You can ask any question about integers in one of those channels. Just go into one, and ask the question you're stuck on. If you don't know how to start, say that, otherwise show your work so far.
#help-10
scarlet gale
topaz tartan
scarlet gale
No, you have to always have both sides separate.
topaz tartan
alr
scarlet gale
Like the left side will be 5a on the third line, but the right side won't be.
It'll be just 5.
topaz tartan
scarlet gale
Yes, that looks good. The only change I'd make is that all the lines right under the ---------- should have an equal sign to separate the sides.
Like 5a = 5 on the third line.
But aside from that, that looks good.
scarlet gale
Also, I assume those as on the right are erased.
topaz tartan
why does 5a / 5 have to be seperate from 5 / 5
scarlet gale
Because you need to know what's on both sides. If you mix them up, you can get things wrong.
5a + 2 = 7
- 2 -2
-----------
5a = 5
÷ 5 ÷ 5
-----------
a = 1
k
scarlet gale
Because they'll be different from each other, so you have to keep them separate so things don't get confusing.
But other than how to write it, the way is to think about how you'd calculate the side with the variable on it.
topaz tartan
alr
scarlet gale
Like here, start with a, then:
1. Multiply by 5
2. Add 2
Then, your steps to isolate it will be backwards and opposite of that:
1. Subtract 2 from both sides
2. Divide both sides by 5
So, that's how you figure out the two steps to do in your two-step problems.
Three-step problems are the same kind of thing.
And so on.
topaz tartan
Like here, start with a, then:
```
1. Multiply by 5
2. Add 2
```
Then, your step...
so the 1st step showed here has nothing to do with my equation
scarlet gale
To get the normal forwards steps, you look only at the side of the equation with the variable in it.
Then, you figure out how you'd calculate it.
Then, the backwards opposite version will deal with your equation.
You do the backwards opposite steps to both sides of your equation.
Then, you get the variable isolated on one side.
topaz tartan
yeah this is...
becoming harder to understand
hold on
can i state the steps again
in this equation
scarlet gale
Sure.
scarlet gale
Yes, that looks good.
topaz tartan
so thats everything to do?
scarlet gale
Yes, those are the steps.
And as you can see, you have a isolated by itself on one side.
Then, you can check your work.
scarlet gale
To do that, all you have to do is write 5a + 2. Then you fill in a with 1.
5(1) + 2
Then you see if that's 7.
Then you know you got the right answer if it is.
Or at least you're more sure.
topaz tartan
ty
scarlet gale
No problem.
scarlet gale
OK, have a good day.
topaz tartan
.close
timid silo
how the heck
obtuse pebbleBOT
how the heck
timid silo
are you supposed to figure out how to do integral of cscx on your own
i just saw the solution
how are u supposed to know that u need to multiply (cscx+cotx) to both numerator and denominator
tough bolt
Did your textbook provide the solution to the integral of sec x?
timid silo
yes, ln abs value secx + tanx i think
tough bolt
The integral of csc x is similar
timid silo
oh. so it's a memorization thing? they just wanted us to figure out the proof basically?
tough bolt
Who are "they"?
timid silo
the people who wrote the textbook
tough bolt
Ok
timid silo
i just am worried if i see a problem that requires i solution like that on a test im just screwed
harsh remnant
Don't you use IBP for cscx?
timid silo
Don't you use IBP for cscx?
is it possible? i did not try it
tough bolt
i just am worried if i see a problem that requires i solution like that on a tes...
That's why analogy is so important in problem solving
harsh remnant
is it possible? i did not try it
I don't exactly remember, but that is what I recall on the top of my head
timid silo
That's why analogy is so important in problem solving
whats analogy mean
tough bolt
Analogous means that something is similar to the other thing
And analogy is the noun of analogous
timid silo
I don't exactly remember, but that is what I recall on the top of my head
just tried it, 95% sure it's impossible
tough bolt
For example, csc^2 x=1+cot^2 x is analogous to sec^2 x=1+tan^2 x
harsh remnant
mb sry
timid silo
For example, csc^2 x=1+cot^2 x is analogous to sec^2 x=1+tan^2 x
oh i see u what u mean
what if it was another problem tho where there was no analogous?
at least not one that was taught
like if it had asked integral secx without teaching cscx first
tough bolt
like if it had asked integral secx without teaching cscx first
If that happens then that's really unlucky
I've only encountered questions like these in challenge problems
timid silo
If that happens then that's really unlucky
hmm.. im curious, do you know if that kind of thing is more common in higher level math? im considering doing math as a major
tough bolt
hmm.. im curious, do you know if that kind of thing is more common in higher lev...
