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#help-10
obtuse pebbleBOT
Available help channel!
half marsh
Hi! I'm trying to show that this function is surjective, could someone help me ?
half marsh
f is surjective if y = f(x), so I have obtained a second degree polynomial and tried to solve the equation using b^2-4ac
i found these two, now I don't know where to go from here
main cedar
More precisely, a function is subjective if for all y in the codomain, there is an x for which y=f(x)
So we can just choose one x out of those two options
half marsh
So we can just choose one x out of those two options
after choosing what do we do with it ?
main cedar
Show that when y is less than or equal to -2, f(x)=y
half marsh
hmm a minute i will try, thank you for helping me
half marsh
Show that when y is less than or equal to -2, f(x)=y
not sure if thats what you meant but i had to solve f(x) inferior equal to -1 for y less than or equal to -2 no ?
viscid delta
i found these two, now I don't know where to go from here
To show surjectivity, you just need to show that there exists an x for which f(x) = y, which you have done here. All that's left is to show that the expression you got here actually exists. So really all that's left is to verify that the radicand (thing inside the square root) is nonnegative. Can you see why it is?
obtuse pebbleBOT
@half marsh Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
vernal crystal
Hello, i've recently been trying to use geogebra, but i'm having trouble filling in arbitrary shapes that are created via intersections, and do not have straight lines. Is there a fill tool or some other algebraic way to fill these easily?
obtuse pebbleBOT
Hello, i've recently been trying to use geogebra, but i'm having trouble filling...
vernal crystal
trying to fill the rounded square in the center. (1)
trying to fill this triangular shape, curved base (2)
trying to fill the circle, but without the triangle (just the segments around the triangle)
vernal crystal
hmm, okay. Is there a way to shade it like this?
not with the checkered pattern
chequered ? dunno the spelling
somber mural
sorry idk. I don't really use geogebra
vernal crystal
alright
thanks anyways, i'll keep this open a while longer to see if anyone else knows
obtuse pebbleBOT
@vernal crystal Has your question been resolved?
vernal crystal
no
@somber mural aha i've done it using some really disgusting arcs and circumcircular sectors
i mean i guess it works, just gotta hide the objects lol
somber mural
fair enough lol
primal oak
haughty coyote
,rotate
warm shaleBOT
haughty coyote
What did you get using u=x^4 ?
haughty coyote
Notice x^3 is almost the derivative of x^4. That's how you get your du
red snow
obtuse pebbleBOT
red snow
i don't understand why a and b =1?
solid sun
-8 = -9 + 1.
solid sun
For all, do you see why?
red snow
ok i think i get it but why is it -8 at the start?
solid sun
Because it’s the problem?
solid sun
Not following?
red snow
i do, maybe like 90%
solid sun
For all, do you see why?
You can split this fraction into two sums.
After you rewrite -8 = -9 + 1.
timid silo
Can someone walk me through this question? I have to prove this trig identity.
sage geode
Can someone walk me through this question? I have to prove this trig identity.
What have you tried so far?
timid silo
What have you tried so far?
I started with the left side
timid silo
Any work you've got to show?
No not really im just overall confused
pine sail
What's 1 - cos^2x?
timid silo
sin(x)^2?
timid silo
sin^2x / cos^2x
pine sail
Yep, try now.
pine sail
,calc 67814637 * 2183897
warm shaleBOT
Result:
1.4810018230039e+14
There you go, now if you have any problems go get a channel of your own.
timid silo
,calc 435126157863762541863543178463174351764563179574289789 * 713854338153368712287628e35653452
warm shaleBOT
Result:
Infinity
mighty geyser
Result:
```
Infinity
```
limited precision
timid silo
sin(x)^2(sin^2x / cos^2x)
. well i got this
haughty coyote
35 million digits is not that easy
Not 35 million, just a few dozen
timid silo
Not 35 million, just a few dozen
Oh yeah because e35653452 are already that many zeroes?
and same for the other right?
so 1-sin^2x(1-sin^2x/1-cos^2x)
haughty coyote
Oh yeah because e35653452 are already that many zeroes?
Technically yes. These digits don't have to be computed. Now because floats have their flaws it probably fails at rounding to the nearest float
timid silo
I see, thanks mateo.
pine sail
and same for the other right?
Well you don't have to solve for the other side? Just make the LHS look like rhs. Or if you wish just solve them in way it makes sense.
timid silo
so 1-sin^2x(1-sin^2x/1-cos^2x)
I fixed it if its right
pine sail
I think solving for the LHS to make it look like the RHS would be easier
Then good, we aren't going to touch your rhs, cause it's fine either way.
pine sail
so 1-sin^2x(1-sin^2x/1-cos^2x)
NO
Where'd 1 - sin^2 x come from?
