seems like yours is a good use case for this ^
#help-13
tropic oxide
high thicket
Thanks! Looking at their courses, would the following do what I want without leaving any major gaps?
Arithmetic
Algebra 1 & 2
Trigonometry
High School Geometry
High School Statistics
Pre-Calculus
Which one would cover mathematical language and formal proofs?
.close
olive epoch
How do i get the surface area?
cedar kilnBOT
How do i get the surface area?
dim sonnet
How do i get the surface area?
pretty sure you have to use integrals for this...
Haven't been into Math for a long time, but this is what i would look for.
olive epoch
Yes u r right. But idk how tbh
cedar kilnBOT
@olive epoch Has your question been resolved?
dim sonnet
Yes u r right. But idk how tbh
midnight void
cedar kilnBOT
midnight void
Because the events are independent it can be formulized as 1/2^5
which is 1/32 but it's telling me that's incorrect
what am I doing wrong
peak minnow
2^5 is the total number of events
and b is asking probability of getting either 5 heads or 5 tails which is 1 each
so 2/32 or 1/16
midnight void
why 2/32 instead of 1/32
oh wait
I understand
because its head or tails,
odds of getting 5 heads is 1/32
odds of getting 5 tails is 1/32
added to get 2/32
I misunderstood
peak minnow
yes
peak minnow
np
smoky sphinx
cedar kilnBOT
smoky sphinx
Idk how to do any of that
crimson sedge
try drawing it out
cedar kilnBOT
@smoky sphinx Has your question been resolved?
crimson delta
ignoring some cases, lcm(a,b) is always at least 2a or 2b
cedar kilnBOT
@brave bough Has your question been resolved?
crimson delta
yes
brave bough
ok thanks
rich talon
i just dont get where the -14 and 3 is coming from. i understand everything else in the problem solution except this.
rich talon
7x = 7(x+2-2) = 7(x+2) - 7\*2
where did u get the -2 in (x+2-2)? ik its centered at x = -2 which = (x+2)
tropic oxide
yes
fast marsh
Need Help
fast marsh
tender vapor
for a slope less than 0, should the y value rise or fall as x increases?
fast marsh
Fall
tender vapor
so immediately you can eliminate the first two
now, what is the y intercept of the line?
for a line of the form y=mx+b, m is the slope and b is the y intercept
fast marsh
I was just typing, sorry.
tender vapor
nw
fast marsh
The Answer is C
Just needed help with elemnating the first two.
Was C correct?
Sorry, sibling got on my phone.
@tender vapor
cedar kilnBOT
@fast marsh Has your question been resolved?
cursive vale
Hi I need help with basically most of a test and I was wondering if I could get help here because if I fail the test then I’ll fail the class as a whole
livid hound
we can help prepare/ review
but not on the actual test
as that would be academic dishonesty
umbral mirage
I divided by x^4, and used l Hopitals rule but I dont know what do after that
strange mauve
What did you get
mental trail
no need for hopital's rule I think
you can just plug in the values after dividing by x^4
cedar kilnBOT
@umbral mirage Has your question been resolved?
cedar kilnBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
tropic oxide
we don't give answers here
rigid lantern
cedar kilnBOT
rigid lantern
help me i have to differienatiate
runic garnet
Which one
rigid lantern
walk me through all
runic garnet
Nah I’ll help w one
rigid lantern
a) i did already check its answer later but lets do b
rigid lantern
find the derivatives
runic garnet
Yea what did u get
rigid lantern
F'(x) = 1 / square root x - 1 over 3 squareroot x^2
do u do quotient rule square thr bottom and apply product rule on the top
@runic garnet
help
runic garnet
F'(x) = 1 / square root x - 1 over 3 squareroot x^2
U mean -1 over 3rd root of x^2?
rigid lantern
ye
runic garnet
Ur answer is correct
rigid lantern
bro real with u i just need to be walked through im lost
rigid lantern
Ur answer is correct
thats final?
rigid lantern
runic garnet
,rccw
wraith daggerBOT
cedar kilnBOT
@rigid lantern Has your question been resolved?
cedar kilnBOT
@rigid lantern Has your question been resolved?
rigid lantern
no mans still need help
cedar kilnBOT
@rigid lantern Has your question been resolved?
rigid lantern
<@&286206848099549185> b and d need help
cedar kilnBOT
@rigid lantern Has your question been resolved?
cedar kilnBOT
@rigid lantern Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
tidal junco
tell me if its possible to calculate the area please
cedar kilnBOT
tell me if its possible to calculate the area please
tidal junco
my teacher tells me im idiot because i know math and cant calculate it
tidal junco
tell me if its possible to calculate the area please
we also use sinuses and stuff to calculate area if this helps
primal socket
tell me if its possible to calculate the area please
What's L?
