#help-23
1 messages · Page 22 of 1
x^-n
And so what's 1/x³?
X^-3
And what's 2/x³?
2^x^3?
$\frac{2}{x^3} = 2^{x^3}$?
Gamer Dio
$\frac{2}{x^3} = 2x^3$?
Gamer Dio
2x^-3?
Yes
Ok
So what's -(2/x³)?
Yes
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Hello, finding x for this equation is not possible, right? Teacher said all the equations given are possible.
yeah
just multiply x by the paranthesis
and continue from there
it becomes this:
5x + 10 -3x = 2x + 14
wait wtf
thats not possible
Lmaooo
yeah lol.. just wanted to make sure 😑😑😑
yup.
Could be a typo, ask your teacher
Wanted to make sure I wasn't doing a big silly.
Yeah will do, thanks
Solved
You're not the OP
i figured
If you're done with the channel, you can close it using .close
Is it me or did the OP leave the server?
hes still here
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Is there a map between set $\mathbb{R}^2$ and set $\mathbb{Q}$?
In other word, given an open set $A \in \mathbb{R}^2$. $A$ is open so by definition this gives a neighborhood $N_r(x)$.
How can we prove that this $N_{r}(x)$ belong to $\bigcup_{a \in \mathbb{Q}} N_r(a)$?

Looks perfectly fine to me \s
soda
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Having trouble finding the raw score with just the mean and the standard deviation
@reef spindle Has your question been resolved?
@reef spindle Has your question been resolved?
<@&286206848099549185>
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a fountain shoots water 9 feet into the air and it lands 6 feet away, what is the equation of the parabola that models this path?
@rugged mason Has your question been resolved?
<@&286206848099549185>
Can you find any points belonging to the parabola?
@rugged mason Has your question been resolved?
no
the question was all that was given
i know the 9 is the value of k
and that's about it
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Hello, I'm in calc 2 and need help on a problem that I'm not sure even involves any calc concepts. The background says: The included angle of the two sides of a constant equal length s of an isosceles triangle is (theta)
Part A of the question asks: Show that the area of the triangle is given by A = (1/2)s^(2)sin(theta)
how am i supposed to show that exactly
@broken saffron Has your question been resolved?
<@&286206848099549185> hello, i was wondering if I could get help for what i believe is a geometry proofs thing
Triangle area = 1/2 x base x height is going to help you here.
Working backwords, you can try to retrofit $ \frac{1}{2} s^2 \sin \theta $ as $\frac{1}{2}$ base x height.
Pond
Redraw the triangle so you have s as the base in the shape most familiar to you.
If s is the base, what's the height?
i apologize, give me one sec to try and think about it with your diagram
oh i think i see why now
so the height is the red line, and using sin(theta) = opp/hyp, but since we want to get the height we can cancel the "/hyp" by just multiplying by s on both sides
since the hypotenuse and base both are s
Yes, the hypotenuse of the right-angled triangle used to define that sin ratio is s, and the line you are looking for is opp, correct.
Just be careful not to confuse the "base" of the right angled triangle used to define the sine ratio with the "base" of the triangle whose area you seek.
wait but then how would i actually go about writing out the proofs or whatnot
Start with that diagram.
Pieces of information you know:
(1) The area of this triangle is 1/2 times the red line segment's length times the blue line segment's length.
(2) The blue line's length is s.
So you just need to find the red line's length.
so then plugging in s we have A = (1/2)bs
However, you also know
(3) The red line segment is the side opposite theta in a right angle triangle with hypotenuse s. Therefore red line / s = so on and so on.
so then i could just do b = s and for height:
h/s = sin(theta), h = s*sin(theta)
then just write out A = (1/2)bh, and replace b and h with their respective values, and it would be as simple as that?