Before parenthesis.
timid silo
Before parenthesis.
so everything after the parenthesis is right?
i would just have to take away the 1-sin^2x and keep it as sin^2x because 1 - cos^2x=sin(x)^2.
pine sail
Well yeah, also write sin^2x or (sin(x))^2
timid silo
Well yeah, also write sin^2x or (sin(x))^2
why over 3?
timid silo
☠️ mb
timid silo
alright
obtuse pebbleBOT
@timid silo Has your question been resolved?
verbal prawn
$$\left(xsin\theta -ycos\theta \right)^2$$
obtuse pebbleBOT
$$\left(xsin\theta -ycos\theta \right)^2$$
warm shaleBOT
Nayan RBLX
verbal prawn
what will be the ans here
restive ridge
show work
somber mural
no one will tell you the answer bro
somber mural
what have you tried
verbal prawn
wait ill send the full question
$$prove:\left(xsin\theta -ycos\theta \right)^2+\left(xcos\theta +ysin\theta \right)^2=x^2+y^2$$
warm shaleBOT
Nayan RBLX
restive ridge
show work
verbal prawn
can I cancel 2xsin * ycos and 2xcos * ysin
somber mural
expand your brackets first
verbal prawn
$$\left(x\sin \left(θ\right)\right)^2-2x\sin \left(θ\right)y\cos \left(θ\right)+\left(y\cos \left(θ\right)\right)^2$$
warm shaleBOT
Nayan RBLX
somber mural
ok good
verbal prawn
now?
verbal prawn
same thing just 2xcos*ysin
verbal prawn
ok so you can do some cancelling yes?
can we actually cut them
cause in the first one there is xsin and second there is xcos
somber mural
cut?
verbal prawn
I meant cancel
somber mural
$x^2sin^2\theta + x^2cos^2\theta + y^2cos^2\theta + y^2sin^2\theta$
warm shaleBOT
IntelligentCake
somber mural
this what you have yes?
somber mural
actually you can
because you have 2xysincos
and -2xysincos
yeah you definitely can
somber mural
yeah
verbal prawn
bruh
somber mural
its all multiplied together
somber mural
yup
somber mural
sure
verbal prawn
$$prove:\left(sin\theta +sec\theta \right)^2+\left(cosec\theta +cos\theta \right)^2=\left(1+sec\theta \cdot cosec\theta \right)^2$$
warm shaleBOT
Nayan RBLX
somber mural
again start by expanding
obtuse pebbleBOT
@verbal prawn Has your question been resolved?
obtuse pebbleBOT
@verbal prawn Has your question been resolved?
autumn tinsel
all values of the real parameter a for which the equation has only one solution
obtuse pebbleBOT
all values of the real parameter a for which the equation has only one solutio...
autumn tinsel
D=0
lusty cedar
do it then?
autumn tinsel
but I get something weird
lusty cedar
like what?
autumn tinsel
i got 4 and -1
lusty cedar
so?
grizzled shore
why must it be 5?
autumn tinsel
why must it be 5?
that is answer
autumn tinsel
-1
lusty cedar
yea i mean
lusty cedar
it is 3
grizzled shore
^
lusty cedar
i think
grizzled shore
it is
lusty cedar
yea
autumn tinsel
all values of the real parameter a for which the equation has only one solutio...
this is equal to zero
lusty cedar
yep
grizzled shore
so -(-6)/2 = 3
lusty cedar
exactly
grizzled shore
this is equal to zero
yes
you find where the discriminant of your first expression to be 0
that's where there's only 1 solution
lusty cedar
it's 2a^2 - 6a + 11
i believe -11
grizzled shore
no that would be -2a^2 + 6a - 11
lusty cedar
hmm
lusty cedar
mine is kinda bit broken
timid silo
it's 2a^2 - 6a + 11
It's -a² + 3a + 4
grizzled shore
It's -a² + 3a + 4
Yep
obtuse pebbleBOT
@autumn tinsel Has your question been resolved?
proper dragon
Hi, let's say here I didn't let x3 = 9t, and instead just did x3 = t as I'm used to.
Would the basis I end up getting still be correct ?