primal socket
Perimeter
tidal junco
thanks
primal socket
Circumference is that of a circle
tidal junco
we say "obwód" for both and use either L or Obw ( for circle we use small l as far as i remember )
tidal junco
tell me if its possible to calculate the area please
anyway, is this possible
hearty arch
yes
you should have three equations to work with, the formula for area, which should be expressed only using b; the formula for perimeter, which you should use to make a substitution into the last formula; the Pythagorean theorem
cedar kilnBOT
@tidal junco Has your question been resolved?
tidal junco
you should have three equations to work with, the formula for area, which should...
can you at least tell me how would first step look like
it looks like that for me
P is area
hearty arch
P is typically used for perimeter, whereas A is typically used for area
that being said, (b^2)/2 - b is spot on for area
tidal junco
P is typically used for perimeter, whereas A is typically used for area
im polish and we use P ("pole") for area
tidal junco
and we dont say "tan" we use "tg" we dont use "cot" we use "ctg"
ok thanks for help
.close
frozen valve
Assume a permutation is provided in a cycle form of $σ = (i_1 … i_k) (j_1 … j_s) … (r_1 … r_l)$
Prove that $σ^-1 = (i_k … i_1) (j_s … j_1) … (r_l … r_1)$
wraith daggerBOT
Davit
frozen valve
have you tried anything
Dont even know where to begin
dawn junco
well you know what permutations are at least right ?
frozen valve
well you know what permutations are at least right ?
Of course
dawn junco
aight so you want to show that sigma^-1 = ....
so that involves showing that sigma * sigma^-1 = sigma^-1 * sigma = identity
just a question, the i, j, k, they're all disjoint or no ?
@frozen valve
frozen valve
sorry for not responding
frozen valve
so that involves showing that sigma * sigma^-1 = sigma^-1 * sigma = identity
yeah right
frozen valve
just a question, the i, j, k, they're all disjoint or no ?
yeah right they are disjoint
given that htey are disjoint, we can consider each property one-by-one right
so prove (i1 .. ij) seperately of (r1 ... rs) and so on
dawn junco
as in, the product of the two [the i's from sigma and the i's from sigma^-1] gives the identity
frozen valve
hmm
so how would i rigorously write this
step 1: given the set is disjoint, we look through each cycle individually
step 2: prove that i * i^-1 yields the identity (epsilion) function
how would i do 2
well i can just say its trivial right
dawn junco
it's not hard to see why yes
winged forum
Anyone can help with q4
cedar kilnBOT
Anyone can help with q4
winged forum
@uncut veldt
cedar kilnBOT
@winged forum Has your question been resolved?
winged forum
<@&286206848099549185>
cedar kilnBOT
@winged forum Has your question been resolved?
velvet mortar
,rotate
wraith daggerBOT
kind sparrow
cedar kilnBOT
kind sparrow
Sorry i do not understand the error
could someone explain it??
because of the squareing in the formula it will always be positive right??
I also tried y(x)=1-(((x-4)^2)/16)
crimson delta
you are writing y(x)
kind sparrow
yeah i noticed, changing it to y_x fixed it
but i did no realy know what they ment with y - on top
notation wise
resolved btw
so this channel is open again
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
daring flame
Hello all, I had a question for solving matrix using gauss elim. I am looking at the textbook answer, and they are dividing the first row by three. When they follow this process, the answer is completely different from mine.
My approach was to swap R3 and R1, but for some reason that gives me a different answer. Could someone help me in understanding when it is appropriate to switch rows? I thought the first row must have a leading 1, so either dividing 3 or interchanging rows would work.
plucky owl
Hello all, I had a question for solving matrix using gauss elim. I am looking at...
It doesn't matter what is done as long as the algebra is correct. In the end, you'll get the same answer
daring flame
I see. I must have messed up my algebra somewhere
Thats good to know then, I thought I was going crazy
plucky owl
So if you swap the rows or divide the row by 3, you'll get up with the same solution
cedar kilnBOT
cedar kilnBOT
I see. I must have messed up my algebra somewhere
Show your work, and if possible, explain where you are stuck.
Available help channel!
oblique prawn
cedar kilnBOT
oblique prawn
is the answer i got at the bottom the net force of the system?
lunar lynx
What is the system? Basically, whenever talking about a system in physics, you explicitly have to mention what is in your system.
cedar kilnBOT
@oblique prawn Has your question been resolved?
junior pebble
cedar kilnBOT
junior pebble
Task: describe the line and surface marked in red
i know partially how to describe a
Its x=3
But i dont know how to describe it starting at y=0 and ending at y=2
<@&286206848099549185>
harsh lake
My best guess is 0 <= y <= 2
junior pebble
My best guess is 0 <= y <= 2
And for b?
harsh lake
And for b?
do you know the line for b?
harsh lake
do you know the line for b?