I learn math purely for fun so I don't know
timid silo
I learn math purely for fun so I don't know
ah even better then, so i assume u try to solve problems first before looking at how to do them right?
like is it something u can develop the ability to do u think or no?
tough bolt
ah even better then, so i assume u try to solve problems first before looking at...
Yes
tough bolt
like is it something u can develop the ability to do u think or no?
Maybe a bit, but I don't think practicing a lot will help
timid silo
hmm ok
tough bolt
For example independently finding out about the product rule or something?
timid silo
yes
tough bolt
I don't think that's ever happened to me before
timid silo
dang...
must have been hard to make those then ig
alr thx a lot of the info
and patience
.close
old bluff
can you please provide a detailed explanation to how to solve this
obtuse pebbleBOT
can you please provide a detailed explanation to how to solve this
old bluff
answer
plain stag
Set up 2 equations modelling the perimeter and area of a rectangle
plain stag
Show your work
old bluff
let me write everything again and show you
dire goblet
@plain stag
plain stag
What?
dire goblet
can u help me
if i need to add subtract multiply or devide
im cheeks at word problems
plain stag
dire goblet
o
old bluff
what do i do after this
plain stag
Solve the quadratic for l
old bluff
i couldnt factor it
thin star
Are you familiar with the quadratic formula?
thin star
Then that
old bluff
oh okay
plain stag
You can always use it if you want
old bluff
i was tryna avoid it lol
plain stag
It is what it is
wintry forum
i need confirmation for my answers on these
obtuse pebbleBOT
i need confirmation for my answers on these
wintry forum
- 49^6 different possibilities or something like that
- 1/6
- 1/52
- 1/4
- 1/64
<@&286206848099549185> ping if anybody responds! thanks 
heyy no rush but i would just like some confirmation on whether or not my answers are correct
<@&286206848099549185>
wide shard
half cairn
Hey, I need a help with this question:
Use the Lagrange multipliers method to determine the maximum points
and/or minimal of the function f(x, y) = x² − 2xy + y², subject to the constraint x² + y² = 1.
obtuse pebbleBOT
Hey, I need a help with this question:
Use the Lagrange multipliers method to d...
@half cairn Has your question been resolved?
half cairn
<@&286206848099549185>
obtuse pebbleBOT
@half cairn Has your question been resolved?
zealous dawn
Hi, I'm trying to solve this problem, and I keep getting the wrong answer due to sign differences. But I can't figure out what's wrong:
obtuse pebbleBOT
Hi, I'm trying to solve this problem, and I keep getting the wrong answer due to...
zealous dawn
Once I take the integral of 1 + sinx, I get x - cosx
then I plug in the values of 3pi/4 and subtract what I get when I plug in the value of 0
urban patrol
did you distribute the negative
zealous dawn
Right, but cos(0) becomes -1 once I distribute the negative
urban patrol
cos(3pi/4) is also negative
urban patrol
Right, but cos(0) becomes -1 once I distribute the negative
you already distributed the negative
zealous dawn
So I get this
urban patrol
this is distributed
zealous dawn
oh I see my mistake lmao
it was all the way at the beginning, thanks
my eyes are too tired
.close
restive ermine
obtuse pebbleBOT
What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
Can you imagine what's happening to the object there?
urban patrol
do you know the difference between velocity and speed
restive ermine
do you know the difference between velocity and speed
Yes
restive ermine
Can you imagine what's happening to the object there?
The furthest I get is I understand that it’s changing direction at one point
When velocity reaches zero
No idea what to build off taht
restive ermine
okay what's the difference
Velocity deals with displacement
While speed is total distance travelled
urban patrol
no
urban patrol
velocity is speed with direction
restive ermine
Oooohh I see
urban patrol
speed is tbe magnitude of velocity
so if youre going 50 mph one way its velocity might be negative but its speed is still the same as if its going in the positive direction
more useful for 2d
restive ermine
I see
timid silo
so if youre going 50 mph one way its velocity might be negative but its speed is...
I am kinda studying the same thing
The answer is D right?
My physics is bad 
urban patrol
I believe there's a typo
I think its supposed to say +x
but yeah D is really the only answer
restive ermine
So here’s the thing
timid silo
Alrighty :3
restive ermine
I also have this up on chegg
It’s not saying d
I honestly agree with y’all
In
Saying it’s d
But cheggs throwing me off a bit
urban patrol
real people > chegg
restive ermine
real people > chegg
Honestly this servers a life saver 🛟
timid silo
Honestly this servers a life saver 🛟
Fr
timid silo
What does chegg say?
