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what happen if x is x<0?
negative things to non-integer exponents very quickly get out of hand
or you could say they go sideways
Where
for example if x=-1 we get (-1)^cos2 which is not defined for real numbers
if you are asking anything else make it clearer
I'm not that advanced but can derivatives exist in the complex world?
well yeah
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Hello, I am having an issue with solving for a variable in the exponent of a fundamental theorem of calculus problem.
what s p(t)
it should be 225/(t+1)^r
what are you having an issue with exactly
225(t+1)^(-r+1)/(-r+1)
everything you ve done seems correct what s the problem
I need to find what r is on the interval 0<x<13
for question d right?
yes
what s the equation you ll use?
to find r
oh you ve written something
this isnt correct
you need to evaluate this at 13 and 0
ok, I will give that a try
So then I have (2925^-r+1)/-r+1, would I then use logarithms to evaluate the rest of it?
I am still having some issues with this
<@&286206848099549185>
Question?
I don't know how to finish this problem, I am trying to solve for r and I don't know how
from here we have 225(t+1)^(-r+1)/(-r+1) evaluated at 13 and 0 is 225*14^(-r+1)/(-r+1) - 225/(-r+1)
correct?
yeah, it should be (2925^-r+1)/-r+1
you just multiplied 225 with 14?
and then subtracted it by 225
no u did 225*(14^...-1)
oh I see
when you have a*b^x it is not (ab)^x
225(14^(-r+1)-1)/(-r+1) is not 225(13^(-r+1)/(-r+1)
which is also NOT (225*13)^(-r+1)/(-r+1)
So, when I get to this point how do I solve for r?
Thank you, I have been having a lot of trouble with this. I appreciate your patience.
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Closed by @surreal topaz
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i feel like there has to be a smart way to do this . but idk how
i tried using taylors expansion but it didnt really help
well you can simply not worry about the cos(1-x^2) factor
since it approaches 1
focus just on the $\frac{\sin( (x-1)(x+1) )}{(x-1)^2}$
yea the issue is sin(x^2-1)/(x-1)^2
Ann
i thought about using sinx/x but i cant ;-;
you can
you'll just end up with...
$\frac{\sin(x^2-1)}{x^2-1} \cdot \frac{x+1}{x-1}$
Ann
and (x+1)/(x-1) explodes so
why would i end up with that?
factor the left term of the denominator
here
you should see things cancelling out to give you the original denominator
i dont really understand why x+1/x-1 appeared though
\begin{align*}
\frac1{(x-1)^2} &= \frac1{(x-1)^2} \frac{x+1}{x+1} \
&= \frac1{(x-1)(x+1)} \frac{x+1}{x-1} \
&= \frac1{x^2 - 1} \frac{x+1}{x-1} \
\end{align*}
$\frac{1}{(x-1)(x-1)} = \frac{x+1}{(x+1)(x-1)(x-1)} = \frac{1}{x^2-1} \cot \frac{x+1}{x-1}$
Ann
also parentheses
\cot 
omnipotententity
why does x-1 become negative in the second step?
There are two copies of (x-1) and one of (x+1) in the denominator, I just used commutativity to swap their order
Why isn’t it like this?
It looks like you're mixing up (x-1)^2 and x^2 - 1
(x-1)^2 = (x-1)(x-1)
x^2 - 1 = (x-1)(x+1)
omg thats true
sorry
nooo i really have to go rn but can i reopen the channel later asking about the same limit? bc im still a bit confused
You can open a different one, and link back to this
that works too, thank you :)
@dawn crest Has your question been resolved?
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!status
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
3
what i tried is i made a sample space
A B C
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1
if number of jobs of B is 0
then those cases would be
2 0 0
0 0 2
1 0 1
and from these probability of getting 2 jobs for A is the first case
so
1/3
but this is wrong
P(X=2 ∩ Y=0)/P(Y = 0)
yup
1 0 1 can occur in two ways
so i think it's better to consider the sample space as {AA, AB, AC, BA, BB, BC, CA, CB, CC}
that way we can be certain the outcomes are equiprobable too
This is what he missed I guess.
and P(X=2 | Y=0) now means "whats the probability both contracts were given to A, given that B didn't get any?"
