#precalculus
1 messages · Page 301 of 1
yes
yeah, (-3,0)
but it achieves a zero for -3
smallest integer n such that 3n > -3 is 0
so whether it means a point or an interval doesn't matter because both the point (-3,0) and the interval (-3,0) are incorrect, as the correct answer is [-3,0)
which might as well be written [-3, -1]
only integers are used here
They are using interval notation but only referring to integers? 
what's wrong with it?
besides, it's unclear whether they mean a point or an interval
Interval notation is almost always used for all real numbers between the endpoints. It only refers to integers in rare cases like in number theory if you don't work with real numbers at all.
it fits the definition, no?
In a precalc or calc course, they really shouldn't do that especially since students already get confused about interval notation
What definition?
of an interval
Do you have a definition?
(-3, 0) = {x in Z : 3<x<0}
You left out a minus, but sure if you define it that way, it is correct. I am saying I think it is bad to define it that way since it is non-standard and a lot of students already think intervals only contain integers when they encounter them so seems like a bad idea to promote that
alright
I agree the answer is [-3, 0) regardless of whether it only means integers or real numbers
thank you all
there is a less sus notation for integer ranges
@willow bear is this right?
no
look at your very last line
the endpoints are (-6,3) and (6,3)
this literally flies in the face of the problem statement
congratulations. you've managed to completely disregard the problem.
?
a : b
defined as {x in Z | a <= x <= b}
@willow bear how about this one? this is from my friend
🙃
i mean it looks correct now
tell your friend she's succeeded at making the computer do it
Our professor gave us permission to use this website to check if our answer is correct
If you have ever programmed python you'd know it's an index slice
but that will not contain 'b' 
Correctamundo
This might not be the fastest route:
That's about 257.12
The mirror in Alan’s flashlight is a paraboloid of revolution. If the mirror is 8 centimeters in
diameter and 4.5 centimeters deep, where should the light bulb be placed so it is at the focus
of the mirror?
I got 0.889cm if rounded to three decimal places.
Can someone check to see if I'm right?
yes, bascially we should find 'a' assuming the equation of parabola to be y^2=4ax
your answer is correct
How do I calculate the % of males that live in a village if I know the number of women/100 males on avg, in cities, and in villages?
Assuming there are only women and men in the village, let's say there are x women per 100 males. Then out of 100+x people, there are 100 males. So, 100/(100+x) * 100 will give you the percentage of men
Thanks
You've given a wrong answer but I had thought I could use it to calculate the correct one. You've calculated the % of males in the villages, whereas the question asks you to do so generally. The idea is that since the avg in this case is 96.7 women/100 males, whereas for the cities it's 99.7/100 and for villages it's 91.6/100 you have to take into account the fact that the number of people that live in the cities is greater
Your question said you knew the average number of women/ 100 men in villages. But okay, so long as you have the correct answer now.
i don't see how luna's statement is wrong
the proportion of men in the village is 100/(100+x) where x is the number of women per hundred men
which you say you know the value of
(# of women living in the cities + # of women living in the villages)/(# of men living in the cities + # of men living in the villages) = 91.6/100
The question asks you to find the # of males living in the villages/( # of males living in the villages + # of males living in the cities)
okay, hold on
do you have a screenshot of the problem?
even if it's in a foreign language.
as-is, your wording is making me confused.
Idk how to calculate how many more men live in the cities than villages if I know these proportions between men and women
should i repeat myself?
...alright looks like my knowledge of russian is not quite enough to decipher the polish
can you translate 96b?
Calculate the % of males aged 20-24 that lived in the villages
so as I had said
The question asks you to find the # of males living in the villages/( # of males living in the villages + # of males living in the cities)
% of males aged 20-24 that lived in the villages
among all males?
You know only that women/100men for the age bracket 20-24 is: avg = 96.7, in cities 99.7, in the villages 91.6
Yeah
so wouldn't this require knowing the number of people in every age group?
Nah, it only asks for the 20-24 age bracket
Where have I said all males
% of males aged 20-24
and then i asked "among all males?" and you said yes.
I meant aged 20-24
Sorry
@willow bear Could you, please?
no
Also, how do I solve:
Show that the following inequality is true
sqrt(2^50 + 1) + sqrt(2^50-1) < 2^26
i have a solution that i am 100% certain you'll be unsatisfied with
...actually no nevermind
there is one that doesn't involve convexity
show $\sqrt{x^2 + 1} + \sqrt{x^2 - 1} < 2x$ for $x \geq 1$, then take $x = 2^{25}$.