Ayo bro the pfp
thin star
Ayo bro the pfp
it's a dog bruv 😂
restive ermine
Pls I had to do a double take
timid silo
I have it in my book as well
Well a bit difft
obtuse pebbleBOT
Available help channel!
placid portal
for the function f(x)= -2x^2 -8x -10, find the slope of the tangent line at x=-10
obtuse pebbleBOT
for the function f(x)= -2x^2 -8x -10, find the slope of the tangent line at x=-1...
placid portal
just worked on one of these and did it correctlly
nocturne minnow
Please close your previous channel using .close before opening a new one
hexed gull
for the function f(x)= -2x^2 -8x -10, find the slope of the tangent line at x=-1...
shall I copy what I mentioned in the last one
placid portal
i did the exact same thing
hexed gull
kk
hexed gull
ys
placid portal
then did -4(-10)-8 which was 32
hexed gull
yes
hexed gull
deltamath was tripping balls
did you copy the right function?
placid portal
yep
hexed gull
because the slope would be 32
placid portal
word for word
lemme see if it doesnt want me to solve and just wants the derivitive
ok weird
i got it wrong but it had ln3 in there for some reason
that makes 0 sense
we were right tho ????
hexed gull
yes, 32 is right
hexed gull
the answer said lim h->0 -2(x+h)^2-ln(3)8x-10?
eh ok
hexed gull
maybe incorrectly copied from another task
placid portal
very odd
hexed gull
the answer said lim h->0 -2(x+h)^2-ln(3)8x-10?
this wouldn't be a good answer either even if it were it, because you'd still solve the limit
placid portal
anyways i need help with derivatives and fractions cause that stuff is hard
-3/4 x^-3 -1/3 x^4 +4/5 x +2/3 x^3
wants f''
so i know how to find f''
im just confused with how you multiply fractions lmao
hexed gull
let me write in desmos so it's less clumsy
k so start with f'
reduce all exponents by 1
and multiply by the exponents
placid portal
(-3x^2) (4x^3) (3x^2)
hexed gull
I'll put the answers here:
placid portal
then so do the fractions come in after that?
hexed gull
placid portal
like i have my f'
hexed gull
like i have my f'
you apply the same for f''
reduce exponents by 1 and multiply by them
placid portal
ok so our f' is different
how d
OHHH
the first one is negative
wild
but still i have no idea how to multiply fractions lmao
hexed gull
which one for instance
placid portal
lets start with the first one
hexed gull
-3/4 x^(-3)
placid portal
for me my f' was -3/4 (-3x^2)
hexed gull
and -3 * -3/4 is 9/4
hexed gull
-3 * -3/4 = (-3*-3)/4 = 9/4
placid portal
alright i got that question then
i have two more of these i need to work through them with you
hexed gull
it's always the same procedure, just make sure not to overlook numerical mistakes
placid portal
for some reason i just have no idea how to multiply fractions
like if you asked me to do -1/2 x^3 i would be at a loss
would that be -1 1/2
hexed gull
like if you asked me to do -1/2 x^3 i would be at a loss
-1/2 x³
becomes 3 * (-1/2) * x²
which is -3/2 x²
sry gotta go call
rotund rune
how do I solve this question?
obtuse pebbleBOT
how do I solve this question?
polar fossil
ooh there's some ambiguity in that question
can the secretary sit next to the president if they're not also sitting next to the vice president?
either way, place those three first and then the rest
because it's a round table, you don't really have a choice for where the first person goes, only where the second one goes
rotund rune
ooh there's some ambiguity in that question
yeah that's where i got confused as well
polar fossil
if i had to pick i'd say it means that the secretary can't sit right between the other two
rotund rune
either way, place those three first and then the rest
wait I don't quite understand what you mean by this
polar fossil
so you start with the first person, let's say we pick where the secretary goes first
doesn't really matter because it's a round table so we only really have 1 choice
then we pick the next person, the president; and we have 7 choices here because there are 7 empty chairs
(depending on how you interpret the problem some of those choices may not be valid)
then the next person is the vice president, 6 choices there (again, some may not be valid)
and then the rest of them similarly
obtuse pebbleBOT
@rotund rune Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
astral bramble
I attempted doing integration by parts but I got stuck in a loop of integration. How could I prevent/solve a problem that seems to have that loop?
