#help-13
428200 messages · Page 525 of 429
I only ruled it out because it doesn't really make sense
yes
OOOOOOOOOOOOOOO
relative rates of growth is in our cal 2 class for some reason
it is mostly
get ready integration is a whole new ball game compared to diff
i have to finish out this section then do integration by parts by tomorrow
oh we're already well into integration
they just have an ungodly scheme for this class
satan made out syllabus
sounds like my calc 1
you de facto have to have a 100% on every homework to do the quiz, and without doing the previous quiz you can't look or see what the next topic is
I swear proff covering calc 2 in calc 1
i wouldn't have minded 1 semester for the entirety of single variable calc
I couldn't care for it much... just gotta take em as a preque
alright so according to symbolab which I'm now learning has a limit function (a funky tool which will help with tomorrow's exam) this evaluates to infinity
which i knew had to be the answer due to nlnx being slow as hell from comp sci
mystery solved boys
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actually let me explain WHY, work-wise this is the answer
constant multiple rule makes it 4/3 * the limit
x^2/xlnx
x^2/xlnx = x/lnx
1/1/x
4/3*inf = inf
hence the function is slower
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Hello so i'm trying to reduce a fraction (i'm not sure that's the correct term)
but the fraction is:
x^2-2xy
2y^2-xy
The answer i got is:
x
y
The correct answer is:
-x
-y
The way i solved:
x(x-2y) x
--------- = ---
y(2y-x) y
-Probably missed an important step or rule but i'm lost to why it's minus and not plus
The correct answer cannot possibly be
Sorry my bad
$\frac{-x}{-y}$
Disorganized
it's just - x/y
That's the correct answer (according to the answers in the book i'm learning)
Ok whats the issue
i got x/y without the minus
You just didn't show one step
Under "way I solved"
2y-x is the negative of what you need
So
Mutliply the denominator by two factors of -1
And distribute one of them INTO the parentheses
This will let you write the parenthesized part backwards
And it will match the factor in the numerator
oh i can't substract what's in the parentheses before i match them to look alike? (bad english sorry)
You didn't try at all in your working out
i didn't know it's not possible if it's not the same
i thought i could just substract what's in the parentheses
i'm sorry if i sound stupid, but what is the way to make (x-2y) into (2y-x)
-1(-2y+x)
Uh uh and that becomes?
Wait no
You are doing it all at once
This is correct
So it's fine
Moving fast then, lol
What you have done
Is multiplied by (-1)(-1)
...which is just 1
But one of those -1's
...
Wait
No.
Let's go slower
Start here
(-1)(x-2y)
Distribute the -1
-x+2y
commute as to explain what i did ?
No, "commutative property", Google it
No, Google "commutative property" so you remember what it is
You need to know it
$a+b = b+ a$
Disorganized
Commutative property (of addition)
Apply this to this
so -x+2y=2y-x
Yes
And this is the factor you needed
BUT
We have changed what we started with
no no you right that was important for me in general
Because we multiplied by -1 in the denominator only
We can fix this
Just by multiplying by another factor of -1
It won't matter if we do it in the numerator or denominator
That will make the net result the same as if we had multiplied by 1 in the first place (since (-1)(-1)=1)
So your denominator could look like
-y(x-2y)
Now you can cancel stuff
i kinda understood what you said, but by what you've said to me about commutative property
isn't y(x-2y) would be the same ?
man I was confused
i just having hard time to realise why it's not x/y and it's - x/y
how did u get x/y
u had
x(x-2y)
y(2y-x)
x-2y <> 2y-x
Yes and just substracted
x-2y = -1* (2y-x)
when u cancel those 2 out, ur left with a negative sign
so it becomes -x/y
alright i got confused with the commutative property i thought that's equal
nope
a-b is not the same as b-a unless both are 0 or the same value
so the reason it doesnt work
is because the commutive property states that a+b=b+a
but a stands for ANY number
that means u have to bring the negative sign into it
so if a is negative thats fine
u would just have (-3)+(4)=4+(-3)
not 3-4 which is what u did
Thank you very much for the explanation
@mental summit Has your question been resolved?