The line you are giving for the two graphs is the domain. and what I gave you was the range of the graph.
ivory finch
cedar kilnBOT
ivory finch
What is this summation with n + 1? is it just the same formula except with n + 2 on the RHS?
ebon pulsar
yes
cedar kilnBOT
@ivory finch Has your question been resolved?
wraith daggerBOT
riemann
fathom mural
cedar kilnBOT
cedar kilnBOT
@fathom mural Has your question been resolved?
vapid scaffold
hello
vapid scaffold
so I have a basic question
so lets say
I have 3^3
if I take the cube root of this I will be left with 3 correct?
ebon pulsar
yh
vapid scaffold
okay
now
if I had like
(3b+1)^3 = 3^3, would it be mathematically valid if I took the cube root of the entire equation to get rid of the exponents?
ebon pulsar
this means:
3b + 1 = 3
ebon pulsar
so b = 1.5
vapid scaffold
but this works out right? (only asking because ive only worked with cases where you just sqaure to get rid of ^2)
ebon pulsar
(x) ^ a = (b+c+d+e+f+g+h+j+k+l+o+l+..................................)^a
this means that:
x = b+c+d+e+f+g+h+j+k+l+o+l+................................
worn geode
How can I find X? A and B are given and all matrices are 3x3
tropic oxide
$A^T X B^{-1} = C$
wraith daggerBOT
Ann
tropic oxide
post-multiply both sides by B and pre-multiply both sides by (A^T)^-1
worn geode
post-multiply both sides by B and pre-multiply both sides by (A^T)^-1
So I get X = (A^T^-1)CB?
tropic oxide
indeed
worn geode
I probably messed up the arithmetic somewhere then
Can I ask another quick one or do I need to open another channel?
tropic oxide
if it's quick it's fine here
generally for different questions you close the first one and then open a different channel
worn geode
Does this look like a vector subspace of R^3?
worn geode
well it looks like it yes
That's the conclusion I got too but this was towards the end and it got...messy
Thanks for the help!
.close
rigid gulch
Can anyone check my work for this problem.
cedar kilnBOT
Can anyone check my work for this problem.
tropic tree
the answer is correct. didn't look through all of your steps though. the e7t give a 7 shift. the rest is linear coeff
cedar kilnBOT
@rigid gulch Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
muted patrol
The table displays the probabilities for each of the six outcomes when rolling a particular unfair die. Suppose that the die is rolled once. Let A be the event that the number rolled is less than 4, and let B be the event that the number rolled is odd. Find P(A ∪ B).
Outcome 1 2 3 4 5 6
Probability .1 .1 .1 .2 .2 .3
muted patrol
im getting two different answers and im not sure which one is correct
idle tusk
what answers are you getting, and how?
cedar kilnBOT
@muted patrol Has your question been resolved?
dire geode
im getting two different answers and im not sure which one is correct
!show
cedar kilnBOT
im getting two different answers and im not sure which one is correct
Show your work, and if possible, explain where you are stuck.
Available help channel!
turbid spoke
how i can sdolve that
cedar kilnBOT
how i can sdolve that
hazy compass
cedar kilnBOT
cedar kilnBOT
@hazy compass Has your question been resolved?
cedar kilnBOT
@hazy compass Has your question been resolved?
cedar kilnBOT
@hazy compass Has your question been resolved?
hazy compass
<@&286206848099549185>
orchid panther
@hazy compass BC and AD would be parallel because FD and BE are parallel
Furthermore since CE=AF and since BE and FD are both parallel youd have FB and DE parallel so BFDE is a parallelogram so BC=AD which means BA is parallel to CD
I think this may work
cedar kilnBOT
@hazy compass Has your question been resolved?
nova snow
B)
i. At the point $P(2,5)$ on the curve $y= ax^2 - 9x + b$ the tangent to the curve is perpendicular to the line $y=x$. Find the values of a and b.
wraith daggerBOT
water beam
nova snow
I'm not sure how to start this
gritty galleon
First find the equation for the tangent
green thorn
**water beam**
do you know how to find the slope of the tangent at a point
nova snow
First find the equation for the tangent
okay
gritty galleon
Can u do that?
tropic oxide
@gritty galleon difficult
gritty galleon
<@717652530061901855> difficult
My approach? That tends to be a thing with me 😅
tropic oxide
finding the eq for the tangent is in fact not necessary. we need to know only its slope (from perp to y=x) & that P lies on our curve -- that's our two eqs.
gritty galleon
I was thinking differentiate the quad. equation, set it equal to tangent equation, sub x=2, y=5
nova snow
so what should be the first step
gritty galleon
so what should be the first step
U should follow Ann's approach
tropic oxide
imo the first step is to write down the statement "(2,5) lies on y=ax^2-9x+b" as an eq in a and b
nova snow
okay
tropic oxide
then write down the second eq in a and b as "the tangent to y=ax^2-9x+b at x=2 has slope -1"
nova snow
wait where does x = 1 come from
nova snow
we know this has slope -1 because?