Ann
you might find it easier to prove $\sqrt{x^2 + 1} - x < x - \sqrt{x^2 - 1}$ instead, which is equivalent.
Ann
I think you are missing a +1 on the LHS of the 2nd red line
Doesn't make sense
Why not?
Try to work it out yourself; I think you will reach the same result
The write-up is fine
IMO 1st redline isn't equal to the 2nd
due to LHS being different
They differ by 1
Oh wait
Nevermind, I didn't see the. 1 there
I'm blind sry
Yeah.it's corrext
Huge thanks
Suppose that the fare is $p.
Try to express the revenue in terms of p
What do they want?
I don't understand the question
Why are so many questions in books super vague
I think, if they give you sin(x)
Then you need to find cos(x) and tan(x).
And if they give you cos(x)
Then you need to find sin(x) and tan(x).
And if they give you tan(x)
Then you need to find sin(x) and cos(x).
How to do such a thing ? I'm stuck
Nevermind
Nope I'm stuck
How to solve such a thing ?
Sin 3/5 is 3 is opposite 5 hyp
Now find cos and tan
Maybe no
What's x?
,nvm sry
I don't understand too
X is the angle
Like you can draw a right angled triangle to solve for the magnitude of the sine cosine tangent, and then find the sign +/- using the ASTC
You have to put between those numbers or choose the right one?
Like for example 7,
sine is positive
cosine is negative
tangent is negative
And you draw a right angled triangle with side lengths 3,4,5
Wait, I can find adjescent with pythagorean theorem
Wait 3s
Found them!
sin x = 3/5, cos x = 4/5, tan x = 3/4
Only one mistake
Fuck
The signs
Yea the signs
This
Using ASTC
I got it
The next one...
Leaves me even more clueless
And this is just chapter 1.3
Tan 2 so is sin 2 cos 1?
Nvm
This is from the book Thomas Calculus
I'm trying this after following KhanAcadamy
But damn these questions are hard, I feel like I barely learnt anything on that website
Khan doesn't explain that good
x is in [0,π/2] hence, sin x and cos x are positive so, sin x=2/√5 and cos x=1/√5.
f o g = f(g(x))
g(x) = x-7
f(g(x)) = f(x-7)
$f(x-7) = \frac{1}{(x-7) - 7} = \frac{1}{x-14}$
Shen
f o g means f(g(x))
f(g(x)) means that instead of f(x) (where x is the input), the input is now g(x).
In this case, g(x) = x - 7, so the new input to f is x - 7.
@hard prairie
I misread the problem, my bad
wait so im right?
but isnt h(x) = 1/(x-7) = (f o g)?
yeah
f(x) = 1/x right
Because that's the function that makes f o g (or f(g(x))) = 1/(x-7)
since f(g(x)) = f(x-7)
yup
and if your plugging x-7 into f then f has to be 1/x
And you need that to be h(x) = 1/(x-7)
nice
so were on the same page right? 😂
ive been pulling my hair out over this single problem for the past 30 min 😂
thanks bro i appreciate the help fr
👍
How do i find the end behaviors of say y = (tan^-1 x)^2 without using a calculator?
<@&286206848099549185>
can you think of any zeros of this function?
try to split apart pieces so theyre easier to think about
im dumb there are no zeros 😄
it may be easiest to think of the function without looking at the +1
then translate it upwards after
It's an exercise in plotting translations and dilations of the graph of a function
Do you know $\log_ax=b$ can become $x=a^b$?
Biscuit
I think that kind of muddles the fact that logarithm is a function with a well defined definition
$\log_a x$ is the unique real number $b$ such that $a^b=x$
Icy001
Much better way to state the definition
We shouldn't be scared to say "such that" in front of high school students
😤
you could do some multiplication-by-the-conjugate tricks, couldn't you?
@acoustic dagger do you even need help with this still?
Nvm I had just woke up
I solved it right away
consider combining your terms into a single fraction
@safe hamlet
its not H
consider combining your terms into a single fraction
what do you have after doing my recommendation
show your work
<@&286206848099549185>
Which triangle? The big one or the little one?
Probably DEF
The domain of a composition of function is the intersection of the functions f and g
The little one
I major in math...
google it if you don;t know.
Also dont know why you felt the need to ask here as opposed to a discussion channel
I'm new here
thank u!