so x(x-2y) = -x(2y-x), right because x^2-2xy=-2xy+x^2
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yes exactly
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that's why @cerulean star brought up commutive property, sorry for not understanding what u meant, thanks both !
u have no question @mental summit
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the top one is in REF and the bottom one is in RREF, yes.
does your textbook forbid non-1 leading coefficients?
show me the definition of REF as given in your textbook.
row echelon form, yes.
orifyoulikeomittingspacesbetweenwordsit'srowechelonformyes
you can and should check it yourself lol
okay then let's walk through the definition step by step
_Recall that a matrix is in row-echelon form if:
- Rows of all zeroes are at the bottom.
- The first entry of any nonzero row is a 1.
- All leading 1's are to the right of the leading 1's in the rows above it._
is it true that in your matrix all rows consisting of nothing but zeros are at the bottom? Y/N
@plain bridge are you still here?
actually you're wrong
it is true that all rows of zeros are at the bottom
precisely because there are no rows of zeros
just to be clear, what you're wrong about is your assessment of the first criterion as false.
the first criterion is true; it's the second which fails, for a reason you have already mentioned.
no, wrong again
how is the top one in ref when there's a leading 5
in the third row
we've been talking about the first one all along.
yes, it is neither.
yes, you were wrong about the reason.
yes
what's wrong with them?
oh, they made a typo
you made it seem like there was something wrong with it fundamentally
there's a typo in either the rref or the parameterized solution
ok then there's a typo in the solutions
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Which one of the above 2 images is the correct version?
First one or second.
I just learned this determinant and properties so I'm not confident
this is the correct form for 2A
But if take value
It doesn't match
Yes
Is not the double
$det(2A) = 2^ndet(A)$
Max..
Where n is the dimension of A
What is dimension?
It is a 2 by 2 matrix
Ohh
So dimension is 2 in this case
That triangle is delta right?
You can prove this for any number lambda using the fact $det(A\times B)= detA \times detB$
Max..
$det(\lambda A) = det((\lambda I) A) = det(\lambda I) \times det(A)$
Max..
So I was having trouble but thanks
And the determinant of I is the product of the diagonal and the product of the diagonal is just $\lambda^n$ and the result follows
Max..
There is I as well there it seems
Yes I = Identity Matrix
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do i just have to plug the numbers in or is a different?
wdym plug numbers in?
your mass defect is when the LHS is not equal to the RHS
you get a difference in mass
wym
im a bit lost
first calculate the total mass on the left
in terms of u
then calculate the total mass on the right in terms of u
find difference
that is your mass defect
Oh okay
how would I do that? Sorry I just wasn't here on this day so I'm clueless
u?
yea
Ohh okay
now for the yield am i suppose to (1.25 x 10^-1) (
what am i suppose to be multiplying it by?
Oh no
1.25 x 10^-1 so i would have to conver this ino kg?
ohh okay
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how many sides has a regular polygon whose number of sides increases by 2 when its angle increases by 9 degrees
I'll be back in a few minutes
sorry but how?
no
@gritty mulch Has your question been resolved?
its a little more convenient to work with exterior angles rather than interior as the angle sums of those is constant
also don't just hang in the background otherwise i'm gonna assume you're not there
start by introducing a variable to denote the number of sides in your original polygon
or don't...
why?
sarcasm
i'm attempting to guide you through the problem
and that takes effort and responses from you
otherwise i'm potentially gonna bail and do other things
ok thank you but i don't know how to write this
pick a letter/symbol or whatever
and use that to represent the number of sides of the original polygon
let "insert choice here" be the number of sides of the original regular polygon
at this point all i really want from you is to pick an appropriate letter/symbol
which is pretty much standard for all questions where you're trying to find an unknown and no variable has been given to you (so you introduce one yourself for ease of calculations)
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very quick q: is (5-x)^2=(x-5)^2
it is right?
yes
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I know (a^b)^x is (a)^b×c but what about
(a^b×k^r)^x
use the product of power rule to move the power of x in, then use power of power rule
@regal fog Has your question been resolved?