I personally got x1 = 4t/9 and x2 = t/9 (with a few different steps, I didn't use the gaussian elimination method to simplify the matrix, I figured using the traditional method was faster.
obtuse pebbleBOT
Hi, let's say here I didn't let x3 = 9t, and instead just did x3 = t as I'm used...
restive ridge
!15m
obtuse pebbleBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
proper dragon
It has been over 15 minutes
Oh nevermind, I misread 9:38 PM as 9:30 PM
That's my bad
obtuse pebbleBOT
@proper dragon Has your question been resolved?
proper dragon
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
chilly cliff
Hi.
chilly cliff
$$\begin{align*}
& \text{I want to find gcd(p, q).}
\
& p(x) = x^{n} - 1
\
& q(x) = x^{n-1} + x^{n-2} + ... + x^{2} + x + 1
\end{align*}$$
The polynomial division is without rest, and the result is p(x) = q(x)(x-1). Does that mean gcd(p, q) = q?
warm shaleBOT
tarık b.
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
chilly cliff
The question is now correctly given in LaTeX.
obtuse pebbleBOT
@chilly cliff Has your question been resolved?
kind hawk
yes
chilly cliff
thanks
dire umbra
How do you get the inverse for something like f(x) = x^2 + 2x on [0, inf>
dire umbra
It seems hard because of the ^2
restive ridge
c is -f(x)
dire umbra
how is c = -f(x)?
dire umbra
0?
restive ridge
-f(x) = 0?
rocky goblet
if c = 0 then where's the f(x)
restive ridge
^
dire umbra
how do you know f(x) = 0?
restive ridge
that's what you just said
rocky goblet
we don't, you're the only one who claimed anything about 0
dire umbra
When?
rocky goblet
we're saying that if f(x) = x^2 + 2x, then x^2 + 2x - f(x) = 0
dire umbra
.
That was more of a question
ax^ 2 + bc + c
in this case x^2 + 2x there is no c
so isnt c than automatically 0?
rocky goblet
well if you're looking at just x^2 + 2x then c is 0
but that's not the quadratic we're suggesting looking at
f(x) = x^2 + 2x implies that x^2 + 2x - f(x) = 0
dire umbra
if you use the quadratic formula dont you need a ''c''?
rocky goblet
so, x^2 + 2x - f(x)
dire umbra
-b +- root(b^2-4ac)/2a
dire umbra
rocky goblet
well you do need to "have" c, but if there isn't a constant term you can just set c = 0 because adding 0 to the quadratic doesn't change anything
dire umbra
Yeah thats what I meant by it
dire umbra
so, x^2 + 2x - f(x)
but why does c become f(x)?
rocky goblet
well again, what else would it be
if you have x^2 + 2x - f(x), what's a, what's b, and what's c
dire umbra
a = 1, b = 2, and c isnt in the formula so c = 0
rocky goblet
c is -f(x)
dire umbra
yeah but what is f(x)?
rocky goblet
that's what we're trying to find out 
or wait no
we're trying to find the inverse of f
so essentially it's just a thing
dire umbra
This was the question How do you get the inverse for something like f(x) = x^2 + 2x on [0, inf>
dire umbra
so essentially it's just *a thing*
a thing??
rocky goblet
i don't really know what word to use to describe it
just treat it like it's an extra variable or something for now
dire umbra
you mean f(x)?
rocky goblet
yes
dire umbra
okay I will do that
dire umbra
Wait
rocky goblet
x^2 + 2x - f(x) = 0
now we solve that, and that gives you an expression for x in terms of f(x)
which is pretty much the inverse of f
dire umbra
x^2 + 2x - f(x) = 0 --> x^2 + 2x - y = 0 --> solve for x a = 1, b = 2, c = 1
x = ( - b + root(b^2 -4ac )) / ( 2a )
x = ( - 2 - root(2^2 -4* 1* 1 ) ) / ( 2 * 1 ) = 6
x = ( - 2 + root(2^2 -4* 1 * 1 ) ) / ( 2 * 1 ) = 8
Is this correct ?
rocky goblet
what's c?
dire umbra
f(x) = c = 1
rocky goblet
how do you know that f(x) = 1?
rocky goblet
that doesn't mean that f(x) = 1, it means that f(x) = y
dire umbra
Well than I do not know
dire umbra
okay
rocky goblet
so, use those values of a, b, c, in the quadratic formula
rocky goblet
well that's fine
we're trying to find the inverse of f
so we want to find x, given f(x)
and that's exactly what we've done
rocky goblet
um
we can though
if we have f(x), we can put that in, and then it's all numbers and we can work out what x is
dire umbra
but we dont have f(x)
rocky goblet
yes we do
because we're trying to find the inverse of f
which means you're given f(x) and have to find x
restive ridge
We only put the numbers in the quadratic formula
f(x) is a number
dire umbra
but what is the constant
restive ridge
-f(x)
dire umbra
I thought it was 1
restive ridge
can't tell if you're trolling
dire umbra
yeah okay
restive ridge
c = -f(x)
dire umbra
okay yeah
but what number does it equal to dont we need a number to solve the quadratic formula
restive ridge
we don't need numbers to "solve" the quadratic formula
dire umbra
How is it gonna be solved than?