tropic oxide
do the things i said & show me the equations you wrote down.
nova snow
well i mean i just wrote down y=ax^2-9x+b has perpm = -1
nova snow
idk what youre asking i just copy pasted what you said into my paper
tropic oxide
the first step is to write down the statement "(2,5) lies on y=ax^2-9x+b" as an equation in a and b
fuck
nova snow
how do I do this as an equation in a and b
tropic oxide
what does it mean for the point (2,5) to lie on a curve described by an equation?
also why tf did you choose not to ask me the FIRST time i told you to do this.
nova snow
I dont know to both of those questions
tropic oxide
what does it mean for the point (2,5) to lie on a curve described by an equation...
it means that if you replace x with 2 and y with 5 the equation remains true
you want to review basic equation graphing
cause not knowing this is a serious serious gap
nova snow
well i know how to graph
tropic oxide
you don't know the basic principle of "point lies on curve <=> point satisfies equation of curve"
nova snow
hmph
tropic oxide
don't hmph me.
nova snow
hmm?*
nova snow
it means that if you replace x with 2 and y with 5 the equation remains true
but how do we know its "true" i mean you could make x and y to be anything
green thorn
but how do we know its "true" i mean you could make x and y to be anything
well each value of x corresponds to a value of y
so its not "any" pair of x and why, its any value of x and then a specific value of y
tropic oxide
but how do we know its "true" i mean you could make x and y to be anything
the curve by definition consists of all those points, and only those points, at which the equation is true
this is cartesian plane 101
nova snow
i havent learnt much of cartesian lore
only that
there is the x and y axis
left is negative
down negative
yknow
that stuff
tropic oxide
absolutely jack shit.
without knowing such basic stuff as what i said.
fuck.
and now it is biting you in the ass.
cedar kilnBOT
@nova snow Has your question been resolved?
wanton sail
well i mean i just wrote down y=ax^2-9x+b has perpm = -1
So do you know how to find the slope of a tangent line?
opaque pivot
B)
i. At the point $P(2,5)$ on the curve $y= ax^2 - 9x + b$ the tangent to the...
You can differentiate to find the slope at x = 1 in terms of a and b, and equate it to -1 since it's perpendicular to a line of slope 1. That's your first equation.
You can set x = 2 and y = 5 to get your second equation in a and b
You have 2 equations and 2 unknowns, so you can solve the simultaneous equations for a and b
cedar kilnBOT
cedar kilnBOT
Available help channel!
rustic steeple
cedar kilnBOT
low halo
I thought if those were intersecting points, it wouldn't matter which equation I chose. but that doesn't seem to be the case
crimson sedge
You just aggressively occupied his channel
low halo
You just aggressively occupied his channel
oh im so sorry. where would I send if I have a questio?
green thorn
oh im so sorry. where would I send if I have a questio?
try #help-2
crimson sedge
Yup
green thorn
its unoccupied
rustic steeple
ouch
rustic steeple
can anyone help with this 😭
crimson sedge
Can you take the photo right way
.
green thorn
,rotate
wraith daggerBOT
green thorn
bruh
wraith daggerBOT
大野雄大 👻
cedar kilnBOT
@rustic steeple Has your question been resolved?
cedar kilnBOT
@rustic steeple Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
rancid pollen
cedar kilnBOT
cedar kilnBOT
@rancid pollen Has your question been resolved?
rancid pollen
<@&286206848099549185>
cedar kilnBOT
@rancid pollen Has your question been resolved?
cedar kilnBOT
@rancid pollen Has your question been resolved?
sand cradle
Show that for $a = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$ and $a = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$, [\gcd(a, b) = p_1^{\min(a_1, b_1)} p_2^{\min(a_2, b_2)} \cdots p_n^{\min(a_n, b_n)}.]
\begin{proof} Let $d$ be an arbitrary divisor of $a$ and $b$. For this to be the case, we need $\nu_{p_i}(d) \leq \nu_{p_i}(a)$ and $\nu_{p_i}(d) \leq \nu_{p_i}(b)$ for all primes $p_i$, where $\nu_{p_i}(c)$ denotes the power of the $p_i$th prime in the prime factorization of $c$. \ To ensure both, we need $\nu_{p_i}(d) \leq \min(\nu_{p_i}(a), \nu_{p_i}(b))$. \Since we want the greatest common divisor, we should maximize $\nu_{p_i}(d)$ for all $p_i$: this is the case when $\nu_{p_i}(d) = \min(\nu_{p_i}(a), \nu_{p_i}(b))$. Thus, we obtain [\gcd(a, b) = p_1^{\min(a_1, b_1)} p_2^{\min(a_2, b_2)} \cdots p_n^{\min(a_n, b_n)}.] \end{proof}
mighty drift
You successfully argued that any divisor is a divisor of that (except for using actual word of division)
I'd just add a sentence to say this divides a and b, to say that the number we find is the gcd
Basically a return to the definition
sand cradle
You successfully argued that any divisor is a divisor of that (except for using ...