Thanks.
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Can someone help me understand why this is incorrect
You multiplied with positive number 3 ... No need to change the inequality sign.
In 2nd last step
Basically you are multiying +3 on both sides. So the inequality doesn't change
Oh okay so if it was X/-3 then we would change signs
if you multiply both sides with -3, yes
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Can anyone tell how it was solved that x,y,z was obtained?
need more information
Never mind, I solved it
.close
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hello can anyone check my answer
$\bar{z}$ denotes the complex conjugate of z?
illuminator3
you can see under the second picture
It looks correct to me 🤔, last line is 4abi + 2a-b = 8i, right?
yeah
but I have no clue what should I do next because the instructions say define every complex number of z for which is true:
Right side and left side must be equal, considering a and b are real numbers
The real part of 8i is 0
What's the real part of the left-hand side of the equation?
@urban plover Has your question been resolved?
@urban plover Has your question been resolved?
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i need to get A valor & B valor
no way ..
Multiply both side by x^2 - 1.
Recall x^2 - 1 = (x+1)(x-1).
The rest follows easily since your dealing with two polynomials that are equal.
I need to solve smth like this?
Oh
um?
yeah exactly why are people here answering in fingquestions
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can you get x+2z=0 and y+z=0?
This ^ is it.
z is the free variable.
So write x = -2z and y = -z.
The solution space is (-2c, -c, c) for any scalar c.
That is one solution.
Did you mean c(-2,-1,1)?
@plain bridge
Yes.
“free non leading variable” trying to parse this.
If the system you have has no solution then it’s not true. But if a solution exist your statement is false.
False I meant.
Yep.
Sure.
rank means number of linear independent columns, which is the columns that have pivot entries, so yes we have one free variable.
Remembering my past LA course, I am kinda of rusty. Currently relearning.
(c,c,c,c,c) yes.
Not sure what you mean by “making our answer just 1”.
Ok then yes.
Are we talking about the same matrix but doing two different elementary operation on them?
If not then yes the system of equation are not the same.
If so then it will have the same solution space not matter what elementary operation you do.
But it possible the two matrix might be row equivalent.
In this case I will refer to my first answer.
Yep.
Unless we know the two different system are row equivalent.
Which we can check by row reducing.
Second I meant.
A solution exist.
@plain bridge
Yep.
x = 1, y = 0, z = 0.
Yep.
x = 1, y = 0, z = 0 if you're using x y and z
I guess it works
Even if we had 1 on the last rows of the last column we still have a solution.
Can you explain why you still think for this case that a solution does not exist?
If we had that.
If the last row was (0,0,0,1) then we have no solution.
Is that what you meant?
np.
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quick question, how do i transform a multiplication to a fraction form?
id like to transform sin(x)*ln(x) to a fraction so i can apply the hospital rule
$y = 1/(1/y))$ where $y=sin(x)$ or $\log(x)$
riemann
riemann
just to confirm that i got it right because i felt lost for a good minute, transform sinx to 1/cscx, multiply by lnx = lnx/csx then put back the sin or just keep as is correct?
basically ln(x)/csc(x) or ln(x)/1/sin(x)
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The letters of promise are arranged randomly in a row. Find how many solutions there are if the letter p and r are seperated by at least three letters
Anyone able to help
It's a permutation since it has order to it
Should I somehow find out when p and r are in 2 letters or less in proximity then subtract that from 7!
🤷♂️
@vast wolf Has your question been resolved?
@vast wolf Has your question been resolved?
@vast wolf I think you should go for the complement here
Take the number of rearrangements that have p and r separated by exactly 0, 1, or 2 letters, and subtract that from all possible rearrangements
Hmm
Maybe there’s something better here
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Hi. I want to prove that a + b = a' + b' mod d.
Please tell me if this is correct:
So let be:
a, b, a', b' in Z and d in Z+
with (a = a mod d) and (b = b' mod d).