A B C
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1
1 0 1
???
1/4??
yes
A getting the first contract and C getting the second contract is not the same as A getting the second contract and C getting first contract
but are we not taking about the number of contracts and not who gets what contract?
1/4 seems correct
thx for the help
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@mossy wedge Has your question been resolved?
find the probability that x is 2 given x less than or equal to 2
@mossy wedge Has your question been resolved?
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TY
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Guys, how do I find the equation of a line for two points with the same x-co-ordinate?
y = y-coordinate
x-co-ord, my bad.
x = x-coordinate
I don’t quite understand, I think. I’m trying to find an equation for a line segment AB where A(2,-2) and B(2,6).
The equation is x = 2. y can be any real number and this equation will be true, including -2 and 6
Right, wow, that solved the whole sum for me. Thanks a lot, man.
@lapis fjord Has your question been resolved?
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the derivative remains the same expression
this is a true statement
the derivative of x^2 is always 2x
the slope of the tangent is the derivative
the derivative changes
at each x
the derivative is a function of x
No, the derivative of x^2 is 2x, and 2x evaluated at 2 is 4
That is false if x is a variable and true if x means “times”
The first or the second?
Yes
4x is an expression, not an equation, and y = 4x does not go through (2,4)
It’s ok
y = 4x does not go through (2,4), but y = 4x - 4 does, and that’s the tangent to y = x^2 at that point
Because then it isn’t a tangent
The derivative tells you the slope, and the point together with it determines the line
What do you mean
what is epsilon
there are no infinitesimals in the reals, maybe you were thinking of h?
you know the slope, and you know the point, so you can get the equation
try shifting right by 2 and see what you get
this is right
it should, did you try?
oh wait, you only have to shift right by 1
look again
the point was that shifting this line right by 1 and shifting down by 4 are exactly the same
Yes!
Should be the same
find the derivative at a point, which is the slope of the tangent, then use the fact that it goes through the point
You can think of it that way
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I still do not understand what was this algebraic manipulation
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what is the answer?
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I just wrote it down and realised that by the definition of function composition this was the same as
T(S(T(v)))
so you can write the composing symbol (the little ball) in both places
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checks another item off mathcord bingo
what
oh I found in on the internet, lmao
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could someone explain these markings? I'm new to constructions in geometry and would help a lot if someone could introduce me to them:
hi
I know the first image shows corresponding and congruent angles, but what is the smaller arc on like HF?
ping @smoky tangle when responding
okay
need explanation on II?
yeah
which means splitting it into two equal smaller angles
yeas
how they did it is
one second
they measured a certain length up BE
and put their compass on that point
and used the compass to draw a curve
which is this one
then they got the same length they went up BE
and went up BF
and put the compass on that point
and drew the other cure
oh okay
does it mark anything?
and then they draw a line from the angle through it
where they cross
which bisects it
equally
thats the image
the question says "which of the following are a construction of an angle bisector?
you can choose more than 1
the answers were 2 and 3
yeah
basc
the guidlines is the same thing as II
theyve gone up the length of the traingle
draw the cruves at the bottom
and bisected it
lenfth of a side?
gone up that for slant 1, marked a cuve, then sllant 2, and marck another
and marked where they intersecty
and drawn a line through that to the angle
which also bisected the base on the way in
it has to be an equilateral triangle
am i right?
@next patrol
yes you are right
sorry for the late reply
@smoky tangle
it would be helpful to try it urself on a piece of paper
@smoky tangle Has your question been resolved?
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is it correct to apply l'hopital like this?
individually to each fraction?
since i rewrote them so that they both equal infin/infin
Looks right.
Youre not doing any calculus up to that point.
Just basic fraction manipulation
yep
yeah
so its fine to apply it like that?
yep.
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What are the necessary steps in doing induction? E. g. If i wanted to prove that the sum of n smallest odd positive integers = n^2 , what would I need to do?
do you know the "induction as stairway to heaven" metaphor
yes, i am only confused with steps involving k
you have your base case and you have your inductive step
So induction has 2 steps, the rest is simply logic. Think of it as dominoes.
1st step : initialisation. You tip the first domino, as the sum of 1 smallest odd positive integers is = 1².