restive ridge
you're asking "f(x) = 2x, how do I solve this? don't we need a number for x?"
restive ridge
c = -f(x)
dire umbra
if you do 2 * f(x) = 2f(x)
restive ridge
2 * f(x) is literally 2f(x)
dire umbra
yes
restive ridge
I give up, someone else help this dude
dire umbra
I dont get you if have 2f(x) how are you gonna solve for x?
x = ( - 2 + root(2^2 -4* 1 * - f(x) ) ) / ( 2 * 1 ) =
youre gonna get stuck if you dont know what number f(x) is
restive ridge
f(x) = something with x
that is the function of x
x = something with f(x)
that is the inverse function of the function
dire umbra
yeah okay I get that
restive ridge
"f(x) = 2x, you're gonna get stuck if you don't know what number x is"
dire umbra
that is the inverse function of the function
Wel when you put it like this I get what you mean
I think I misunderstood you
So okay after you put everyhting in this formula
x = ( - 2 + root(2^2 -4* 1 * - f(x) ) ) / ( 2 * 1 )
x = ( - 2 + root(4- 4f(x) ) ) / 2 =
What happens after this
rocky goblet
nothing does
that's the answer
dire umbra
I meant completely
rocky goblet
you can't "completely" solve it because there are multiple possible values of x
dire umbra
yeah thats what I mean that f(x) isnt a number
I follow you now
can I divide by 2?
for the denominator and numerator
rocky goblet
oh yeah actually i guess you can do that
dire umbra
x = - 1 + root(2 -2f(x)
rocky goblet
you can then turn that into something about the numbers within the root, but you need to be careful
dire umbra
yeah that was wrong
dire umbra
x = - 1 + root(2f(x) =
this should be right
dire umbra
but how can you divide 4f(x) than
rocky goblet
well we're dividing it by 4
dire umbra
why 4?
rocky goblet
2*2
dire umbra
from where does the 2nd 2 come from?
rocky goblet
because it's the square root
sqrt(x)/y = sqrt(x/y^2)
because (sqrt(x)/y)^2 = sqrt(x)^2/y^2 = x/y^2
dire umbra
Let me write it out
x = ( - 2 + root(4- 4f(x) ) ) / 2 =
x = (( - 2 + root(4- 4f(x) ) )) / 2 / 2/2 =
x = - 2 + root(4 ) root(- 4f(x) ) / 2 =
- 1 + 2/2 root(- 4f(x) ) / 2 =
1 + (root(- 4f(x) ) /2 ) =
Is this right till the end?
rocky goblet
ok wait hang on
x = ( - 2 + root(4- 4f(x) ) ) / 2 =
this isn't actually quite right, it should be 4 + 4f(x)
since it's -4ac and c is -f(x) so double negative
dire umbra
ahh yeah so it becomes positive
rocky goblet
aside from that: you're correct that you can divide by 2/2 (since that's just 1), but between the second and third steps you've changed a lot of stuff and i can't really understand what most of it is supposed to mean
why is root(4-4f(x)) equal to root(4)root(-4f(x))? 
dire umbra
well I took them apart?
rocky goblet
you can't do that though
dire umbra
isnt that possible?
rocky goblet
you can't take apart sqrt(a+b)
rocky goblet
think about, like, sqrt(2)
that's irrational
but it's also sqrt(1+1)
so yeah, there isn't really a generally useful thing you can do with sqrt(a+b)
only sqrt(ab), which is sqrt(a)sqrt(b)
what you can do with sqrt(4+4f(x)), is take out a factor of 4
4+4f(x) = 4(1+f(x))
rocky goblet
yep
2sqrt(1+f(x))
so now we have x = (-2 + 2sqrt(1+f(x))) / 2
can you see anything else we can do from there?
dire umbra
-1 + (2sqrt(1+f(x))) / 2 )
rocky goblet
yep, you can do that
anything else? 
dire umbra
and than
-1 + (2 * (1+f(x)) ^ 1/2 / 2 )
after that
-1 + 2/2 * (1+f(x)) ^ 1/2 / 2 )
-1 + 1 * (1+f(x)) ^ 1/2 / 2 )
-1 + (1+f(x)) ^ 1/2 / 2 )
and from here I dont know
rocky goblet
-1 + 2/2 * (1+f(x)) ^ 1/2 / 2 )
how did 2 turn into 2/2?