obtuse pebbleBOT
I attempted doing integration by parts but I got stuck in a loop of integration....
timid silo
the fact that it loops is what will help you here :p
polar fossil
actually no, call it L because you can actually see that as i'm typing
astral bramble
the fact that it loops is what will help you here :p
how so?
polar fossil
you'll have L = (bunch of stuff) - cL
astral bramble
you'll have L = (bunch of stuff) - cL
would I then solve like a normal algebra problem?
polar fossil
well, you're trying to find a value for L right? since that was the original goal?
astral bramble
well, you're trying to find a value for L right? since that was the original goa...
ah okay
polar fossil
or rather an expression for L in terms of x
astral bramble
or rather an expression for L in terms of x
so i got to here with the algebra, but when i plug in the answer to the homework software it is saying the answer must be linear with c
polar fossil
so i got to here with the algebra, but when i plug in the answer to the homework...
c is some constant, it's up to you to figure out what that constant is
your IBP seems flawed here
i should see coefficients of 4, 5, and 20 somewhere around here due to chain rule
astral bramble
i should see coefficients of 4, 5, and 20 somewhere around here due to chain rul...
chain rule? why would I be taking the chain rule of this function if im trying to find the integral?
polar fossil
can you show me your IBP steps?
polar fossil
chain rule? why would I be taking the chain rule of this function if im trying t...
(i meant u-substitution, the counterpart to the chain rule)
astral bramble
(i meant u-substitution, the counterpart to the chain rule)
ahh okok ima try that and Ill get back to you with whatever i get
astral bramble
can you show me your IBP steps?
polar fossil
lol
you said:
I attempted doing integration by parts but I got stuck in a loop of integration
can you show me that loop?
feel free to take a picture of your work rather than mspainting it up
astral bramble
feel free to take a picture of your work rather than mspainting it up
funnily enough ive been using a digital notebook that works similar to mspaint to do most of my work since doing it on paper gets way too chaotic way too quickly
astral bramble
you said:
> I attempted doing integration by parts but I got stuck in a loop of ...
warm shaleBOT
hayley!
polar fossil
so in this case it looks like your $u$ is $u = \sin4x$? and you're taking $\dd{v} = e^{5x}\dd{x}$?
warm shaleBOT
hayley!
astral bramble
so in this case it looks like your $u$ is $u = \sin4x$? and you're taking $\dd{v...
yep
polar fossil
ok, so then what are du and v?
astral bramble
ok, so then what are du and v?
i just was taught to use f and g instead of u and v
polar fossil
f and g is fine too
that looks better
i'd probably write that as $\f15e^{5x}\sin4x - \f45\int e^{5x}\cos4x\dd{x}$
warm shaleBOT
hayley!
astral bramble
i'd probably write that as $\f15e^{5x}\sin4x - \f45\int e^{5x}\cos4x\dd{x}$
so would I have to do another IBP but this time use the e^5xcos(4x)?
polar fossil
yep
polar fossil
there we go!
well almost
that 4/5 needs to distribute into both terms
are you doing this with a mouse or do you have a stylus of some kind?
astral bramble
are you doing this with a mouse or do you have a stylus of some kind?
a stylus but i just have insanely messy handwriting mixed with a shaky hand
polar fossil
ah ok
astral bramble
that 4/5 needs to distribute into both terms
the 4/5 as in the one present in the lowest function or the 4/5 in the one above the lowest?
polar fossil
$L = \f15 e^{5x}\sin(4x) + \f45\lp \f15e^{5x}\cos(4x) - \lp -\f45\int e^{5x}\sin(4x)\dd{x} \rp \rp$
warm shaleBOT
hayley!
polar fossil
is what i think you wrote
astral bramble
is what i think you wrote
yep
polar fossil
$L = \f15 e^{5x}\sin(4x) + \blue{\f45}\lp \f15e^{5x}\cos(4x) - \lp -\f45\int e^{5x}\sin(4x)\dd{x} \rp \rp$
warm shaleBOT
hayley!
polar fossil
the 4/5 in blue is the one that needs to distribute to both terms there
astral bramble
the 4/5 in blue is the one that needs to distribute to both terms there
ooh ok so it would become 4/25 and -16/25?
polar fossil
yeah exactly
i always lose a bit of sanity when dealing with these fractions for this particular integral
so then what you end up with is like
astral bramble
so then what you end up with is like
wouldn't there still be that last integral to contend with?