2nd step : "Induction". You make sure that for each domino tipped, the next one will tip as well. So, knowing that the sum of n smallest positive integers is n², we need to prove the sum of n+1 smallest positive integers is (n+1)²
those are the only two "steps" that the method of induction itself is concerned with
how you prove either one is a different matter entirely
but it does help to have your statement written down formally
in your case $\sum_{i=1}^n (2n-1) = n^2$
Ann
Induction only has 2 steps
Once you know that the first domino is tipped, and that for each domino tipped, the next one is tipped, for sure, all the dominoes will be tipped in the end
So in your case :
We easily can see that for $n=1$, $\sum_{k=1}^12k-1 = 1$
rafilou2003
*(2i - 1)
And 1² = 1
Now for the induction step
Let n be any positive integer
Let's suppose we know that $\sum_{k=1}^n(2k-1) = n²$
rafilou2003
We need to prove the following : $\sum_{k=1}^{n+1}(2k-1) = (n+1)²$
rafilou2003
And now we do some calculations :
$\sum_{k=1}^{n+1}(2k-1) = \sum_{k=1}^n(2k-1) + 2(n+1)-1$
rafilou2003
rafilou2003
So $\sum_{k=1}^{n+1}(2k-1) = n² + 2n + 1 = (n+1)²$
rafilou2003
Which is the proof that we needed for "if a domino is tipped, the next domino is tipped as well"
Which concludes the proof of this statement
because n+1 can become the new n ?
Yes in a certain way
Closed by @tall talon
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just started learning non-homogeneous odes, not really sure how to get the complementary and particular solution thanks
Well did you do the substitution
is t a constant ?
No
nope.
whaat
ive only learnt constants (for example a times d^2y/dx^2 + b times dy/dx + c times y) are c1 e^rx + c2 e^rx
sorry for bad notation
not sure how to do this type of question since its x^2 times the second derivative, not a constant
I need help 😭
!help
Please read #❓how-to-get-help
this channel is occupied
I'm seriously lost
bruh get your own channel
@flat roost Has your question been resolved?
is this an euler cauchy equation
@flat roost start with the substitution, after you perform the substitution (remember the chain rule for the differentials) share what you have and we can go from there. You'll generally use an ansatz to get the particular solution, but there are other more systemic methods as well, such as variation of parameters. To get the general solution you just add the homogeneous solution to the particular solution.
whats an ansatz
A fancy word for guess
i dont get how chain rule can be used here though?
You need to convert dy/dx into dy/dt
That would be how I approach it.
i see
so now we are only trying to have y and t?
Exactly
ohh i see
thanks, ill try that
wait but
dy/dt x dt/dx
which one do i do?
i mean i dont even know what dy/dt is
but i do know dt/dx which would return something with x
To be fair you don't know what dy/dx is either
thats why im chain ruling
Yup
which would give dy/dt times dt/dx
Yup
So dt/dx = 1/x = e^(-t)
I will need to double check it, but that looks about correct other than you dropped the t somehow on the rhs of the equation
@flat roost
This is for the complimentary part(?)
Ok
like the c1e^rx + c2e^rx
Thanks
Unfortunately, it seems like I'll need to be afk for a while.
thats alright
wolfram says otherwise 😭
can someone please help me double check, tysm
@flat roost Has your question been resolved?
should be =t instead of =0
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does this sum mean something like, assuming n is "4":
1/4 + 1/4 + 1/4 + 1/4
or
1/1 + 1/2 + 1/3 + 1/4
bruv
this is sign of sigma
it means multiply
so it means
1/1
+1/2
+1/3
+1/4
hope it helped :)
but usually the variable used is the index
nah its not
wtf
i wait wait
this is the whole image
it's approaching infinity
idk if it's still considered constant
yes
positive integers
from k to n
I got it, thanks
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Yes
what's the answer in the book
I'd say -2 is right
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hi i'm trying to apply the rule to rationalize the expression and getting tripped up. how can i arrange the denominator properly to start
i'm getting really close but mixing up on algebra somewhere
or you can just use (a+b)(a-b)=a²-b²
$\frac{x+3}{4-\sqrt{2x+22}}\cdot\frac{4+\sqrt{2x+22}}{4+\sqrt{2x+22}}$
biscuityxd
i keep getting a sign wrong somewhere
need to factor in such a way that i wind up w (x+3) in the denom
$\frac{(x+3)(4+\sqrt{2x+22})}{4²-(\sqrt{2x+22})²}$
biscuityxd
yeah so here is where i'm getting mixed up. let's say the squared and the square root cancel each other out? then do i have to multiply a negative one through that expression? the implied negative one out front of it???