dire umbra
I just divided it by 2
rocky goblet
why?
dire umbra
-1 + (2sqrt(1+f(x))) / 2 )
rocky goblet
you didn't remove division by 2 from anywhere else
dire umbra
-1 + (2sqrt(1+f(x))) / 2 )
here the whole sqrt was gonna be divided by 2
after it became this -1 + (2 * (1+f(x)) ^ 1/2 / 2 )
I just wrote the 2 that was on the end with other 2 because that also needs to be divided
rocky goblet
...when you multiply two things and then divide it by something you don't divide both things
otherwise you'd have 2 = (2 * 2) / 2 = 2/2 * 2/2 = 1 * 1 = 1
dire umbra
first it was this
if you have this 2sqrt(3) / 2
and than you make it 2 * 3^ 1/2 /2
you still need to divide the 2 by 2 and 3 ^ 1/2 by 2?
rocky goblet
no
rocky goblet
otherwise you'd have 2 = (2 * 2) / 2 = 2/2 * 2/2 = 1 * 1 = 1
.
2sqrt(3) / 2 is just sqrt(3)
(xy)/z = (x/z)y = x(y/z)
division is the same as multiplication, it's just backwards
so dividing by 2 is the same thing as multiplying by 1/2
dire umbra
so the 2 just goes away
dire umbra
but the number in front of the sqrt
needs to be the same as the one you divide with or not?
rocky goblet
well if you want to just get rid of it, then yes
dire umbra
2sqrt(1+f(x))/3 = sqrt(1+f(x))
rocky goblet
try thinking about a concrete example and see what happens
dire umbra
2sqrt(1+f(x))/3 = sqrt(1+f(x))
like this is imppossible
rocky goblet
is 2sqrt(9) / 2 equal to sqrt(9)?
is 3sqrt(9) / 2 equal to sqrt(9)?
rocky goblet
3sqrt(1+f(x))/3 = sqrt(1+f(x))
that works
again, think about an example and see if it works
is 3sqrt(4)/3 equal to sqrt(4)
dire umbra
no? its 2
rocky goblet
...and what's sqrt(4)
dire umbra
2
rocky goblet
...so they are equal then
rocky goblet
yep
dire umbra
but because it was given that [0 , inf>
only -1 + sqrt(1+f(x)) is right
Or am I wrong
rocky goblet
well -1 - sqrt(1+f(x)) is probably going to be something negative
like if f(x) = 24, you get -1 - sqrt(25) which is -6
so yeah we want the -1 + sqrt(1+f(x))
dire umbra
Okay one more thing
I thougt for the inverse
you just swap y and x and than solve for y
why wasnt it possible in this case?
dire umbra
really? x = y^2 + 2y doesnt seem the same
rocky goblet
i mean it's the same equation, just different letters
it doesn't matter what symbol you give to each thing, if you're consistent with it you'll always get the exact same results
dire umbra
Well I tried with that and it didnt work for me
x = y^2 + 2y
sqrt(x) = sqrt(y + 2y)
sqrt(x) = sqrt(3y)
rocky goblet
x = y^2 + 2y
y^2 + 2y - x = 0
y = (-2 +- sqrt(2^2 - 4*1*-x))/2
...
dire umbra
nevermind it seems I did it wrong
rocky goblet
you turned y^2 + 2y into y + 2y
dire umbra
its sqrt(x) = y(sqrt(2y)
rocky goblet
that's not correct either
dire umbra
owh
rocky goblet
square roots just don't like addition
it's sqrt(x) = sqrt(y^2 + 2y) and there is no way to simplify that
square roots only behave nicely with multiplication
dire umbra
it's sqrt(x) = sqrt(y^2 + 2y) and there is no way to simplify that
So this is the final form
rocky goblet
Well I tried with that and it didnt work for me
x = y^2 + 2y
sqrt(x) = sqrt(y ...
in this case, that was the wrong way to do it anyway, you're trying to get y in terms of x so you want the quadratic formula
y^2 + 2y - x = 0, then solve from there
dire umbra
x = y^2 + 2y
y^2 + 2y - x = 0
y = (-2 +- sqrt(2^2 - 4\*1\*-x))/2
...
how did you go from 2 to 3rd step? how did you add the 2y with the y^2?
rocky goblet
from second to third step? -b +- sqrt(b^2 - 4ac) / 2a
dire umbra
ah so the quadratic formula
rocky goblet
yep
rocky goblet
i don't really know how you'd solve this without the quadratic formula
other than using some other tactic for solving quadratics
the problem is to solve a quadratic so
dire umbra
I understand so whenever you got a function with a square and another unknows so x^2 + x or y^2 + y you always quadratic formula?