Doesn’t seem you got an answer yet. The domain is all x values that give a real number output. When do you not get a real number output for f(x)?
is the 5 my vertical stretch?
the 5 is inside the function
so what does the 5 do
sorry my brian is frozen.. is it horizontal stretch?
hellooo
follow professional mathematicians
quick question
what de frick happened to the 3/4
It’s reciprocal, 4/3, was mutiplied to both sides
3/4 is irrelevant to the zero product
oooooh
whether 3/4(x+4)^2 is 0 solely depends on (x+4)^2
multiplied both sides by the reciprocal
Whats the zero product
Or instead of 4/3 multiplied to both sides, 3/4 were divided from both sides
the product of stuff that results in zero
Ah, ty
((e^ipi) + 1)^2
@uncut mulch is this also considered a 0 product
$-\log(5 \cdot 6) = -(\log(5) + \log(6)) \neq -\log(5) + \log(6)$
Ann
and $-\log(5) - \log(6) \neq -(\log(5) - \log(6))$
Ann
this is just basic algebra
if you thought it had anything to do with logarithms you were mistaken
So, you want to find values of x where the function is undefined first, so either where the denominator is 0 or where you are taking the square root of a negative number. Here there are no square roots and so you want to solve the equation 9x - 5 = 0 where 9x-5 is the denominator. This will then be the x input where the function is undefined and it's domain will be all other inputs other than this.
how do I take the domain and range of a composite function given the graphs of the functions
15000=1750e^(.3t)
thank u..
which is more difficult?
pre-cal or cal 1
probably calc 1... i thoguht so at least
ayooo
guys i need help with identities
can someone explain how the book got this answer
they rewrote sin3x as sin of 2x+x and apply the sin(a+b) formula
then the one for 2sinxcosx which is indeed sin2x
u just gotta know your identites here
not that hard tbh
Alright I'm having a bit of trouble with this problem, I'm supposed to look at the graph and answer what the X-ints are and then confirm my answer with the equation
But the thing is
When I plug in zero, it works for 3 just fine but for some reason on the other one it just doesn't work
You'd get X=5
So I don't know if I'm just doing something real dumb or what
ab = c
generally doesn't mean that a=c or b=c
(if c isn't 0)
it would also be much simpler to plug in your x values into the given quadratic equation to see if you get y=0
that would be much easier. and i also found the answer to what i was trying to do, apparently i was supposed to put the entire equation on the other side
so it
it'd be -x^2+2x+3
for some reason i guess
yes
thxx
So guys this problem is asking me to solve for all real values of x
\sqrt{2x+1}+1=x
I think I'm supposed to add 1 on both sides
Then square both sides to get rid of the radical
I have 3w + 2l = 1000
so far
And then I wrote out A(w)=w * (1000-3w/2)
does this seem like a decent first step towards finding the perimeter so I can solve for the area?
Find such real numbers a and b so that the expression 10a^2 + b^2 + 6ab + 4a + 1 acquires the smallest value.
Got to:
(3a+b)^2 + (a+1)^2 + 2a and don't know what to do now
√(2x+1)=x-1
square both sides and solve the quadratic equation.
btw that is not really calc
<@&286206848099549185>
@civic furnace pls post questions 1 at a time
@stuck lark why>
rules don't say anything about it
also
have you really fucking deleted all my questions
could you please restore them
i don't have them written anywhere else
pls don't try to find loopholes here. spamming questions as you did is obnoxious
nobody has found an issue with it before
including @willow bear
aren't moderators supposed to moderate according to the rules?
i was writing a reply to the previous points but forget it; you've escalated your rules lawyering to a point where i find you in bad faith. goodbye
b&?
b&
good, they were giving off bad vibes at nearly every turn anyway
Hey guys I have a question, I saw it on a paper and I cannot solve it, anyone up for help?
Hey guys is there any way to closely estimate sinx in terms of basic equations
what do you mean by "basic equations"?
taylor series maybe? not really algebra
expressed under addition, subtraction, multiplication, division, fractions, the brackets, exponents, roots
what's the series
$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$
Ann
this is the taylor series for sin(x)
for the exact value the series goes on forever
but if all you want is an approximation you can get away with only calculating finitely-many terms
is this what you were looking for?
its also better for angles close to 0 displacement
yup
so if youre in a small angle the so called "small angle approximation" $\sin (x) \approx x$ works more or less fine
jan Niku
engineer moment

note to use radians
certified engineer
theres a great 3b1b video on this exact topic
Pre calc makes me want to cry
How do we get rid of log(y)=x if the base isn't mentioned?