Isn't that included when I say "since we want the gcd, we should maximize v_p_i(d) for all p_i"?
sand cradle
well it does state that it's the greatest, doesn't it?
sand cradle
Only implicitly
oh, ok
sand cradle
Does that state you have a divisor? And that it's the greatest ?
How do we show it's really the greatest, then?
mighty drift
This isn't computer science, this is math. Give the correctness far more importance than the algorithm's intuition
mighty drift
How do we show it's really the greatest, then?
You already showed that any divisor has lower valuations. So it's very short to say that therefore any divisor divides that number
Then just show it's a divisor
sand cradle
Then just show it's a divisor
Do you mean show the gcd is a divisor?
mighty drift
Kinda important yeah
sand cradle
"Let d be an arbitrary divisor of a and b. For this to be the case ..."
We already constructed the gcd after the requirements for it to be a divisor of a and b
mighty drift
Yeah, so divisor => this form
Not that this form => divisor
sand cradle
Yeah, so divisor => this form
Not that this form => divisor
What do you mean?
mighty drift
Then just show it's a divisor
This
sand cradle
This
But
We already constructed the gcd after the requirements for it to be a divisor of a and b
mighty drift
You constructed what turns out to be the gcd
You didn't prove it's the gcd
sand cradle
Yeah, we didn't prove it's the gcd yet, but we do know that whatever we have must be a divisor of a and b, since it fulfills the requirements v_p_i(d) <= v_p_i(a) and v_p_i(d) <= v_p_i(b) for all p_i
sand cradle
You constructed what turns out to be the gcd
You didn't prove it's the gcd
We can show that all divisors of a and b in common also divide what we have, if that's what you mean
And from that, follow that it's really the gcd
mighty drift
And from that, follow that it's really the gcd
Yes
sand cradle
Yes
How'd we do that though? We'd need the common divisors of a and b in the first place
mighty drift
You have them
You just forgot to state that what you have are equivalences rather than implications
sand cradle
You have them
You just forgot to state that what you have are equivalences rathe...
Well, the common divisors are p_1^min(a_1, b_1), p_2^min(a_2, b_2), ...
And those are divisors of the gcd after Euclid's lemma.
So we are done?
mighty drift
Well, the common divisors are p_1^min(a_1, b_1), p_2^min(a_2, b_2), ...
And thos...
no that's the gcd, not any common divisor
just write it out properly once and for all
sand cradle
no that's the gcd, not any common divisor
So just p_1, p_2, ..., p_n?
With exponents up to min(a_i, b_i)
sand cradle
just write it out properly once and for all
Those are all divisors of gcd, since v_p_i(x) <= v_p_i(gcd) where x is any common divisor
mighty drift
so every divisor divides it, and it divides a and b, so it's the gcd
sand cradle
so every divisor divides it, and it divides a and b, so it's the gcd
Great, thanks.
For the v_p_i(x) <= v_p_i(gcd) for all p_i where x is a common divisor, what would be the justification? That just holds after our construction, right?
mighty drift
just write it out properly once and for all
it was already in your previous proof
cedar kilnBOT
cedar kilnBOT
Available help channel!
stark dirge
stark dirge
how do I do the multplication of all these logs?
I tried getting log 4 in base 4 from 3 to an exponential form
log 2^2 (3)
and then do 2 log 3
same with log 8 (7)
spice kraken
cedar kilnBOT
@stark dirge Has your question been resolved?
remote glacier
.reopen
timber stone
Not stricly maths, but can anyone have a look at a code thats supposed to be removing elements equal to k from a linked list?
fair geyser
are you like embarrassed?
you didn't show it, so it looks like you're embarrassed, so there's kinda no upside if looking embarrassed is embarrassing
atomic pecan
Y is correct but why can't I just put it back in any equation?
woeful cloak
atomic pecan
I need to find
x y z
But here z will be -192/4
Which is not correct
So I don't know what did I do wrong
I foundbit
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
golden finch
$\begin{pmatrix}
1 & -1 & 0 & 3\
0 & 3 & 1 & -5\
2 & 7 & 3 & -9
\end{pmatrix}$
wraith daggerBOT
Penca53
golden finch
Given the above matrix, determine, if there exists, a solution (x1, x2, x3, x4) for AX=0 perpendicular to (1, 0, 0, 0).