alright well thank you for the help, i really appreciate it
.close
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Tips for 39pls
try u = sqrt(1/x -2)
yeah got the same thing
u forgot the dx at end tho
and rmb once u intergrated it, convert all u back to x and add constant
i dont think there is, usually w sqrts like these letting u as the sqrt part makes it easy
i dont think trig sub is suitable here and u cant rly change the sqrt part by factorizing or expanding
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Can someone explain to my why the group ${e, x, y, z, t, u}, *$ such that $x^2 = y^2 = z^2 = t^2 = u^2 = e$, $x * y =z$ doesn't define a group? This shouldn't be possible because a group of order 6 can be proven to have an element of order 3, but I'm unable to reach a contradiction
Yuese
(Note that it also follows that $x * t = u$ because that's the only thing it $\textit{can}$ still equal)
Yuese
Ah, I think I got it, so I'll still leave the room open for someone to confirm: the contradiction is that in this case, ${e, x, y, z}$ would be a subgroup, but this can't be the case because it's order (4) isn't a divisor of the order of the original group (6) which contradicts Lagrange's theorem
Yuese
@full horizon Has your question been resolved?
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.close
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where is my mistake?
.close
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How would one find the value set of this function -2x^2 + 4x + 8?
The definition set is easy, all real values
But you could derive this function and find the value set that way to see where the derivative is zero and thats where u could find min and max
but isnt there any easier way
completing the square is the other way I think about.
But differentiation sounds easier
yea i guess completing the square is the way to go on easier functions like this
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I am trying to figure out why x + 1/2 is the same thing as 2x + 1
Cross multiplication?
one is just a scalar factor of the other
(2x+1)/2
But when I plug in values such as 6 + 1/2 it doesn't make sense
6 + 1/2 doesn't equal 13
oh so it's not cross multiplying
it equals 13/2
ye
why are we adding a power to x?
noo I'm talking about the relevenat cubic
that you were factoring
oh sorry, I'm just looking at the light blue text on the right
I should have cropped that better
Lemme resend the screenshot
But if 12+1 is correct for the numerator.. why do I need to add the denominator? /2
if 2x + 1 doesn't need the denominator
huh
I was using cross multiplication which is just wrong
I assumed that was the correct way but I don't even know if I can do that
I'm just trying to get from (x + 1/2) to (2x + 1)
why
and plug in other values to do the same thing
to understand why they are the same thing
and how to get there
but in my example with 6 + 1/2 it doesn't work unless i add a denominator of 2 for some reason
i didn't add a denominator for 2x + 1
they're the same thing whwen they equal 0
the y coordinate is set to 0
so you find the x coordinate
multiply by 2
but it only works because it's 0 on the RHS
you can't exclude that
as 2x0 = 0
no I'm saying ignore the above, we don't know it's 0
Hmmm
I don't understand
Ty
Does it matter if a factor includes 1/2?
Do you try to make factors whole numbers always
it doesn't matter
iit only matters if it doesn't expand to the cubic
which is why it is sometimes necessary to multiply by a scalar
@fickle trail Has your question been resolved?
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Is the maximum less than or equal to every element in the set?
Yeah my bad
But it is ≥ every other element right?
Is there any smaller element that's also ≥ every other element?
Then it's the supremum
In the field of reals, yes
In the field of rationals, the supremum doesn't exist
Well, the supremum is a number
So if the number isn't in the field, it doesn't exist
99.99% of the time though the field is the reals
So yeah, the supremum would be ³√29
Yep
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hello I feel like somethings a bit off with my answers
can someone let me know if they’re wrong and some point in the correct direction
3y²-3 for fy
For fyy it will be 6y
Yeah so in the next step it's just normal differentiation keeping the other variable constant
what about fxy?
Fyx would be 0
0
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Easy just place x in the denominator and you get vertical asymptote x,=0
@lethal scaffold Has your question been resolved?
well the question is requiring to make a function then describe certain things about it
like this
or this
@lethal scaffold Has your question been resolved?
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I’ve gotten to this point
But I’m not sure how to write it in terms of an inequality
I searched online and it said x<-3/4
But when I put it into original equality I got 0 so I’m not sure how that works
How would I do this without doing all that eg in an exam
you could use quadratic formula
Right and it would get me x =-3/4 and x=2 right?