Then let be k, k', k'' in Z
(a = a' mod d) <=> (a = a' + kd)
(b = b' mod d) <=> (b = b' + k'd)
So
a+b = a' mod d + b' mod d
= a' + kd + b' + k'd
= a' + b' + kd + k'd
= (a' + b' + k''d)
= a' + b' mod d
Is thiss proof correct?
@balmy beacon Has your question been resolved?
assuming that k''=k+k', the proof looks good
just as a side note, the "a' mod d + b' mod d" isn't strictly necessary
a+b = a' + kd + b' + k'd is sufficient
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Ooo fancy question
I like it
Build systems of 180-a
180-b
Bla bla
And you will get it
cant i just use exterior angles of polygon add to 360? i know that if i can get to x-180 + a + b + c = 360 then x= 540-(a+b+c) but i dont know what that last angle is
your method does work tho so ill just use that ig
.close
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How to increase your chances of winning at games of chance?
you don't
LMFAOO
But like, be sure to get a gain at least
not every game is gameable, and it wouldn't be the same strategy anyway
I know I'm going to have to calculate the probability of winning the games and see if it's over 50% etc but I don't know where to start
It's fine I think, cuz i can choose the game
Coin flip game I see
it doesn't make sense to do that no
Do what?
oh you mean you are going to calculate it before choosing the game and then just play it?
Yes, it can be a benefit over a certain number of rounds or in general
U got another game to compare?
i meant like, it doesn't matter if your chance to win is 58% if you pay x to win 1.75x
something like that
But, if the winrate is 58% if I play a lot of times I'm sure to make a profit
no, that depends on what the prize is and what you pay to play
Ye but in most of game u pay less than the profit right
of course you pay less than the prize, it's not enough
The probability of winning is 53% and we pay 2€ we can win 3€ if we play 1000 times we are sure to make a profit I mean
So, u make profit right
How u lose money
are you used to people saying the opposite of what they want to say lmao
english not my native language haha
your lawyer in court wouldn't be too happy about that
just count it
you pay 2€ 1000 times
you win 530 times, -2000 paid, 1590 won
wtf am i that dumb
Ye i mean ur right
forgot that the gain must be above x2 at least
slightly below x2
50% and x2 means you can play for fun, it's free
53% and x2 means you make profit, and it can be a bit below x2, still profitable
Oh ye
depend of the winrate
so how can u know the profit u need to make with the winrate
when you win you gain (prize − cost), this happens 53% of the time, rest of the time you lose the money, you can write an equation
0.53((multiplier * x) - x) = 0.47x
when it's that, you don't lose
1.89x is below 2x
so you want the multiplier to be 1 / winrate
,w 1/0.53
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well i constructed line CA thats literally it but im pretty sure i cant say its parallel to qp cause it wants me to prove that so im not sure
@craggy anchor Has your question been resolved?
<@&286206848099549185>
do you know the requirements for proving similar triangles
well i usually just use equiangular
<@&286206848099549185>
BQ/BC = BP/BA = 1/2
does that prove that the triangles are similar?
that doesnt prove that pq is similar to ca does it?
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will i have to find the distance between every point
or the gradient
since 2 sides
need to be equal
you can either find the gradients of your lines and show that they're equal or you can find the distances and show those are equal
it's entirey up to you
you can do both if you're really paranoid
Ohh right , since if the gradients are the same it gotta be parallel and parallelograms have parallel opposite sides
didnt think i could use the gradient for that tbh
ty for help
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how do I prove that OBB' is congruent to AXX'?
OABX is the original parallelogram, with a shear applied to it by some multiple of OA
@bold snow can u prove that the 2 triangles are similar?
U can easily prove congruency after proving the similarity
I need to prove that BB' is the same as XX' to check the similarity
No u dont
So angle BOB' = XAX' , because a shear was applied and the incline should be equal
Since OB is Parallel to AX , and BX is the transversal, we can say that angle OBB' = angle AXX'
So from AA property, the two triangles OBB' and AXX' are similar
Now since it's a parallelogram, OB = AX
So the respective sides are in a ratio 1:1
Therefore triangle AXX' is congruent to triangle OBB'
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can you phrase this more formally?