I wans the log on the other side
i think in most texts log is treated as base e
it's base 10 by standard notation
so will it be y=10^x?
i believe base 10 is usually written as lg instead of log
hmmmmm
this is confusing now
actually maybe it depends on where
yeah
from what i learnt log is base e but fair enough i know people who use log as base 10
I learnt log = base 10
and use ln for base e
I'm kinda dum when it comes to these notations
Can someone write me a written solution to this answer please. It would be greatly appreciated.
any ol solution should be okay why this one?
is there something youre confused about in the solutions to these in general?
can i teach myself precalc
im currently a sophomere taking alg II
i want to take ap calc next year so i can take multivar in my senior year
for sure! Everything you need is already on youtube
hi, idk if i'm chatting in the right text channel but i really need help with my hw. i'm not asking for answers, i really want to understand the lesson and i wanna know how to solve for the answer or something. my teacher cancelled our class for this week so yeah i'm really having a hard time understanding this.
you're substituting one function into the other
so i'm gonna substitute f (x) into g (x)?
(f o g)(x) = f[g(x)]
So you're gonna substitute the function g into the x for function f
ohh okay, are there any video tutorials on youtube for that?
Yes search up composite functions.
Try it yourself first though. f[g(x)] = f(x-2) =
okayy, thank you very much your help :))
hi can someone please explain why in the answers they don't first differentiate the equation to get the velocity?
You are given a function that outputs velocity. Why would you differentiate that?
yeah just realised
thanks
what's the difference between average velocity and acceleration
How do I find the zeros of x^3+2x^2+4?
it doesn't look like the zeros can be readily expressed
are you asked explicitly to find the zeros?
I need to sketch (x^3+2x^2+4)/(2x^2+1)
hm.
well i can tell you from graphing that x^3 + 2x^2 + 4 appears to have only one real root and it's a nasty decimal
yah ill just say I used a graphing utility to find it
How this happen explain me lol
?
Heyo anyone open to help me out?
consider the second derivative
this is wrong since cross multiplying it gives different values
thats the derivative ... d/dx[2sqrt(x)] = 1/sqrt(x)
i mean thats not how u write it nor in precalc ...
im curious where u got that from
What is the x-intercept? I got x=-1
yes
x=-1 might bring quite destruction
nah
It would be undefined
factor out some stuff
oh pfttt
continuously extend it bruv who cares 💀
New here
Is there a way to find a zero of this function without testing all the possible roots? It takes way too long to test them all.
You could find one root, then perform polynomial division and factor the quotient to find the other roots.
is it possible to locate the center or create a standard equation given only the "vertex" and asymptotes? If yes, what should I do?
Consider the following polynomial.
F(x)=x^3+x^2−22x−40
Use synthetic division to identify integer bounds of the real zeros. Find the least upper bound and the greatest lower bound guaranteed by the Upper and Lower Bounds of Zeros theorem.
Upper Bound:
Lower Bound:
What is the part you are having trouble with? The theorem explains what you are dividing by. Are you unsure how to do synthetic division?
i couldn't figure it out
with solving it the equation and found no solution
Did you identify the least upper and greatest lower bounds first?
no
i need that help
okay
what i should do
What does the theorem say to do
the same
What are a_0, a_1, ..., a_n?
can you send a pic of your work? it'd help a lot
That's the question... not your work
oh i don't have it
so you have made no attempt on your own to try and do the problem...?
um do u need help on that..?
@rare flame ^^
try to solve the problem first
i did it online but not in my notes
i believe it is Positive Real Zeros: 1 , Negative Real Zeros: 2 or 0
can u show your work
f(x) is not the product of a magical number f and x
it's not
@rare flame this is an easy problem, get a pencil and paper and hack it out, the computer is going to slow you down
okay
Look at the theorem, it tells you what to do, and it is all you need to do
I have a question that I asked 3 hours ago but still not answered pls help-22
It's a cambridge question btw
@viscid thistle when you bump, quote your original post
You dont have anything else in this channel from today
It's in help channel no. 22
If sum of first p terms is q and sum of first q terms is p then what is the sum of first (p+q) terms?
my efforts...
sum of first p terms = (p/2)(2a+(p−1)d)=q
and sum of first q terms = (q/2)(2a+(q-1)d)=p
Idk how to proceed further
I guess i need to find "a"(first term) and "d" (common difference)
sum of first p terms of what?
is the sequence you're summing actually known to be arithmetic?
is sin^2(x) the same as saying sin(x)^2 ?
$\sin^2(x)=[\sin(x)]^2$
Mosh
huh. does that apply for all the functions?
ralphwiggum_imindanger.jpg
Sooo I still can't dm ?
no
Why? What can make you say yes ?
My choice to not let you dm. And nothing.