see i don't know how i would know that or how to memorize that
seems counterintuitive or arbitrary
just memorize this
my issue is w the algebra
well, you can just rough work to just write the denominator out step by step
like
4²-(2x+22)
16-(2x+22)
16-2x-22
if it were instead addition, one could simply remove the parenthesis at the second step? no changing of signs?
if it's addtion
like 4+(3+2)
we don't have to change sign inside brackets
oh i see it now. minus the quantity 2x and 22
just remember
-(2x+22)
is
negative of ( postive 2x + positive 22)
the 22 being positive would make no sense
you'd be ignoring the fact that you're subtracting it
alright thanks. you'd shudder if you knew what grade and program i'm in 😉
i might be stuck on the next if you're still around
open a new channel for new question for better performance 😆
remember to close this one first ;P
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I need help finding angle EDC the topic is about cyclic quadriliaterals
Deduce whatever angles you can
Since EF is parallel to CH, angle HCF = angle CFE = 46 °
oh
angle FBD = 90° - angle FBG = 90° - 58° = 32°
i see it now, thanks so much
You're welcome 🙂
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how do you do the power of -2
,tex .exp rules
Hayley
so
$\frac{1}{x^2}$
putridplanet
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i'm sorry?
@scenic wing why did you send me a message just containing the word "booba"?


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Do similar triangles have to have all the same angle measures?
Yes
Yeah, else they wouldn't be similar
Thats basically the definition of similar triangles
And if the have the same sides length they are called congruent
but at the same time if all three sides have the same ratio does it also determine they also have the same angle measures?
Yes
If all angles are the same the triangles are similar. If the ratio of the sides are the same they are also similar. Or if the ratio of two sides are the same and the corner in between those sides is the same then they are also similar
okay thanks my friend
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how do I differenciate 1^2-x^2 into (1-x)(1+x)
is there a method or is it just memorising
a^2 - b^2 = (a-b)(a+b)
it's called
difference of squares
so it's basically memorise
it isn't that hard, let's be honest
ah ye
you can search for short multiplication formulas
there should be a list or something like that
what do you guys use to do maths when on paper
with many examples
The factorization is "obvious". Consider f(x) = 1-x^2. This is a quadratic polynomial with 2 roots. So either see what they are directly, x = +-1 or use the quadratic formula. In either case, take your pick of a factorization theorem to get (1-x)(1+x).
yeah I'll try to remember them
Difference of squares is the shortcut here. Anyone that would know it without it would probably have memorized difference of squares long ago
necessarily do examples, not just learn them by heart
Also if you need need a visualization, here's why the diff of squares works:
$\\ (a + b)(a - b)$
$\= a^2 -ab + ab - b^2$
$\= a^2 - b^2\\$
So now when you run across two squares (including an implicit $1^2$) being subtracted, then you can use the difference of squares trick.
mellowdramallama
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I want some opinions
Yes
like can I get away with writing y' entirely in the exam
and not be risk of mark deduction
Yes
well usually
"usually"
more like
Right
No provided y is a function of x
to just write y' instead of the full thing
y’ is fine notation
if I have time I'll write the full thing ig
but if I'm running out of time I'll do y'
tq
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✅
yes
it depends on your professor. Most don't seem to care but I would just clear it with them. Symbollically they both mean the same thing (as long as y' is with respect to x)
true
but I'm in a situation where I might not know who is marking the paper
but yeah I'll ask him if I can
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or also write "Let y be a function of x"
that may help avoid confusion but kinda ruins the whole time saving point 
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I am done with this, I just have it on paper. One second while I type my question
Okay so my solution is $$y=\frac{1}{2}-\frac{e^{-2x}}{2}$$ for $0\leq x \leq 3$ and $$y=ce^{-2x}$$ for $x>3$
austinu
I am wondering about part (c)
is the reason that this is not a "solution" because it isn't necessarily continous
depending on the choice of (c) this function may not be continous over the interval of our solution
so we can't really say it is a "solution" by the strict definition
that requires both y(x) and y'(x).... to be continous over the interval of solution
does that sound about right?