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Hi, I am not sure how to solve these problems, but i attempted and got 6591 for the second one. Can someone check it?
ye
Try to write them in base forms first
Do you know how to write square root as a power
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What exactly is he doing here, what does he mean by "taking x * h^(n-1) n times" and "selecting h from atleast two factors"
This seems like another simple thing worded confusingly which is what always gives me the most headaches
Are you familiar with it?
Well, i forgot what the two numbers ontop of eachother mean but i know what the sigma means
Maybe that's a sign i should reread the previous book
It'll be easier to retain everything
Those are binomial coeffizients
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Can I get some help with this
Find the y intercept and the gradient
What have you tried
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I'm writing a short paper where I analytically prove a few theorems of elementary arithmetic about the divisibility relation n | m \↔️ ∃p(m = np), prior to introducing division with remainder and/or its related algorithms, as a proof of concept for a friend. right now I'm trying to prove a lemma of Euclid's lemma, to in turn prove the uniqueness part of the fundamental theorem of arithmetic
the lemma is that, given nonzero integers m,n,p,q where p,q are coprime and mq = np (i.e. m/n = p/q), there is an integer k such that m = pk and n = qp (i.e. m,n are multiples of p,q by the same integer). I know that, if q | n, then the theorem follows (because then n = qk for some k, in which case mq = pkq, and so m = pk), so in turn I just have to prove that q | n, but I don't know where do I even begin
so far I've proven that divisibility is a pre-order over the nonzero integers having 1 as a minimal element over the positive integers, that every composite number is a product of primes (though not the uniqueness of this product, of course), and that the integers resulting from the division of m,n by their gcd are coprime (i.e. that if m,n are nonzero, and if m = gcd(m, n)p and n = gcd(m, n)q, then gcd(p, q) = 1)
To prove q|n, use mq=np. This means q|np, but since q and p are coprime, it means (Gauss's lemma) that q | n
note that I'm trying to prove this lemma in order to prove Euclid's (Gauss') lemma
Oh
so that would be circular
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Hi guys, what equation should I use to build the following graph? The second screenshot is the closest I found so far
you need a fourier series there I think
nvm
you can use something like
$y = -\frac{2a}{\pi}\arctan(\cot\frac{\pi x}{p})$
illuminator3
with a being the amplitude and p being the period
such things are called sawtooth waves
also messing with $a\cdot\text{arctan}(\tan(b(x-c)) + d$ will get you there
ΣAC
well, I think I can work with that
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They gave you two great answers
you can also use the floor() function, but I'm pretty sure that's not what you are meant to use
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Am I doing this right?
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i have one that makes no sense to me, and yes its math not science
a radioactive material has a half-life of 20 Years
how radioactive is the material after 80 years, if the value is set to 1 at the year 0?
<@&286206848099549185> idk if the question is broken or im dumb
What have you tried?
Just different ways of trying to understand it
Well, how radioactive is the material after 20 years?
Literely no idea
I dont know where to subtract from
Like halflife
It doesent tell me how much there is, it just says that the value is 1 at year 0
What does halflife mean?
That it looses its radioactive power
By how much?
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log(x^2) = 2log(x)
But also it does meet l'hopital conditions
Recheck what those are in your notes or textbook
I thought log of infinity was undefined
I thought incorrectly I suppose lmao
its inf/inf
nevermind, thanks @dire geode
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guys can anyone telll me how is (b,c) = (b,-c)?
@vernal palm
oh damn
@vernal palm btw one remark is also given
idk if its related to it
@vernal palm^
idk lol sorry
can u plz help me with this
any common divisor of b and c is also a common divisor of b and -c, and vice versa
"the gcd is always the same" is too vague
then what's the most appropriate statement for it?