I'm also more than happy to dm mods if you continue to pester me
I know... how can I change your mind about it?
Nothing, now stop pestering me.
What.... ok
can anyone help me for the love of gof
i know that i could just use the table to find it, but i need help understanding why those values on the table correspond to the values in the question.
hi, i dont get how the system got these answers... the only one that i get is part C. bcuz 360 - 315 = 45..
this looks wrong?
i mean the drawing
the length of that vector isn't sqrt 2
@small tree
ignoring the drawing the rest of it is right
oh, a point
nevermind 😄
well you found a point on the line,
and then the sqrt 2 is just from like
the distance formula
$d = \sqrt{ x^2 + y^2} = \sqrt{1 + 1} = \sqrt 2$
jan Niku
Can someone help?
Problem:
Find the first three terms of the geometric sequence whose sum is 7/16 and product is 1/312
I don't seem to go anywhere with this problem, i've tried
x/y + x + xy = 7/16
x/y * x * xy = 1/312
But still I don't get the right answer
If x = ³√1521/78, how do i find y?
How do you get that x value?
x/y * x * xy = x^3 = 1/312
To get y plug the value of x into the other equation
That's where I am stuck at, I can't get the exact value for y, im left with a polynomial answer
Can you further simplify this equation
I can't read it.
Lemme try again
Do you need help with this or...?
Yes
yes
That was replying to Alejandro. You specified you needed help, I just can't read yours.
2 . cos²x . cos²y + 2 . sin²x . sin²y
What sort of help? Do you know where to begin?
There
Ah - ok. I can read that. 🙂 Let me see...
not really
I just wanted to know if we can further simplify this 2 . cos²x . cos²y + 2 . sin²x . sin²y
Hmm..how about trig identities?
Something to do with that yeah
Give me a moment to try something out
ok
Start by plugging x+h into your function
ok
Hm, I'm not seeing identities. The only identity I can think of cos^2x = 1 - sin^2x. You could factor out 2 from both terms and try something with that - I don't see any, though. Perhaps someone else will?
Once you've got that, subtract f(x) from it, then put it all over h
yeah i was thinking about factoring aswell
I would definitely factor out the 2 - that's the easiest one to see.
Does that make sense?
a transformation image of the graph of y=f(x) is represented by the equation y+2=-2f(1/2(x+3)). The point (-9,2) lies on the image graph. What are the coordinates of the corresponding point on the graph y=f(x)?
i got (-21,-6) but now that i look at the question again i think i got the wording confused
did it maybe ask me to do something else?
im confused
i don’t get it
Give me a moment to draw something that may help - to get you started.
Now you just have to simplify
appreciate it
a
how do i do the same thing, but with f(x) = 3 - 2x + x^2
when i did it, the answer became zero, as only h was left
^via nth root theorem. I guessed the root was -3 and guessed right. Then I guessed it had multiplicity-2 (repeated root) and was correct again. Last two roots had to be imaginary to get x^2 + 1
needed a binomial where there is no middle term and the constant comes out positive. That's imaginary conjugates.
or more generally, complex conjugates
no, check it:
(a+bi)(a-bi) = a^2 -bi + bi -b^2 * i^2 = a^2 + b^2 (always)
and in fact, (x-i)(x+i) = x^2 - i^2 = x^2 -(-1) = x^2 + 1
Subtract 2pi from IV, what do you get?
11pi/6 - 2pi= -pi/6? I don't get it please explain
Oh, I thought the first one, I, was -pi/6. Looks like 2 of the answers work, -pi/6 and 7pi/6. But that was wrong for some reason
Hello everyone I would like to get ur suggestions regarding to the materials and Topics I've to go through in Pre-Calculus .
can you help me with factoring?
The domain of $sin^{-1}(x) is [-\pi, \pi]$
not discordmod
that isn't right
@viscid thistle do you still need help with this?
yes
use trig rules to solve
are you there?
i can write it out but would rather write it out if you would eventually see it
just ping me
If y’all can just answer a relatively simple answer for me I’d appreciate it. (Ill put it down in this mssg as well) currently I’m trying to find the roots of f(x)=x^4-4x^3+2x^2+x+4, a 5 term function degree 4, I see what the roots are but would it be possible to solve for the answer without graphing it on a graphing device? Because the roots are decimals so im lost how id solve on paper.
Its's worse than just decimal: they aren't rational (via rational root test)
You can try and estimate using intermediate value theorem, but you really want to use Newton's Method, which is introduced in Calculus.
Okay I see let me dig into that, thx for the suggestion.