polar fossil
$\green{\int e^{5x}\sin(4x)\dd{x}} = \f15 e^{5x}\sin(4x) + \f4{25} e^{5x}\cos(4x) + \f{16}{25}\green{\int e^{5x}\sin(4x)\dd{x}}$
warm shaleBOT
hayley!
polar fossil
because we started with the thing on the left, did two rounds of IBP, and ended with the thing on the right
which has a copy of the thing that we started with in it
astral bramble
because we started with the thing on the left, did two rounds of IBP, and ended ...
oooh and then c is accounted for with the 16/25
bruh calc 2 is wack
polar fossil
this is a particularly weird integral
because you like
find its value in terms of itself
and then wrap that back around to find an actual solution
astral bramble
this is a particularly weird integral
yeah my professor has a really nasty habit of making the homework much more difficult compared to classwork
polar fossil
that's how it usually goes :D
astral bramble
that's how it usually goes :D
lmao tru
polar fossil
?
polar fossil
uh yeah because you didn't solve correctly
look at this
the green thing is what you want to solve for right
astral bramble
the green thing is what you want to solve for right
yes
polar fossil
great so solve for it
$\green L = \f15 e^{5x}\sin(4x) + \f4{25} e^{5x}\cos(4x) + \f{16}{25}\green L$
warm shaleBOT
hayley!
astral bramble
$\green L = \f15 e^{5x}\sin(4x) + \f4{25} e^{5x}\cos(4x) + \f{16}{25}\green L$
so the L's would just cancel out? or would I move the 16/26L over?
polar fossil
you'd move the 16/25 L over
astral bramble
astral bramble
now distribute that 25/9 and some stuff cancels
like so?
astral bramble
sure (i'd write that as 5/9 and 4/9 instead though
okay now do i need to solve for the -cL all the way from the beginning?
polar fossil
no you're good
when i told you the cL thing
i thought that you had already gotten to this point
you see how that fits the L = (bunch of stuff) + cL
astral bramble
no you're good
well then something is wrong because its giving that same "there is always more than one possibility" error again
polar fossil
it'll be a sign error
astral bramble
it'll be a sign error
sign error? like two answers one with a plus at the beginning and one with a minus at the beginning?
polar fossil
this negative sign became a plus sign somehow
astral bramble
your final answer should involve the number 41
like this?
polar fossil
pretty sure that's right
rain loom
.close
obtuse pebbleBOT
Available help channel!
timid silo
hi can someone explain how to know whether the hyperbola is vertical/horizontal?
fervent cradle
if the equation has like x^2 - y^2 = 1 then you can sorta see that (1, 0) is a possibility but (0, 1) isn't, so it'll look like the first case, intersecting the x-axis
if the equation has like y^2 - x^2 =1 then you can see that (1, 0) isn't a possibility but (0, 1) is
brisk arrow
They have similar properties as “Inverse functions”
timid silo
ohhh
okay thank you!
hi if the equation is y^2/16-x^2=1
what's the b^2?
would it be 0?
<@&286206848099549185>
warm canopy
The b² would be 1
timid silo
thank you!
timid silo
hi
timid silo
can someone explain what this is
c units left and right of c??
how do I solve for that?
oh nvm i got it
dire plinth
how do i find the equation from an exponential graph?
obtuse pebbleBOT
how do i find the equation from an exponential graph?
cloud belfry
like ke^(mx) format?
dire plinth
y = a^x
atomic hornet
look at x=1
queen sparrow
dire plinth
huh sorry im confused
like say ur given a graph and have to put it into form below
y = a^(x-k) + h
firstly wouldn't i only need to solve for k and h, and secondly how would i find them
oh wait id also have to find a
obtuse pebbleBOT
@dire plinth Has your question been resolved?
dire plinth
.close
quaint acorn
obtuse pebbleBOT
quaint acorn
I dont know where to start
cloud belfry
thats (x2-1)x
quaint acorn
what?
cloud belfry
bro
royal basin
I dont know where to start
do you know how to shift a graph 1 unit to the right
royal basin
thats (x2-1)x
missing the ^ there...
cloud belfry
sry
quaint acorn
But x^3-x-1 does not equal the result that we want??
quaint acorn
could you please show me
cloud belfry
(x-1)^3-(x-1)
quaint acorn
thats not even correct
u have no idea what you are talking about, if you don't have any idea what you are talking about please leave instead of confusing people and putting them in a worse scenario.