The Gauss reduced matrix is
$\begin{pmatrix}
1 & -1 & 0 & 3\
0 & 3 & 1 & -5\
0 & 0 & 0 & 0
\end{pmatrix}$
wraith daggerBOT
Penca53
golden finch
And this is the system of equations:
$\begin{cases} x_1 - x_2 + 3x_4 = 0 \ 3x_2 + x_3 - 5x_4 = 0\end{cases}$
wraith daggerBOT
Penca53
golden finch
Now I don't know how to proceed (finding all the solutions, and then finding a solution perpendicular with the vector above)
cedar kilnBOT
@golden finch Has your question been resolved?
cedar kilnBOT
@golden finch Has your question been resolved?
cedar kilnBOT
Available help channel!
opal jacinth
I need help solving this Permutations question
cedar kilnBOT
I need help solving this Permutations question
crimson sedge
I need help solving this Permutations question
how many ways can m men be arranged
opal jacinth
m! ways
opal jacinth
m places?
crimson sedge
no
opal jacinth
(m+1)!
crimson sedge
nop
opal jacinth
okay
crimson sedge
connect those two together
opal jacinth
wair
opal jacinth
oh
OH
m+1-n
OH
no
wait
im confused
m! ways for the men
(m+1) placws
for the wwomen
i dont get it
wait
wair
i think
i got it
its
crimson sedge
like u wanna assign those m+1 places to the n women
crimson sedge
yess
crimson sedge
no
crimson sedge
YEP
opal jacinth
thanks!
crimson sedge
ayee
opal jacinth
hehe
crimson sedge
good job
opal jacinth
.close
opal pebble
- Maximum Area A rectangle is bounded by the x-axis and
the semicircle
y = √25 − x2
(see figure). What length and width should the rectangle have
so that its area is a maximum? - Maximum Area Find the dimensions of the largest
rectangle that can be inscribed in a semicircle of radius r (see
Exercise 27).
idle tusk
i gotchu
idle tusk
can you derive the formula for area of a rectanlge inscribed in a semicircle?
hint: use coordinates and the equation for y
opal pebble
i dont know what formula to use
all i can think of is x^2 + y^2 = r^2 but idk how thats goona help here
opal pebble
is x0 the radius?
idle tusk
no
idle tusk
is x0 the radius?
it's the half-width of your rectangle
change it and you'll see different rectangles
opal pebble
but how does this help me solve it using maximums
cause i gotta find the biggest possible rectangle
idle tusk
cause i gotta find the biggest possible rectangle
well, let's find a formula for the area of that rectangle and then maximize it
suppose x_0 is the half-width of our rectangle (it's symmetric around y-axis)
then the width of the rectangle is 2x_0
and the height is y(x_0)
now, what's the general formula for an area of a rectangle?
idle tusk
yep
idle tusk
yes
or more formally 2x y(x)
now, substitute the definition for y there
and differentiate to find the maximum
opal pebble
whats the definition of y
idle tusk
problem 27
idle tusk
no but thats what i did for 27
broski
"(see Exercise 27)"
oh
radius r
nvm
still, it's similar
if you have x^2 + y^2 = r^2
y^2 = r^2 - x^2
you can from here find the expression for y
opal pebble
would i substitue that into 2xy?
idle tusk
$2x\sqrt{r^2 - x^2}$
wraith daggerBOT
artemetra
opal pebble
yeah that
idle tusk
correct
opal pebble
but when i take derivative idk what to do with r^2
opal pebble
OH
idle tusk
you are differentiating with respect to x
opal pebble
when i set it to 0
i got x =
(r^2 / 2 ) all square rooted
sqrt{r^2 / 2}
sqrt{r^2 / 2}$
idk how to use the bot
idle tusk
$\sqrt{\frac{r^2}{2}}$
wraith daggerBOT
artemetra
idle tusk
this?
idle tusk
sqrt{r^2 / 2}
this is good enough for me to understand btw
you don't have to use LaTeX
opal pebble
yes
and than i substitued that into the original y = sqrt(r^2-x^2)
and i got the same value for y
so im pretty sure the answer is $\sqrt{\frac{r^2}{2}}$ , $\sqrt{\frac{r^2}{2}}$
wraith daggerBOT
gzus
idle tusk
hm
opal pebble
ay there we go
idle tusk
let me check
wraith daggerBOT
artemetra
idle tusk
np
opal pebble
.close
green vapor
kinda stuck on the induction proof for the claim
Claim: Let $n\geq 2$, and let $A_1,A_2,\ldots A_n$ be sets from some universal set $U$. For all $n\geq 2$, $\overline{\bigcup\limits_{i=1}^{n}A_i}=\bigcap\limits_{j=1}^{n}\overline{A_j}$
wraith daggerBOT
42
cedar kilnBOT
@green vapor Has your question been resolved?