sounds about right
if that’s how “solution” was defined then i agree
but it depends on how they defined a sokution
well, a solution has to satisfy the equation over the interval of said solution, and also the solution and its nth derivatives (for an nth order equation) have to be continous over the interval of said solution
so this piece-wise function of mine satisfies the equation, but the continous part depends on my choice of c
I think it should be impossible that both the function and the derivative are cont. feels like it
well I can easily verify tht
they require different choice of c
there should only be 1 c that makes it continous
then I can plug that in and check if the derivative is continous
it probably wont be
got it
ty guys
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i need help
look im not smart like everyone else i struggle with math

Send what you need help with and we'll be able to help
whats confusing about it to you
i dont understadn how to solve it
look i dont learn by reading text i learn by someone being there
and telling me how to do it
@normal heath Has your question been resolved?
first thing we need to do is determine how many solutions these systems have
do you know how to do that?
!help
Please read #❓how-to-get-help
that image says AVAILABLE help channel
this isnt your channel
you're in an occupied one
@normal heath Has your question been resolved?
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I'm having trouble figuring out how to apply this formula
To this equation
For further context I am talking numerically so
I have the values for Psi(Z) along a horizontal line in the complex plane described by x + is where x goes from -inf to inf and s is a fixed constant
I know there are poles at Z = Pe^(i*theta)
Which mess up my attempt at creating a contour integral around r (r is real here)
@ornate ore Has your question been resolved?
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How can I tell which has rotational symmetry? They both don't look like equilaterals
full image: i removed c and d because i knew they had rotational symmetry unless the star doesn't?
the star does if it's regular
if you have to choose a single answer, B must be equilateral
so then it's gonna be A
oh
but this is a bad question with bad art
the star looks kinda stretched too
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why do they convert it to ln
so whenever you use the change of base the num and denom become natural logs
you can change to any valid base you want
ok so why did they make it ln here
part of the question was to get a rounded value
which requires you to use common log or ln in your calc, they've chosen ln
when i put this in my calculator i dont get the same answer
what are you getting
0.092
,w calculate ln(9/7)/(6*ln9)
can you show exactly what you're putting into your calc
$ln(9/7)/6*ln(9)$
putridplanet
order of operations
paranthesis are your best friend
you want to divide by "6ln(9)"
entering that, you're only dividing by 6 and then multiplying by ln(9)
i used a bunch and it worked
cool
@scenic wing Has your question been resolved?
no, this is for the complimentary solution
i wanna know if the quadratic is correct tho
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i brute forced it and got the answer
but surely theres another way to do it
(a^2 - b^2) =(a+b)(a-b)
ik that but how do i apply it to the question
Do this for every term
ohh wait
1-(1/2)^2 =?
i see
but then i have to go multiply 3/4 by 8/9 and so on and so forth
is there any way to make that part easier
3/9=?
1/3
Do similar cancellations for all the terms
you can factor the numerators
Welp that's not what brute force means but ok
oh ok mb
keep stuff in factored form
leave (1-1/2)(1+1/2) as (1/2)(3/2)
etc
that is what i did
oh ok
then you didnt do the method provided by riemann
ohh no i multiplied them
but cancelled it out
nvm
im just dumb dont mind me
how do i close the channel
Type .close
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.close
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help
There are typos
one is open and one is closed
what does that mean
just different notation
Oh i havent seen that one
the fourth option does not include 2
is the fourth one saying everything but -2 and 2
and 2nd one is saying only -2 and 2
?