The absolute value of gcd is always the same?
perhaps "multiplying either argument of gcd by -1 doesn't change the value"
or "gcd doesn't depend on the sign of its inputs"
oh
Guys is it the general form of gcd in which the largen number will always be in the first place and be considered as the value of b in b=aq+r ?
@tropic oxide
I mean why is it that 4840 = b and 1512 = to a like yea it's quite common to divide the bigger number with the smaller one but still I should have some statement or prove in order to make it true right?
......
what do you mean?
are you looking for a theorem that says "The evaluation of a gcd MUST proceed with the bigger number first and the smaller number second, and in NO OTHER WAY"?
Kinda
well there is no such theorem because such imperatives are very very rare in math
i think you're confusing the gcd function itself with the Euclidean algorithm that evaluates it in a certain way.
My question is that why in b=aq+r, b=4840 and a=1512
what "lemme"?
Tbh yea I am kind of confused with gcd and Euclidean algorithm
Division algorithm or like what bruuhhhhhh
Now it's freaking confusing
there should be a clear difference between "what gcd is" and "how to find gcd"
Ok
one method of finding the gcd of two numbers is to run the euclidean algorithm on them, recursively doing division-with-remainder until the remainder turns to zero at some point
To find gcd I can use the Euclidean algorithm, using b=aq+r where the gcd of a and b can be represented as ax+by also which is Stated in thee bézouts theorem, right?
...no, i never mentioned bézout's theorem...
again you are confusing two things that are very different: finding the gcd itself vs. representing gcd(a,b) as a linear combination of a and b
No I mean there is a part of the question that requires the bezouts theorem
It's just the half question
im trying to take you through the underlying theory
Ok so u mean bézouts theorem help us to represent the gcd of A and B in a linear combination of a and b?
bézout's theorem states that such a representation EXISTS
but it does not "help" us beyond that
the theorem itself gives us no method of computing it
And you Euclidean algorithm helps us to find the gcd of two number say a and b
True
Hmm
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Guys can anyone explain me what are the definitions for bézouts theorem, division algorithm and eucliding algorithm and what is the use of gcd in all of them
do you mean bezouts identity?
Yep
Like I am a bit confused as all of this uses the gcd
if we're just considering integers: if gcd(a,b)=c then there exists numbers x and y such that ax+by=c
and assuming you're referring to the Euclidean algorithm
the euclidean algorithm is used for finding the GCD
Well one more question why in this question the lhs is 4840 and not 1512?
Haha sure
that is part of bezouts identity
No I mean in the part where the Euclidean algorithm is used
we are trying to find the remainder when 4840 is divided by 1512
i.e. 4840 mod 1512
since the remainder is 304
next, we do 1512 mod 304
and so on
yes
that is the process of the euclidean algorithm
4840=3(1512)+304
so 4840 mod 1512 gives 304
Oh
@frozen tendon Has your question been resolved?
did you have another question?
Nope
But just one thing
The larger number is always supposed to be on the left side right?
@torpid fulcrum
@frozen tendon Has your question been resolved?
sorry, was away for a bit
the order doesn't matter
but if you start with the smaller number on the left
the next step will swap the two numbers
for example if we had 425 and 623
425 mod 623 = 425
so in the next step we do 623 mod 425
@frozen tendon
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hello, for this problem i have all inequalities set up for exponents of base 3 and 4, but cannot think of a way to compare exponents of base 3 and 4.
i will show you my reasoning so far:
3^3^4 > 3^4^3, because 3^4^3 = 3^64 and 3^3^4 = 3^81
3^4^4 > 3^3^4 just by looking as it as both are the same, except that for 3^4^4 its exponents are greater
so we have the inequality 3^4^4 > 3^3^4 > 3^4^3
4^3^4 > 4^3^3 because 4^3^4 has greater exponents
4^3^4 > 3^3^4 because 4^3^4 is of a greater base
now i know i must change bases of these exponents because there appears to be no other way to figure this problem out
i have done a bit of research but cannot find anything conclusive about how to change the bases of these exponents
Of thats really what you want to use
But you could just intuitively look at the numbers
yeah, but that only got me so far
@fast turret Has your question been resolved?