cedar kilnBOT
green vapor
.reopen
cedar kilnBOT
✅
green vapor
<@&286206848099549185>
cedar kilnBOT
cedar kilnBOT
Available help channel!
dapper nacelle
what kind of distribution is this shape?
cedar kilnBOT
what kind of distribution is this shape?
dapper nacelle
cedar kilnBOT
@dapper nacelle Has your question been resolved?
covert knot
Is this the area of a trapezoid? I have tried a ton of combinations of the area formula
covert knot
It looks like height is 15 base is 4 and the other base is 12 then I plug it in and get 120
zenith sail
The bases are the parallel sides
In this case the vertical sides
The height is the distance between them
The height doesn't necessarily have to be vertical, despite its name
covert knot
So 108?
3 is the base a 15 is base b and 12 is the height
Because the math software says I’m wrong
zenith sail
The bases are right, 3 and 15
The height is not 12 tho
The two bases are only separated by 4 units
tired inlet
Could someone help me differentiate this
cedar kilnBOT
Could someone help me differentiate this
tired inlet
$$\frac{d}{dx}\left(\frac{x^3-\pi }{x}\right)$$
wraith daggerBOT
Lex1729
tired inlet
Can I use quotient rule?
livid hound
you "could" but its a bit overkill
lethal ivy
Yeah.. think power rule heh
wraith daggerBOT
Lex1729
lethal ivy
yeah!
so now can you apply the power rule to those 2 things
(i would say you could)
tired inlet
The answer I get after differentiating is $$2x+\frac{\pi }{x^2}$$
wraith daggerBOT
Lex1729
tired inlet
However apparently it's wrong...
lethal ivy
well what is the answer? :))
dire geode
,w differentiate x^2 - pi/x
lethal ivy
haha
wraith daggerBOT
dire geode
However apparently it's wrong...
show why you think that's wrong
lethal ivy
well i think the ans is probs in a different form
tired inlet
well i think the ans is probs in a different form
Oh never mind yeah it was in another form
But why do I get a different answer when I use quotient rule?
If f = x^3 - pi
and g = x
ornate spindle
What's the different answer you get?
tired inlet
$$\frac{\left(x\right)\cdot \left(3x^2\right)-\left(x^3\right)\cdot \left(1\right)}{x^2}$$
wraith daggerBOT
Lex1729
dire geode
where did pi go
ornate spindle
You lost your pi somewhere along the way
tired inlet
$$\frac{\left(x\right)\cdot \left(3x^2\right)-\left(x^3-\pi \right)\cdot \left(1\right)}{x^2}$$
wraith daggerBOT
Lex1729
tired inlet
Alright I see now, thanks for clarifying.
lethal ivy
yayy
cedar kilnBOT
Available help channel!
tired inlet
Could someone help me differentiate this?
cedar kilnBOT
Could someone help me differentiate this?
tired inlet
$$\left(\cot \left(x\right)+x\right)^2e^{3x}$$
wraith daggerBOT
Lex1729
tired inlet
Here is my working so far:
f = (co(x) + x)^2
f' = 2(cot(x) + x) * (-cosec^2(x) + 1)
g = e^(3x)
g' = 3e^(3x)
Just not sure how to further simplify this
ocean mural
product rule?
ocean mural
oh ok
since its f'(x)g(x) + g'(x)f(x) one thing u can definitely factor out is the e^3x
tired inlet
$$\left(\cot \left(x\right)+x\right)^2\cdot \left(3e^{3x}\right)+\left(e^{3x}\right)\cdot 2\left(\cot \left(x\right)+x\right)\cdot \left(-\operatorname{cosec}^2\left(x\right)+1\right)$$
wraith daggerBOT
Lex1729
tired inlet
How exactly would I go about simplifying it though?
the right side of the equation to be specific
<@&286206848099549185>
ocean mural
maybe u can use the pythagorean identity
the third one
1 - csc^2 = cot^2
im not sure if u can get much simplet than that
cedar kilnBOT
@tired inlet Has your question been resolved?
livid hound
yes, but don't use x for multiplication
can be simplified further though through factorisation and aforementioned trig identities
tired inlet
Oh ok
Also one more thing
Is this correct?
Not entirely sure if It's correct.