rocky goblet
there might sometimes be a quicker way, but the quadratic formula will always work for solving quadratics
rocky goblet
no those aren't quadratics
dire umbra
but just replace the ^2 in thhe formula with ^3
rocky goblet
that doesn't work
for ^3 and ^4 there are equations that do it but they're a lot more complicated
for ^5 and above it's completely impossible, just roots isn't enough
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
robust field
HI i really need help with this i dont 100%5 understand and the answer i got was c but i could be wrong
obtuse pebbleBOT
HI i really need help with this i dont 100%5 understand and the answer i got was...
verbal prawn
you said you would expand
Bruh I did.
bold bane
@robust field The average rate of change is the change in y divided by the change in x. In this case, the temperature is the y value and time is the x value.
$\frac{\Delta y}{\Delta x} = \frac{\Delta Temperature}{\Delta time}$
warm shaleBOT
Kookiemon
gritty coral
Hello there, I was working on this Bernoulli IVP ODE and I am not sure where I messed up in regards to my work on it
obtuse pebbleBOT
Hello there, I was working on this Bernoulli IVP ODE and I am not sure where I m...
gritty coral
This is my work (I apologize in advance for the poor handwriting)
This is what Wolfram said I should have gotten, but I wasn't able to follow along with the solution it gave:
obtuse pebbleBOT
@gritty coral Has your question been resolved?
tranquil arch
$$\frac{dy}{y^3+2y}=dt$$
warm shaleBOT
秋水
obtuse pebbleBOT
@gritty coral Has your question been resolved?
gritty coral
Thank you, I'll try the problem by separating the terms so that all of the terms with y are on the left hand side
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
obtuse acorn
How to do 4 and 5?
obtuse pebbleBOT
How to do 4 and 5?
warm shaleBOT
What the hell am I doing here?
obtuse acorn
ok thx
obtuse acorn
For common base.
Where did i go wrong here?
obtuse acorn
Why don't you take all the common log base to one side of the equation?
Idk i just learnt this topic and im confused af
hexed agate
Where did i go wrong here?
Ok your understanding is good but how did log2 2x^2 become 2 log2 2x?
obtuse acorn
Ok your understanding is good but how did log2 2x^2 become 2 log2 2x?
power law
hexed agate
If it was log2 4x^2 it would have worked
obtuse acorn
like shift it from right to left
hexed agate
The exponent 2 is only on the x, not 2
obtuse acorn
but why cant it work in this case
hexed agate
The exponent 2 is only on the x, not 2
.
hexed agate
Power
obtuse acorn
so if the power is on the log then it works
hexed agate
(2x)^2 =/= 2x^2
obtuse acorn
but if its only on x it doesnt work
hexed agate
(2x)^2 =/= 2x^2
.
obtuse acorn
(2x)^2 =/= 2x^2
so the left equation is what im doing wrongly rn
hexed agate
Let's do it step by step ok?
obtuse acorn
sure
obtuse acorn
i shifted -log2 x to the right
so that i can merge it with log2 2x using product law
obtuse acorn
yup
hexed agate
Yeah there's the mistake
See
If it was something like log (2*5^2) can you move the power in front?
obtuse acorn
ye
obtuse acorn
wait what does the * mean
hexed agate
Multiplication
hexed agate
It's log2 2x^2
If you were to move at the front, it had to be log2 (2x)^2
2x as a whole
Not just x
hexed agate
So just leave it as log2 2x^2
obtuse acorn
okie
hexed agate
The RHS you can make it (1-x)^2
obtuse acorn
ok
hexed agate
So log2 ((1-x)^2/2x^2) = 3
obtuse acorn
ok lemme try
obtuse acorn
So log2 ((1-x)^2/2x^2) = 3
so from here i can just revert back to index form?
hexed agate
Ye sir
obtuse acorn
2^3= ((1-x)^2/2x^2)
hexed agate
Yes sir
obtuse acorn
thx!
hexed agate
It was just that one small mistake
obtuse acorn
It was just that one small mistake
thats how i lose marks in exams -_-
swift dirge
Hello, Im solving past questions from a competition and need clarification to the answer of this question
obtuse pebbleBOT
Hello, Im solving past questions from a competition and need clarification to th...
swift dirge
earnest elk
What do you need clarification with?
swift dirge
the answer
earnest elk
What is the given answer?
swift dirge
oh
earnest elk
Try using Venn diagrams or try letting the number of students who like both the sports m.
timid silo
i guess the question is incomplete
timid silo
because you see there's no information about total students or the number of students who like both of the sports
earnest elk
Correct me if I am wrong but to have the maximum number of total students, you have to have the minimum possible number of students who like both
The minimum would be 0, this will give a max value too
zenith raft
yea I think the question makes sense
earnest elk
Which competition is this though?
zenith raft
you can consider any number of students and any number of them liking both sports
swift dirge
Which competition is this though?
national olympiad '21
earnest elk
Which country Olympiad?
swift dirge
bgd
earnest elk
Correct me if I am wrong but to have the maximum number of total students, you h...