😭
@upbeat hornet
i actually hate math so much
i believe its the same as (-2,2)
2nd: All numbers from -2 to 2 (-2 and 2 included)
4th: All numbers from -2 to 2 (-2 and 2 excluded)
and what about the 1st and 3rd options, what does it mean when [U] vs ] U [
[] includes endpoints
oh so its the same
]U[ would mean the end points are included
ok thanks
i dont like that notation
isn’t it the opposite?
i wouldnt think so?
cause like [0,2[U[4,6] 2 isnt included right but 4 is
cause the [] is faced away
omitting too much making it hard to read
() is superior notation for exlusive ranges
no

i thought it was infinite domain
im dumb asf
desmos checks that out too but theres no option for ]-inf,inf[
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How would I solve something like $\frac{\dd y}{\dd x}=e^{y}$

XxMrFancyu2xX
Kinda new to diff. eq so would really appreciate a kindergartner explanation 
ain't this separable tho
that's solvable by separation. Divide both sides by e^y and "multiply" both sides by dx
then take the integral of both sides
it becomes $\int e^{-y} dy = \int dx$
mellowdramallama
ohhh ok that makes sense, I only did the dx to both sides shenanigans 🫠
thanks Mellow 
anything for you
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Bob goes to run. From his starting point, he runs due east at 10 ft per sec for 250 ft. He then turns and runs north at 12 ft per sec for 400 ft. He then turns and runs west at 9 ft per sec for 90 ft.
Express the (straight-line) distance from Bob to his starting point as a function of t, the number of seconds since he started.
That's the question and solution from here - https://www.freemathhelp.com/forum/threads/word-problem-find-distance-as-a-function-of-time.72575/
Bob goes to run. From his starting point, he runs due east at 10 ft per sec for 250 ft. He then turns and runs north at 12 ft per sec for 400 ft. He then turns and runs west at 9 ft per sec for 90 ft.
Express the (straight-line) distance from Bob to his starting point as a function of t, the...
@dapper bison Has your question been resolved?
@dapper bison Has your question been resolved?
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isnt the last step wrong?
Yeah it should say -k
wait but how is it rearranged liked that
P_0/P = e^kh
P/P_0 = e^-kh
P = P_0 e^-kh
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State The Value Of Digit 3 in number 121134₅ in base ten
a)15
b)25
c)50
d)123
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can anyone show me how to find range of this function:
f(x)= x|x|
The range?
Yes
Idk
The domain is R, right?
It’s most likely just all ranges
Yes
Write the function piecewise, it should make it easier to see
All y values are in the range
I just told you
invert it and find the domain of the inverse
I dont know inverse of function
I don't really see where the confusion is
I got range = R-{-3}
whatt
,w range of 3x/(28 - x)
where did that come from
Correct
Does this method work on all types of function to find the range
invertible functions
any linear/linear should work
for quadratic/quadratic or any quadratic you'd need to use the discriminant
Does this work on under root func.
if you can write x in terms of y it works
well the modulus doesn't really have an inverse does it
there is no antimodulus
For modulus functions you need to just use the piecewise definition
Yeah |x| has no inverse although the function that you had earlier involving the modulus did
So every modulus function have domain =R and range =R??
But you will want to use the piecewise definition to find its inverse as well
does |x| have range R
Yes
uh
Can |x| be equal to -1?
Yes
what, no it cannot
give us a real number whose absolute value is -1
for what value of x
Sorry positive R
also let this be on record
Nonnegative would be more precise
indeed
Why??
Why
What's |0|
Because it includes 0 whereas positive R does not
0
Is 0 positive
And we want 0 to be in the image because |0| = 0
Nonnegative what ?? Real no.
Nonnegative R
No
I think?
it isn't
Saying the domain is positive R would not include 0
but the modulus function does have 0 in the range and domain
Which is why Lonely Bean insisted on non-negative R
So, the question was can i use the method u said of writing x in terms of y work on modulus function
??
neonperseus
(You could write x in terms of y but the relation that you get would not be a function)

Is there way to solve these question to find domain of mod. ( Without graph)
Piecewise definition I told you
Then can you give one example question?
To find range and domain
unless there is a restriction (a fraction, a root, log, or sth else), the domain of a mod is R
Find the range of $f(x) = x + |x - 1| + |x - 3|$
neonperseus
domain is R but I asked for the range
you have to what
write the piecewise definition
Write the function seperately for when x < 1, 1 < x < 3, and x > 3

you got this