<@&286206848099549185>
tender cypress
<@&286206848099549185>
lilac relic
uh
that one is kinda interesting
so i think we should find a pattern of it
so suppose there are n circles, and n is a even number
we have:
area = (area_n - area_n-1) + (area_n-2 - area_n-3) + ... + (area_2 - area_1)
i mean even area
no
circle area
and then since the radius is increasing by 1
we have:
area = (pin^2 - pi(n-1)^2) + ... + (pi2^2 - pi1^2)
since area = 2023pi, we have
2023pi = (pi * n^2 - pi * (n-1)^2) + ... + (pi * 2^2 - pi * 1^2)
devide everything by pi to make it easier to understand, we got
2023 = n^2 - (n-1)^2 + (n-2)^2 - (n-3)^2 + ... + 2^2 - 1^2
simplify:
2023 = n^2 - (n^2 - 2n + 1) + (n^2 - 4n + 4) - (n^2 - 6n + 9) + ... + 4 - 1
2023 = n^2 - n^2 + 2n - 1 + n^2 - 4n + 4 - n^2 + 6n - 9 + ... + 4 - 1
we found that most of the terms will be eliminated
for the term n^2, it will basically eliminated.
for the term 2n,4n,6n,8n,..., only 4n will left, others eliminated
so
2023 = 4n - 1 + 4 - 9 + 16 - 25 + 36 - ...
and the last term should be a negative number
ok i gave up
hope someone else can clean up the mess
cedar kilnBOT
@tender cypress Has your question been resolved?
sage bobcat
cedar kilnBOT
sage bobcat
Dont know what went wrong here
cedar kilnBOT
@sage bobcat Has your question been resolved?
sage bobcat
<@&286206848099549185>
green thorn
think i might need a higher res pic
orchid inlet
green thorn
wow good job reading that
yeah you cant flip a sum of fractions
you have to make them into one fraction then flip
green thorn
My algebra skills are so bad wth
youre fine
orchid inlet
orchid inlet
How did I mess up the algebra again 😭
IDK IF YOU USED PYTHAGORAS OR EQUATION OF A CIRCLE, BUT EITHER WAY 10 IS SUPPOSED TO BE SQUARED THERE.
PYTHAGORAS:
HYP^2 = OPP^2 + ADJ^2
C^2 = A^2 + B^2
EQUATION OF CIRCLE:
X^2 +Y^2 = (RADIUS)^2
cedar kilnBOT
cedar kilnBOT
Available help channel!
grizzled sapphire
prove that gcd(a+b-1, a^2-ab+b^2) = 1 for all positive integers a,b.
grizzled sapphire
sorry i havent done math in a really long time and i cant figure this out
slate lintel
i think that's gcd
green thorn
prove that gcd(a+b-1, a^2-ab+b^2) = 1 for all positive integers a,b.
is this gcd
grizzled sapphire
is this gcd
yeah it's gcd
cedar kilnBOT
@grizzled sapphire Has your question been resolved?
cedar kilnBOT
@grizzled sapphire Has your question been resolved?
cedar kilnBOT
@grizzled sapphire Has your question been resolved?
daring peak
cedar kilnBOT
daring peak
how do i start this?
idle tusk
if you are asked to find ln^2(1+x) as polynomial of degree 4, that means that you don't have to square the x^3 term (as that would be degree 6)
that means to find ln^2(1+x) you just square (x-1/2 x^2)
daring peak
so to find ln^2(1+x) i just have to square ln(1+x)??
idle tusk
yes
you are approximating it
and it's saying that you need to approximate it by a polynomial of degree 4
therefore you need to square the sum of only the first 2 terms
because the rest will be O(x^6)
daring peak
ok that makes sense. so when ive done that how do i do the second part?
just sub in the values?
idle tusk
yep
daring peak
.close
tepid scarab
Could someone please help me on how I work out this question?
sand cradle
Could someone please help me on how I work out this question?
Is this an exam?
sand cradle
No. I’m just practicing questions
Alright, do you have some idea for (a), (i), then?
tepid scarab
I know that there is an equation Ae^kt is this what I would use?
sand cradle
I know that there is an equation Ae^kt is this what I would use?
Look at model A
Does that look like an exponential function?
tepid scarab
hmm no
sand cradle
Yeah
tepid scarab
I don’t think so
tepid scarab
A linear graph
sand cradle
Yeah
I assume you aren't supposed to put your rules on this and read the volume off at t = 3, so you should set up an equation for it.
You know it's a linear equation
And you have two points
For g(x) = mx + c, you can right away tell c because of P(0, 16000)
tepid scarab
Yeah so now I need to work out the gradient?
tepid scarab
C is 16000
sand cradle
Yeah
tepid scarab
How would I figure out the gradient? I mean if you could start me off
I’m new to this topic
sand cradle
Recall that it's the change in y divided by the change in x ("rise over run")
So $m = \frac{y_2 - y_1}{x_2 - x_1}$.
tepid scarab
So 9000/4
sand cradle
tepid scarab
Ohhh
sand cradle
9000-16000/4-0
Yes
tepid scarab
-1750
sand cradle
9000-16000/4-0
Be careful, put brackets, since if you would type this into a calculator and it wouldn't be 4 - 0, your calculator would read it as ((9000 - 16000)/4) - 0