How would I put this into the inequality form
Online says x<-3/4 x>2
factorising is more appropriate here
plug in 0, 0 and see if it satisfies the inequality
How do I get this?
graph the parabola
and see when it is above the x axis
this will tell you the values for which it is >0
Is there any other way to do it?
you have already calculated the intercepts
you can plug in values in each interval that the intercepts split the real line into
but that's is more tedious than simply graphing
you already know the intercepts presumably from solving the quadratic =0
or perhaps just factoring
Yes
so just draw a rough shape of the parabola
and determine when it is above the x axis
x axis
not y
if it's shaded inside or outside the parabola
shading?
Is that how you would do it in an exam?
that's how i would
as in something like this
Is there any other way to do it instead of graphing it
instead of graphing it, you can just plug in the coordinates 0, 0 to see what part you shade
inside or outside
take the example above
if you plug x = 0 and y = 0 in the inequality, it wouldn't become true
after you evaluated it
no that's not really what we're looking for
the original inequality has no y in it
we are not concerned about the y values
y is only there so it shows up
desmos doesn't graph it without the y being there
it's just an example as to what shaded parabolas look like
I’ve got the two x values, now how do I put it into the inequality form
you don't shade parabolas for these inequalities
the parabola tells you whether the values are greater than or less than 0
I also know what the inequalities are I just don’t understand how to get that answer
when it's above the x axis then the quadratic is >0
X<-3/4 x>2
yes
Yes but how can I do this without using the parabola
you should draw the parabola
Is that the only way?
think about it like 2 < x < -3/4
that the intercepts split the real line into
but it is more tedious as i have said
you want to draw the parabola
Right okay, then how does that become x>2
you know the intercepts
so you know when the quadratic changes signs
i.e when it switches from being <0 to >0
so you just have to check whether it's >0 or <0
you do this by plugging in a value in the interval you want to check
so say for example you plug in x=0
then you just get -6
so you know that on the interval -3/4 < x < 2 the quadratic is negative
so that means outside that interval it'll be non-negative
Right sorry i understand partially up to the last two parts
If I plug in x=2
I get 0
@lucid atlas Has your question been resolved?
Okay this whole thing confused me
I’ve got it now but without the plotting on the graph thing
I did it with a bit of trial and error
Thank you very much for your time anyway!
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I don’t understand how this got 9.33?
Help pls
6/18 simplifies to 1/3 because both 6 and 18 are divisible by 6
when you divide both 6 and 18 by 3 you get 1 and 3 respectively
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, my teacher explained a identical question on the board but i learned nothing from it, im supposed to expect a question like so on the test
the trajectory of the bounce angle will be equal to the 90 - the angle of impact
ur supposed to use trigonometry to find this angle
and then subtract what you got from 90
then you subtract THAT by 180
and draw a line at that angle
@keen agate Has your question been resolved?
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hm
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Sorry, using iPad and Discord, it wants to create a new channel when I add screenshot
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hi can someone help me with this?
i did part a i and ii
but im not sure how to do b
my idea is to do by cases, like
cases where all letters are distinct + cases when there are 2 Es or 2Ls + cases where there are 2 Es and 2Ls + cases where there are 3 Es
you are talking about iii?
yes
so itll go like
(6C4 * 4!) + 2 * [(3C2 * 5C2 * 3!) / 2!] + [3C2(select Es) * 4!] / 2!*2! + [6C1 * 2!]
i'm not sure about all of them
its u
hi
hi
\
last one i would have had 5c1 for the choice of the letter that isnt E and also 4C1 for its place inside the codeword
i dont get what you mean
oh like
so for the last one
when there are 3 Es
i selected the last one with 5c1
wait
oh it should be 6c1
wait
it cant be E
but theres alr 3 Es
J E W L R Y
E L
E
i selected all the Es so i just counted everything else
yeah so there are 5 possible letters remaining
wer'e not doing probabilities
where each letter has equal chance of being drawn
so its 5C1
why by 2! ?
