people discussing the necessity of precalc

and since no one asked I’m going to weigh in anyway and say if you took algebra II you don’t need precalc lamo

now I’m not trying to infuriate anyone

But this is my opinion

I was able to grasp calculus without actually taking algebra 2 lol, I just knew a lot of math

[cloud cedar]

i've been struggling on this question

was wondering if anyone could help

[vale pewter]

what's the amplitude

[blazing raven]

0.5 😃

[vale pewter]

what's the frequency

[raven birch]

well it's doing three rotations on the unit circle to hit two extremums cuz 6pi=3*2pi

[cloud cedar]

so

<@&286206848099549185>

[slender river]

mate

bad pingus

ask teh question first

[dense zealot]

Hallo

U ping meh

I sue u

[cloud cedar]

lol

[dense zealot]

@slender riverits ok I can do this

[cloud cedar]

question is above ^

[slender river]

ok

[dense zealot]

Simple!

[slender river]

oh WAIT i got screwed

[dense zealot]

Look at the highest point

What is the y value

[cloud cedar]

.5

[dense zealot]

ok so what's the highest point normally

Of a sin/cos graph

[slender river]

apologies for being harsh so soon

[obsidian monolith]

Yue:

[dense zealot]

Yue just let me do it :(

[solemn tiger]

[dense zealot]

don't waste ur time

I have time to waste unlike u

[cloud cedar]

.5 or 1 ?

[dense zealot]

Anyways what yue said is all you need to do the problem, the main issue is do you understand
A) why this works
B) what the parent graphs are
C) have any trig experience

graphs a sin graph

y = sin(x)

[cloud cedar]

yeah

max is 1 then

[dense zealot]

Ok from 1 to .5

[cloud cedar]

1/2

[dense zealot]

That's what you will put in front

It's your amplitude

[cloud cedar]

ok so 1/2sin()x

[dense zealot]

Next, what is the period

Do you know how to find one?

[cloud cedar]

no

[dense zealot]

Ok so

A period is how long it takes for the sinusoidal graph to repeat

Sinusoidal just means graph of a trig function

[cloud cedar]

so like

[dense zealot]

so how long does this take to repeat?

[cloud cedar]

from one relative max to the next

[dense zealot]

yes

[cloud cedar]

the picture on the graph doesn't show it ?

[dense zealot]

Well graphs sinx

without any added information

[cloud cedar]

5

[dense zealot]

You will see we can see it repeats every up thing and then down thing

6pi

If I were to continue this graph you would see

that it appears in blocks every 6pi

Anyways, the number you put in the equation is 2pi/period

So 2pi/6pi is 1/3

[cloud cedar]

is 6pi the period for all sin functiosn then

[dense zealot]

No

Only this case

The normal period is 2pi

So the equation will be (1/2)sin((1/3)x)

[cloud cedar]

i don't understand how the normal period is 2pi

when i graph y=sin x

it's a distance of 6pi?

[slender river]

=pup graph sinx

[dense zealot]

I highly recommend you watch a Khan academy video on sinusoidal eauations/graphs

[granite stirrup]

Do more with Wolfram|Alpha pro

Query made by

@slender river

[dense zealot]

See, 2pi

Just go on Khan academy

[cloud cedar]

alright

[dense zealot]

I think it will help you quite a bit

[cloud cedar]

so the first number will always be the amplitude and the second is period

[dense zealot]

And then after that are 2 shifts, look at what yue posted. also type t!rep @dense zealot

[solemn tiger]

After x changes by 2pi units, the function repeats itself.

[dense zealot]

self advertising

[solemn tiger]

[dense zealot]

Lol

[cloud cedar]

doing in this channel alright

[dense zealot]

Ur a helper

Smh

[tidal galleon]

What is this equation called:

$ \frac{1}{(x+a)(x+b)} = \frac{1}{b-a}(\frac{1}{x+a}-\frac{1}{x+b}) $

[granite stirrup]

Command disabled

The sever owner has disabled that command in this location.

[tidal galleon]

well ...

[frozen needle]

,tex like this

[obsidian monolith]

Tuong:

[fallen cloud]

$ \frac{1}{(x+a)(x+b)} = \frac{1}{b-a}(\frac{1}{x+b}-\frac{1}{x+b}) $

[serene heath]

\frac

[obsidian monolith]

Yeat:

[serene heath]

[frozen needle]

\left \right tbh

[serene heath]

smh

casuals amma rite tuong?

[frozen needle]

( ^_^)

[tidal galleon]

so what is it called?

it says: we use it in nth derivatives.

[serene heath]

partial fraction decomposition

also why u got 2 bs on the right

[frozen needle]

typo probably

[tidal galleon]

yes typo its a

thanks

[obsidian monolith]

Akane Kawahara:

[viscid thistle]

What does the notation <BAC mean?

[fallen cloud]

the angle of BAC

in this case BAC is arctan(12/5)

or arcsin(12/13)

[viscid thistle]

So the first letter indicates the angle we are taking? Like <CAB would equal 90 degrees?

[fallen cloud]

no the middle letter is

[viscid thistle]

Okay that was my confusion

so <ACB is 90

[fallen cloud]

yep

[viscid thistle]

Sweet!

Thanks

[fallen cloud]

no problem

[viscid thistle]

Need some guidance

Should I try to learn how the sum and angle identities are derived or just accept them and move on?

@fallen cloud

[slender river]

depends on purpose

[fallen cloud]

Hm, I find that I learn best by deriving things. But yeah lookup a proof. (it doesn't need anything beyond precalc)

[slender river]

but yeah what yeat said it's good to

if you're trying to learn sure but if you just want to do quic mafs then you can go without

[rare zephyr]

What's the proof for sin a/line A = sin b/line B?

[atomic zodiac]

draw it

[cloud cedar]

can someone confirm if it would be 3/2

[clever inlet]

yes

that would be correct

[cloud cedar]

thank you

[swift glacier]

Would it be legal to solve sec(x) + 1 = tan(x) by squaring the entire equation?

[kind lotus]

no

[swift glacier]

ok cool

How would you precede then? Cause there is no identities possible

[limber bone]

you want to prove sec(x) + 1 = tan(x)?

[swift glacier]

No, I want to solve for x

[solar abyss]

you may be able to square both sides

[swift glacier]

That is allowed, right?

[solar abyss]

yes, but i think you gotta verify your answers that u get at the end

because not all of them might work

[swift glacier]

Oooh, okay, thanks

[slender river]

uhh that sounds like a fishy identity tbh

for these problems try not to do operations to both sides

stick to one side and do operations that do not change the value of the stuff youre working with

like multiplying by 1

and stuff

oh wait what the fuck i cant read

youre solving for x?

yeah whatever yall said is fine

smh i feel dumb

[dense zealot]

@slender river it's ok to be dumb

[rare zephyr]

Correct Answer: 4

My answer: Out of place

I stopped there cause that's where reailty starts to break

Any help?

[serene heath]

try writing 2 as log base 2 of 4

nvm you did something similar

your working is correct

just solve the last equation

@rare zephyr

expand the right side

and combine with the left

if gives you x=4

[rare zephyr]

Ok I'll try your second method

[serene heath]

yes

that equation is correct

just solve it for x

[rare zephyr]

Damn I actually got the answer

I guess the reason why I've failed so many times is probably because of careless mistake

Now that I look back at the failed attempts

[serene heath]

nice

[swift glacier]

Is there anyway to simplify cot( π/7 - π/2)?

I know it's not a cofunction, cause the π/2 is in the back, but if I tried to do the tangent sum formula, tan(π/2) is undefined