Oh this is a bmo1 problem I’ve literally done this before
hey, thanks for chiming in. could you explain the remaining part of the solution please?
i understand that to prove 3^256 is greater than 4^81 (without using a calculator) i would have to either find a common base or use some exponential/logarithmic rule which i do not know of
imma try and figure it out using a logarithmic rule rq
one sec
No no this isn’t necessary
We don’t need bases or exponents to be the same to make comparisons
We could have both the exponent and the base of a number be less than another
4^81 = 2^162 and clearly 2^162 > 3^256
very true actually
You can do a similar thing to fit the remaining numbers together
yeah like with 4^3^3
Yea
do you come from the uk then?
ahh right im in y11
i really should be studying for science and history but this is way more fun lol
besides after this week exams will be done
ill have 3 months to do just competitive maths
well i really like the subject so it’s fun but there’s also the advantage that it will look really good for maths related degree courses and jobs and such
but it’s mostly recreational for me
do you do the olympiads or do you just look at the questions?
That’s cool, I did the most recent bmo1 although I only started preparing like a week in advance
2016/17 had some pretty doable problems imo
Q1 - Q3
It’s decent to pick up skills tbf
I didn’t even know about this stuff in year 11 so I just did a level stuff instead, I kinda wish I did tbh
@fast turret Has your question been resolved?
yeah i can’t wait for college personally, but this is rlly fun and will be really beneficial for solving questions at college
@fast turret Has your question been resolved?
@fast turret college is fun. I became heavily interested in maths whilst at college
I found further maths so much more interesting than normal A level maths
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f(x)=x²-4x+8
g(x)=-|-x-2t|+9
h(x)=f(x) (f(x)<g(x))
g(x) (f(x)>=g(x))
for some t y=2k and h(x) has 3 different intersection points
find every integer k
$f(x) = x^2-4x+8$
IntelligentCake
$g(x) = -|-x-2t|+9$
IntelligentCake
constant
value of constant??
idk
i edited the question
idk what to call t
maybe its not called constant
<@&286206848099549185>
ur prob right
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420 hah
Kurama
Im trying to generate some contradiction here
Not sure what the next step should be
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@crude scroll Has your question been resolved?
You will need to provide more information
@crude scroll Has your question been resolved?
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”Circle: (x+2)² + (y-3)² = 20 Line: y = -⅓x + 3. Give me the coordinates of the points were they intersect”
what do you get if you substitute y = -⅓x + 3 into (x+2)² + (y-3)² = 20
@lethal moat Has your question been resolved?
just substitute it
What do you need help with
he just wants the answer
Oh psh
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.reopen
✅
Answer + wanna learn how to do it
. That's how you do it
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this calculation seems wrong, shouldn't the power of 10 be -1
It shouldn't be -5, but it shouldn't be -1 either
@delicate ember Has your question been resolved?
,calc 1.7 * 5/14.9
Result:
0.57046979865772
should be 5.7 * 10^-2 surely
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Would this be present value formula what formula would I use to solve this
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Let’s just figure it out @crimson sedge
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can I find the gradient for both first
then use m1 m2=-1
then I can use simultaneous eqn to check parallel?
cuz u want it to become a mx+c
it must be in y = mx plus b form
mb
put it into y = mx + c form
and then
use this formula
y2 - y1/ x2-x1
3/4
ok good
if the gradients are the same than they are parralel
now for l2
and then its 9/6
-9/6
so therefore they are not parallel
cuz using m1 m2=-1
no
now to check for perpendicular, m1m2 =-1
its not true as well
thats for perpendicular lines
didnt read the question
so you are correct @pastel sonnet
okok
thx
I was over complicating stuff earlier
thats why I was a little confused
I thought about inplict differentiate the first one
implicit*
but I didn’t see I could just change sides lol
.close
Closed by @pastel sonnet
Use .reopen if this was a mistake.
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