there are 4 spots for the non E letter
4!/3!
ahhhh
yeah its not the end
oh wut
oh..
and im lazy
yes
would it work if i choose the Es first
wait
oh wait
i dont have to
cuz theyre all the same
its just about knowing their place
4!/2!2!
chose 2 spots among 4
xor*
you have to chose their spots though
(6C4 * 4!) + 2 * [(3C2 * 5C2 * 3!) / 2!]
ill permute them later
hmm
so i need to choose others
so
you didnt get rid of anything
do it from the beginning i think
yeah
then uh
then you have chosen all the letters
ah wait
there are 4 spots in your word
but the Es are equivalent
hmm
the EEE is wrong
its 6C1 * 4 = 24
right
ah wait
bruh
5C1
bruh
bruh
got ideas for b?
lets place the two parents first
total number of ways when mr lee sits with mrs lee - number of ways mr lee sits with mrs lee and the children are all sat together
why -?
only the children right?
yyea
it would need to be
total number of ways when mr lee sits with mrs lee - number of ways (mr lee sits with mrs lee are sat together) and (the 3 children are sat together)
oh
LOL
yea
i did
oopsie
lets do it
so total number of wyas with mr lee and mrs lee sitting toegther
its uh
so i fix the parents
then just permute the rest
so for the rest it would be 8!
with parents 9!
then because its a circle
so (9-1)!
so 8!
they are 8 in total
wut
oh nvm
lmao
wait so 8! is correct?
heh
oh right
then for the
parents together and children are sat together case
its uh
parents is one group
children is one group
so 7! in total
but circle so 6!
dont be confused
the children themsleves dont form a circle
they are separated by the parents
that are sat together
wait wat
separated by parents?
i thought im finding if the children are sat together
yea
in both cases the parents are sat together
yes
so you treat the childrens not as a circle
i didnt tho
.
everything?
okay
so there are 5 other circlees
so 5!
then there are two "groups"
children and parents
so 7!
then because its a circle
6!
oh ok
its right then
yeah
how many was that?
72000
yeah they got plenty of room
this kind of thing is always tedious
sadly this isnt recursive
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Hello I was wondering if someone could verify if this problem is correct?
This is probability and I worked on it with a friend but am not sure if it's correct or not
it looks right to me
@untold wadi Has your question been resolved?
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Okay
🫡
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@low jungle Has your question been resolved?
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Can someone explain why it goes from 144 to 169?
Arent you supposed to set it equal to 0?
set what to 0
the right side
no
then apparently not
do you know the equation for a circle?
x^2 + y^2 + 2gx + 2fy + c = 0
or more common is $(x-h)^2 + (y-k)^2 = r^2$
maximo
actually it may not be more common but it’s more helpful here
anyhow, we want to get the equation you’re given into that form
to do so we want to get (x-something)^2
since we have x^2 - 6x, that something has to be a 3
since (x-3)^2 = x^2 - 6x + 9
so it’ll look like this:
x^2 - 6x =
x^2 - 6x + 0 =
x^2 - 6x + (9 - 9) =
x^2 - 6x + 9 - 9 =
(x-3)^2 - 9
do you understand how i got here so far
Where does the 6x go here? I can see the x^2 becomes x and the 9 becomes 3. But what about the 6x?
think thats the part im stuck on
(a+b)^2 = a^2 + 2ab + b^2
(a+b)^2 =/= a^2 + b^2
sometimes it works but far from always
nono
you can’t
that’s the point
so we get (x-3)^2 = x^2 - 6x + 9
but we start with only x^2 - 6x
do we add 9 and subtract 9
we + (9 - 9)
Ok I think im starting to understand
I just need to look at it more
wait @hot thistle why are we adding the 9's? I get its converting x to square form, but why 9 specifically?
it’s the number that lets us write (x-something)^2
for example if we had x^2 + 2x
we want to add 1
so we get (x+1)^2 - 1
well we want a number so that when we expand it out we get -6x
so since (x-a)^2 = x^2 -2ax + a^2
we want -6x = -2ax
so a = 3
so we add and subtract a^2=9
ok its like 12am so ill have another crack at it in the morning, getting close though. Thank you!