One other idea though
If you get lucky, a simple guess at the complex solutions will give you a quadratic
So maybe see if i and -i are roots
Using synthetic division
(I mean, wolfram will tell you)
(Nope.)
I just used the newtons law and i was able to show my work for this, Had to learn what derivatives were but I think I got the hang of newtons law. I am not understanding how I’d use i and -i with synthetic division in your other idea honestly my teacher last yr barely skimmed over i last yr.
Newtons law?!?! Lmfao newtons method
I used the rational root theorem you mentioned as well on my next problem to guess it completely without adevice so i appreciate your help
Oh good!
you can do synthetic division with complex numbers and I was just saying "hey, we know the real solutions are ugly, but maybe the complex solutions aren't? guess +-i". But the complex solutions aren't those anyway, so nevermind.
how do i convert this into logarithmic form?
wdym by that? the problem is that the fraction is confusing me
You may have to show your work to make it clear what you want and have a chance for other people to help you 😄
guys i need someone to help me rnw plz
@viscid thistle apologies for not being able to respond earlier after you replied to me when i was already asleep.
i take it you still need help solving the equation $\frac{4}{\sin(x)} + \sin(x) + 5 = 0$, and you have made zero progress so far?
Ann
Hi so with the ambiguous case with law of sines, how do you determine the number of solutions?
What I'm getting, is that if the sum is less than 180, then there are two solutions, but if it is more than 180 it is only one solution
anyone here knows where I can do math exercises with step by step solutions? preferably math exercises of functions, domain, asymptotes, interceptions, derivatives etc
How would you solve:
ax + atan(x) = (a*pi)/2
a = Any real number
Solve for X, no estimates
In range 0 < x < pi/2
All me and the other students were able to find is that
There are 2 solutions
x1 + x2 = pi
x1 = pi - x2
x2 = pi - x1
tan(x1)/tan(x2) = -1
and
tan(x2)/tan(x1) = -1
(and obv tan(x1) = -tan(x2))
Aside from these observations I don't think we found anything else
x1 is about: 0.710462737775517
x2 is about: 2.43112991581428
i need help with 57 and 59!
Hello! Quick one. How would I find the average rate of change using a slope function? Thanks lads.
Hello!! My name is Koyaaa and I need help with my homework I'm stuck and it was due 15 minutes ago 🥺 please help
Can anyone help please
Ahh 😧 what happened
Ok I think I see what your saying @viscid thistle
sorry I was on my phone for that.
No worries!!
No-one click on these links! <@&268886789983436800> Can you please remove?
<@&268886789983436800>
a = 3 and b = 9 and c = -9 and d = 9
So I got this problem involving inequalities
if the solution to this is -2<=x<=2
What is k?
I tried squaring both sides and going from there but thought that there was no way that that's the intended way to do it
So I checked the solution
And it was just like 2 lines long
The above condition is satisfied when the line y=kx-2 passes the point (2, 2)
2=2k-2
=> k=2
I drew it on desmos
To get a feel for what it looks like
It makes a lot of sense
But what I want to know is how do I know that it should be like this
What's -x/9=x/5?
Gave me 90 but idk how
<@&286206848099549185>
Been getting 14x

can you show the procedure that yields this answer of 90?
A little blurry since I'm on mobild bht
mobile but
You can see the question and A the answer choice i said it was..
There's no answer choice that is 0 lol
can you show the original question? to make sure we're not missing something
,rotate
Thank you so much and sorry if I wasn't clear on my question or came off a little rude
aight, mystery solved. no problem lol

Hello! I need some help with my precal homework. My teacher sent me an example but I’m too embarrassed to ask her any questions again.
So what I'm thinking is this
Log5(128)xLog128(2)=Log5(2)
5^3=125~128
2^7=128
So 3/7=Log5(2)
lol
Log5(75/8)=Log5(75)-Log5(8)=Log5(25)+Log5(3)-3Log5(2)
Log5(3)=Log27(3)Log5(27)~2/3 (Since 27 is kinda close to 25)
So Log5(75/8)=2+2/3-3x3/7
Oh wait they already gave you Log5(8) and Log5(3)
Those are just circles, I think
(x-1)^2+(y+2)^2=26
(x-6)^2+(y-3)^2=16
Think of it like this
P(x,y)
P1(1,-2)
The vector PP1 = (x-1,y+2) has a length sqrt((x-1)^2+(y+2)^2)
You want that length to be equal to sqrt(26)
In mathematics, physics and engineering, a Euclidean vector or simply a vector (sometimes called a geometric vector or spatial vector) is a geometric object that has magnitude (or length) and direction. Vectors can be added to other vectors according to vector algebra. A Euclidean vector is frequently represented by a ray (a directed line segmen...