Do you understand now?
earnest elk
national olympiad '21
Which grade level?
swift dirge
Which grade level?
higher secondary
swift dirge
Do you understand now?
im not sure
earnest elk
Okay so can you tell me what part of the question you need clarification with?
obtuse pebbleBOT
@swift dirge Has your question been resolved?
sand hare
Hello. I have a question about Hamiltion cycle. Let's say I have a cycle like (1,2,3,4,1). What is its length? Is it 4 or 5? I don't know if we count a repeating digit
obtuse pebbleBOT
Hello. I have a question about Hamiltion cycle. Let's say I have a cycle like (1...
main cedar
for any cycles (including Hamiltonian cycles) the length is referring to the number of unique vertices
so it would be 4 in your case
peak hamlet
Q12
When I solve for m using the quad formula I get an non real number
I can’t tell whether this is correct as the textbook I am using doesn’t have an answer for this question
visual onyx
I can’t tell whether this is correct as the textbook I am using doesn’t have an ...
non-real means that it doesn't exist
peak hamlet
Yeah it’s imaginary
obtuse pebbleBOT
timid silo
Did I just delete the channel?
no
visual onyx
it can still work for a while ig
peak hamlet
Alright
peak hamlet
Ok
visual onyx
maybe take a good use of the discriminant
timid silo
open another one might help
timid silo
it still didn't get closed smh
visual onyx
.close
timid silo
ig it will be better if we just discuss it here if it gets closed
peak hamlet
Alright
warm shaleBOT
d^infinity sin(x) / d^infinity x
peak hamlet
Yes
timid silo
hmm ok
obtuse pebbleBOT
Available help channel!
gentle flower
hiya
peak hamlet
Hey
gentle flower
whats the problem :D
summer spire
do you have a question?
peak hamlet
Yes but it’s above, I accidentally deleted my original post so this channel got closed
peak hamlet
.
timid silo
well use the discrimant
summer spire
^
timid silo
Alright
for the LHS to equal something [only talking about reals here]
4(193-c)=<1
here c is the value we want it to equal
peak hamlet
for the LHS to equal something [only talking about reals here]
4(193-c)=<1
h...
How did u get 4(193-c)=<1?
timid silo
How did u get 4(193-c)=<1?
discrimant
warm shaleBOT
timid silo
m^2-14m+193=c
=> m^2-14m+(193-c) =0
Dis=1-4(193-c)
which can't be negative
hence 4(193-c)<=1
peak hamlet
discrimant
Wouldn’t that be (-14)^2-4(1)(193)?
timid silo
Wouldn’t that be (-14)^2-4(1)(193)?
I just realized I messed up
peak hamlet
Ohh
peak hamlet
I’m not sure how you did that
Why is 14^2 on the rhs? And is (4(193-c)) meant to be the discriminat?
If anything wouldn’t it be (discriminant < 0)
strong sinew
ig it's better to use the fact the parabola opens upwards
so root2>anything > root1
would be valid value for m
strong sinew
solve root1 and root2
and u will notice that the parabola actually never reaches 0 in reals
strong sinew
Where is c coming from?
in the discriminant method?
peak hamlet
Okok c = 193
warm shaleBOT
idk7
peak hamlet
HOLY SHIT
IM AN IDIOT
THERE IS NO X
HOW AM I JUAT REALSING THIS
Roots are both = i
Alright thanks for the help I’m an idiot
.close
cloud dove
$$\lim{t \to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}$$
obtuse pebbleBOT
$$\lim{t \to 0} \frac{1}{t\sqrt{1+t}} - \frac{1}{t}$$
warm shaleBOT
meek
royal basin
$\lim_{t \to 0}$
warm shaleBOT
Ann
cloud dove
this is better tbh
royal basin
$\lim_{t \to 0} \left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)$
warm shaleBOT
Ann
royal basin
there, got the whole thing
cloud dove
I got it simplified to (1-sqrt(1+t))/(t * sqrt (1+t))
And
I'm not entirely sure how to get any non undefined value
Because i graphed it to make sure that it's possible and it does approach a value at 0
royal basin
multiply by the conjugate now
timid silo
multiply up and down by a conjugate of sorts 
cloud dove
OH GREAT
Got the answer finally
cuz then it became (-t)/(t^2+t+tsqrt(t+1))
factored out the t
-1/t+1+sqrt(t+1)
then just use the direct substitution prop. by substituting c, which is 0, for t
and i get -1/2
thanks for the advice
didn't consider using the conjugate!