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For 7 part (g), I believe the statement is true. If I was going to prove that it was true, could I use a contrapositive or maybe a contradiction?
the biggest number in S(x) is always x
Right
so if S(a) = S(b) their biggest numbers would also be equal, and thus a = b
In terms of a contrapositive, right?
you could say that if a weren't equal to b and S(a) did equal S(b) that would imply that a = b, so the statement that a is not equal to b and S(a) is equal to S(b) could never be true
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hello little boys
how do i move y^2 to the other side and x also
to get y^2=ax
i mean like whats the metho to rearrange it
it just shows they rearranging it
multiply both side by 40?
okay is that the only method
thats the easiest
so if it was 40 in the original equation i would multiply both sides by 1/40?
yes
yea or just divide by 40
that would be 40/1 times 1/40 and how does that work
like i dont understand the logic behind it could u explain pls
multiplYING 2 fractions
40/1 * 1/40 = (40 * 1) / (1 * 40) = 40/40 = 1
sorry, i forgotten latex already, so I can't give proper format
when multiplying fractions do i cross multiply?
wait
oh
do i
when multiplying 2 fractions do i cross multiply or multiply straight through
multiplying numberators by eachother
and denominators by achother
I am sorry, I am not a native english speaker
ok i will send photo
I am not too familiar with the terms cross or straight multiplication
sure that'll help
yeah
when do u cross multiply
because thats something i remember but dont know when to use it
i am not too sure... I assume if you cross multiply, you divide
you're welcome, if you do not have anymore question, you can close the channel by ".close"
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@timber briar Has your question been resolved?
@timber briar Has your question been resolved?
@timber briar Has your question been resolved?
@timber briar Has your question been resolved?
Where have you seen that cube before?
I'm just gonna throw the words menger sponge into this thread although I doubt you mean those
Something about this reminds me of fractals tho I don't know what
This looks very similar to a merger sponge
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b) onwards
i have a strange solution
no clue how to proceed
proofs are not my strong suit
hm factorised u get
2x(x+2)(x+1)
yea?
2x(x+1)(x+2)
imma reorder that
so for any arbitrary int a
like if we take a mod 3 we would get either 0,1,2
yeah
we can use case by case
if its 0 then its div by 3
if its 1 then a+2 is div by 3
if its 2 then a+1 is div by 3
n that covers all the cases
can you explain this bit?
yes
um
like cuz
if u take mod n of some int
ull get stuff from
0,1,2,..., n-1
n would reduce back to 0
like a cycle
so in this case mod 3 will just be
0,1,2
i don't think they are supposed to use mod to explain the proposition
oh?
i dont remember using mod in the textbook
poopy mod
u can just do the same thing
but with
a = 3k
or
a = 3k + 1
or
a = 3k + 2
for some k
right
for some integer k
mhm
yea
so i can start with
then
smt like this from there
a = 3k, a= 3k + 1 a = 3k+2
each is a case
umm
well
like
if
a = 3k
then a is div by 3
if
a=3k+1
then a+2 = 3k+1+2
=3(k+1)
and k+1 is an int
thus a+2 is div by 3
last case,
if a=3k+2
then
a+1=3k+2+1
a + 1 3k + 1
right...
and those r the 3 cases
essentially the same as mod lol
um
i mean
it was kinda
intuitive
that 1 of em wld b divisible
so just add some dummy var to
um
idk
beef it up
yeah but proving it is hard lmao
alright
complete explaination anw
or d
its similar to the b
where u got
a,a+1 and a+2
so 1 of em wld div by 3
so therell b nth in the denom
n itll div n give u an int
from part b
and as x in N
when div 3 u get some int >=0
then
multiplication is preserved in N
i got most of that until this bit
closed under?
right closed under multiplication