oh but how did you get 1 and -2 values
That's point P1
I think if you can multiply u by a constant c so that it's equal to v then they are collinear
meaning
cu=v
(-8c,5c)=(x,-2)
-8c=x
5c=-2
<=> x/-8=c=-2/5
I hate arithmetics
ok thanks
because in the domain [0,90] the cos function is an increasing function, so cos(30) < cos(40)
does anyone know how to find values a, b, and c given three points (0,4),(1,14),(2,20) and the exponential equation S(t)=ae^bt +c?
decreasing*
S(0) = ae^b(0) + c = 4
^ like this
ah yes i’ve gone from there but still no success- thank you though, that’s the right start
a + c = 4 ..... i
ae^b + c = 14 .....ii
ae^(2b) + c = 20 ....iii
subtract i from ii & ii from iii
then divide the result of the two
thank you so much i think this is exactly what i was stuck on :)
How would I approach this question? I tried using trig identities and the squeeze theorem but still can't quite get it:
Try using l'hospital
I am not allowed to use it but thanks for the suggestion
Ann
anyway, may i suggest multiplying and dividing by x?
so you have $\lim_{x \to 0} \frac{\sin(29x)}{x} \cdot \frac{x}{\tan(2x)}$
Ann
and what would be the next step? 
well, are you able to find the limit of sin(29x)/x as x -> 0?
well yes, it would still be in Indeterminate form
can someone explain to me how to solve this?
RaghavJeyan
@quiet marsh
would this simplify to 6=ae^b(2)-ae^b, if so I likely did something wrong in the previous parts.
yes, $ae^{2b} - ae^b \neq b$.
Ann
...wait, what
what even happened there
did you try to add on the left side but subtract on the right?
looks like this is nonsense in more ways than one.
@oak drum do you happen to have the original problem statement that this is the solution for?
i’m not even sure what happened i’ve got it in a huge mess 3: it’s a multi step problem where i am to use three points to find an exponential model, i can get a screenshot of it
here's what I was given and I also have substituted in the three points I was also given into S(t)=ae^bt+c, ill show what ive done there as well, i may have simplified them incorrectly
problem resolved, i met with my professor privately today 
guys whats the F(x) here??
You mean f(x)?
no the integral
Use the power rule for integral
wait do i just have to add +1 to every n
The sum notation is the general form for some N-degree polynomial
alright thank you
rewrite cscx*tanx in terms of sin and cos
yeah lol but how
nvm i got it
does someone here know about perpendircular lines
and general ecuation
i really can not do resolve this one
it says finds the k value so that these general equations are perpendicular
do you know the relation between the slopes of perpendicular lines?
hello, can you please explain, these 2 to me.. like why is option "d" the answer to 1- tan(theta)/1+tan(theta).. and also the orange one thank you!!
I’m having trouble understanding why -1<=cos x<=1 becomes -|x|<=xcos x<= |x|
Actually I understand it now nvm
hi
anyone active here rn?
I m interested to know why wolfram thought of resolving that limit instead of the given one, as in, how did it come up with the 1/4 term and +3 term?
is this precalc?
i suppose?
limits of series are done before continuous limits so i thought it should be
yeah this is just a geometric series, so it's fine in precalc
I also wouldn't personally follow what Wolfram gave, the sum $\sum_{i=0}^n(\frac{-1}{3})^n$ has a known explicit formula
Mosh
that is?
Yes
👍
Oh
But if you are allowing duplicate roots and not considering multiplicity, then your previous answer is wrong
Oh
Hm
So if it’s a duplicate root and not considered multiplicity, a 5th degree can also be said to have 4 roots??
And a 4th degree can only have 3 real roots when there’s a duplicate root and not considered multiplicity
Ohh
No, a 4th degree polynomial can only have 4,3,2,1 or 0 roots if you don't consider multiplicity, and 4,2,0 roots if you do (by which I mean a repeated root is counted twice)
👍
its infinite gp
1 -1/3 + 1/9 - 1/27 + ..... here r is -1/3
and a is 1
the forumla for gp is a(1-r^n)/(1-r) when r < 1, r^n where n -> infinity is basically zero so the formula for some of infinite gp will be (a - ar^n)/(1-r) => a/(1-r) [as ar^n -> 0]
How do I use the squeeze theorem to evaluate the limit as x->0 of x^2sin(1/x)
-1 <= sin(1/x) <= 1, and x^2 >= 0
Combine those to get bounds for your function whose limits are the same
Hey guys,
Anyone from Kuwait?