.close
quiet drum
obtuse pebbleBOT
quiet drum
are you allowed to cancel out if its from the same side??
idk how they cancelled out the srqrt3's
somber mural
just carry out the multiplication on the top
why do you need to multiply the bottom
quiet drum
wtv u do to the top u do to the bottom?
like its kinda weird they multiplied only one side with 3
somber mural
they haven't
quiet drum
where does the x3 come in
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
late tide
obtuse pebbleBOT
late tide
can someone help please?
royal basin
P is an odd function and P(2) = 5. What is P(-2)?
?
royal basin
i am rephrasing the problem in a way that is hopefully more obvious
unless you happen to not know what an odd function or an odd polynomial is, in which case you could've and should've said so explicitly along with the question.
late tide
i still don’t know sorry
royal basin
do you know what an odd function is? Y/N
late tide
not really
royal basin
right
late tide
i know f(-x) - f(x) is that related
royal basin
but just for the future, if there's a word or phrase in the problem that eludes your understanding
then you really really really should say that right away
royal basin
it's f(-x) = -f(x). that's the definition of an odd function. (or more specifically it's the statement "f is an odd function" so you will have to adjust it to have the name match up)
timid silo
use the remainder therorm which states that when p(x)[some polynomial] is divided by x-a the remainder is p(a)
and the fact that p(-x)=-p(x)
if
p is an odd function
royal basin
use the remainder therorm which states that when p(x)[some polynomial] is divided by x-a the remainder is p(a)
and the fact that p(-x)=-p(x)
i already said this
in my rephrasing of the problem
timid silo
oh
late tide
so is it just 5?
royal basin
no, it's not 5.
timid silo
I don't think u said about remainder theorom
late tide
OH wait -5?
timid silo
OH wait -5?
ye
late tide
ahh ok i see
timid silo
coz p(-2)=-p(2)=-5
mighty ether
can someone explain what the hell is going on at IV?
obtuse pebbleBOT
can someone explain what the hell is going on at IV?
mighty ether
where the heck did they get that matrix from
royal basin
have you already done part (iii)?
royal basin
and do i understand correctly that you are confused as to why $S^n = MD^nM^{-1}$?
warm shaleBOT
Ann
mighty ether
um i just dont know where they got that matrix from on the right
mine looks completely different but yes i also dont fully understand
royal basin
they did the calculations as they wrote
mighty ether
why s^n = all the stuff on the right although i do get diagonalisation
royal basin
ok so you have $S = MDM^{-1}$, yes?
warm shaleBOT
Ann
royal basin
you found M and D such that this works
royal basin
...
mighty ether
okay my inverse might be wrong but I used that method where you swap the top left and bottom right entries
and divide the whole thing by the determinant etc
my friend also did it though so ill send both inverses
this was my inverse for S
mighty ether
1/2 -3/2
-3/2 1/2
and this was my friensd
this is everything that she's done
yeah i suppose, I've barely done this type of problem so i'm really wading through it atm
royal basin
your friend never actually wrote down what she had for S^-1. at least not explicitly.
mighty ether
i think she just wrote it as m^-1
royal basin
no
mighty ether
royal basin
M^-1 is M^-1 and S^-1 is S^-1.
mighty ether
she said this is her inverse
royal basin
while the two matrices are somewhat closely related, they must not be confused for each other.
yes, this is the inverse of M.
it is NOT the inverse of S.
mighty ether
yeah i see what you're saying actually
royal basin
,w {{1/4, 3/4},{3/4,1/4}} * {{1/2, -3/2},{-3/2, 1/2}}
warm shaleBOT
royal basin
okay so it looks like you made a sign error in calculating your inverse
and you ended up with -1 times what it should be
mighty ether
but when we're diagonalising , aren't we supposed to find the inverse of our vector matrix?
d = x ^-1 A x
royal basin
hhhhhhhhhhh
mighty ether
where x is the eigenvector matrix?
royal basin
jsghsjdfhkjkhjh.d;hd;.h;.;l;;l;
sorry im trying to talk about two different things at the same time
one is the computational level, i.e. the equivalent of arithmetic mistakes
the other is the conceptual level, i.e. WHAT you were supposed to calculate in the first place
mighty ether
okay got it
royal basin
and to be