Hello. I am having trouble understanding how to find the difference quotient of a function; apparently, f(x) = (see attached image) and not 5/4h^2, when the difference quotient of f is f(x + h) - f(x) / h. Does anyone know why?
What is the main problem?
We can't tell how to do it if you give an answer but no question
Are you saying that the difference quotient is $$-\frac{20}{(4x+4h-3)(4x-3)}$$
Omega Warrior
and not $$\frac{5}{4h^2}$$
Omega Warrior
@round hemlock Yes, exactly.
Hold on. I will post the question
Alright
Omega Warrior
you should get $$\frac{\frac{5}{4(x+h)-3}-{\frac{5}{4x-3}}{\frac{5}{4h-3}}$$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text>
\par
<*> 538528087764107274.tex
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```
ok nvm it didnt work
Actually, I think I saw my mistake
oh
I only subtracted the denominator (4x - 3) and not all of f(x), which is 5/4x-3
yeah
Tehe
you need to take the entire function and substitute each part in
so x for x, x+h for x, etc.
sure
I'm in Calculus, so I get anything in Precal
Ok, so, number 87 in the picture I posted has an answer of 2x + h - 1. However, I got 2x + h. Do you know where the -1 came from?
I'm doing the same but for precalc (I and II) :l
Ohh no. I am only doing one year, too; I am taking precalc I now and precalc II in the spring
on 87 you should get $$\frac{((x+h)^2-(x+h)+4)-(x^2-2x+4)}{h^2-2h+4}$$
Omega Warrior
@obtuse dagger
when you substitute all of the values
and then simplifying you get:
What?
$$\frac{((x^2+2xh+h^2)-(x+h)+4)-(x^2-2x+4)}{h^2-2h+4}$$
Omega Warrior
doing further simplfying gives $$\frac{((2xh+h^2)-(-x+h))}{h^2-2h+4}$$
Omega Warrior
You solved this differently than I did
Hmm
I got the answer right, but I was just missing a -1 :\
oh
wait
i realized i made a mistake
thats why its not solving right
$$\frac{((x+h)^2-2(x+h)+4)-(x^2-2x+4)}{h^2-2h+4}$$
Omega Warrior
this is what it actually was
$$\frac{((x^2+2xh+h^2)-2(x+h)+4)-(x^2-2x+4)}{h^2-2h+4}$$
Omega Warrior
$$\frac{((2xh+h^2)-h)}{h^2-2h+4}$$
Omega Warrior
wait
I DID IT WRONG AGAIN
AAAAAAAAAAAAAAAAAAA NOTATION SUCKS
anyway
$$\frac{((x^2+2xh+h^2)-(x+h)+4)-(x^2-x+4)}{h^2-h+4}$$
Omega Warrior
I think you are miscalculaing the denominator. It's just h, remember ?
ohhhhh
$$\frac{((x^2+2xh+h^2)-(x+h)+4)-(x^2-x+4)}{h}$$
Omega Warrior
$$\frac{((2xh+h^2)-h)}{h}$$
Omega Warrior
and thus, 2x+h-1
@obtuse dagger there solved it
Sorry about that, I wasn't taking a good look at the problem
Ha! You did 😏
yep
I should have written the problem down first, I wasn't looking at the question
No problem
I am going to bed now, though
I'm most likely in a different timezone than you lol, so that's why
No problem and goodnight
Trad: the right soports of the sides of a triangle are:...
Find the a,b,c
I tried to the distance but it didn't work
What can i do to find the a,b and c
find the points where each of the graphs intersect each other
where AB and AC intersect is Point A, where AB and BC intersect is Point B and where AC and BC intersect is Point C
no problem
I didn't think about do that
@short pilot anything else?
Nope, thank u fr
you're welcome 😄
"next state"?
That would be postcalculus 
sory, is next stage, next level
Calculus
that would be the next stage of calculus
hello
can someone help me its not precalucls
but i couldnt have permission to send in chats
<@&286206848099549185>
do 90*10
ty
do not give out answers.
@willow bear why is that a rule of something
hmmm why tho it doesn’t seem ethical wrong
we are not chegg
well yeah of course this a improper forum board
but what is ethically wrong with giving someone an answer that they want
There wasn’t any other way you could have said it
Unless like you say that 0.1x=90
Idk
you're always welcome to ping Moderators to convince them to rewrite the rules
yes, say it like that at least
Ah alright, will take it into consideration for basic problems