@royal basin
royal basin
yes you add a -1
you do not "add a -1".
royal basin
But x^3-x-1 does not equal the result that we want??
that's shifting 1 unit down.
quaint acorn
no all inside i meant
royal basin
shifting 1 unit right means going from y = f(x) [= x^3 - x for you] to y = f(x-1)
quaint acorn
ahhh okay, thanks heaps
royal basin
so namo was in fact right to say it'll be y = (x-1)^3 - (x-1).
royal basin
thats not even correct
and you were wrong to deride it as wrong here.
signal light
Hello can someone explain why my Calculator says that 0^(-2) = infinite?
warm canopy
what do you expect it to say?
solar trellis
Hello can someone explain why my Calculator says that 0^(-2) = infinite?
because the programmer decided as such.
signal light
when i typ in 0^(-1) it says not definable.
warm canopy
it is interpreting it as a limit
its interpreting $0^{-2}$ as $\lim_{x \to 0} \frac{1}{x^2} = \infty$
warm shaleBOT
ΣΑCu
warm canopy
but $\lim_{x \to 0}\frac{1}{x}$ does not exist, which is how its interpreting $0^{-1}$
warm shaleBOT
ΣΑCu
signal light
ah Thank u
wispy patio
help me understand the solution
obtuse pebbleBOT
help me understand the solution
sage geode
Okay so let's think about what a1 + a2 should be equal to given that a1 and a2 are additive inverses
Since a1 is an additive inverse, adding it to any number does nothing to the number, right?
wispy patio
yea
sage geode
And a2 is not an exception to that rule
wispy patio
makes it 0
sage geode
Meaning a2 + a1 = a2
sage geode
makes it 0
No adding the additive inverse doesn't make the number 0, it will just yield that exact number
sage geode
Meaning a2 + a1 = a2
Here a1 was added to a2, but a1 is an additive identity, meaning adding it to any number does nothing to that number, meaning the sum is equal to a2
sage geode
But a2 is also an additive identity, so we also have a1 + a2 = a1
sage geode
a1 + a2 and a2 + a1 must be the same according to commutativity, right?
wispy patio
alright thanks bro that's all i needed 2 understand
sage geode
You might as well note that the only property that we used was commutativity, so this doesn't apply to just real numbers
sage geode
Alright, good luck
wispy patio
.close
round dagger
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the
following may be considered as universal set (s) for all the three sets A, B and C
(i) {0, 1, 2, 3, 4, 5, 6}
(ii) φ
(iii) {0,1,2,3,4,5,6,7,8,9,10}
(iv) {1,2,3,4,5,6,7,8} _____________________________________________________________________________________________ and also what is the Venn diagram?
obtuse pebbleBOT
Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of th...
nocturne sundial
<@&286206848099549185> help me quick
Are we allowed to choose only 1 answer?
nocturne sundial
I'm asking if there is only one answer or are there multiple correct answers?
round dagger
yes
nocturne sundial
That does not answer my question LOL
nocturne sundial
A Venn Diagram is a visual representation of a collection of elements and how they relate to one another.
I'll send a picture, I'm sure you've seen them before.
round dagger
ok
round dagger
ok
nocturne sundial
The image is sending still
round dagger
. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?
nocturne sundial
The sideways U represents a "subset," which is another way of saying "the elements inside this set are also inside another set."
Tell me, are the elements inside of A also inside B?
round dagger
yes they are
round dagger
ohh now i get it
nocturne sundial
Now, for A U B. That's known as the union of sets, and it's basically just the combined elements of both sets.
round dagger
but what is the the opposite of a subset
nocturne sundial
Hmm, good question
Not a subet, I guess LOL
For example
A = {1, 2}, B = {3, 4}
round dagger
it looks like an upside sown U
round dagger
what is that
nocturne sundial
It represents how many elements of each set are in common with each other
What are the common elements of A and B in your problem?
By common, I mean shared
round dagger
ab
nocturne sundial
Yeah
nocturne sundial
Draw me two circles and label them A and B
But make sure the circles overlap in the center
nocturne sundial
Good, it should look like this
round dagger
ok i did
nocturne sundial
We're going to have to assign the elements "a" and "b" to a section of this diagram
Where do you think they should go?
nocturne sundial
It's true that a can be found in the first circle and b can be found in the other, but Venn Diagrams seek to display the relationship between the overlapping parts of each set.
round dagger
ok
nocturne sundial
There's a more accurate place we could put a and b to demonstrate this relationship.
Try again, where do you think a and b should go?
To make it as accurate as possible
Remember, the overlapping part of the venn diagram repsents the intersection of the two sets
nocturne sundial
It's okay, you're on the right track.
round dagger
ok bro i will come back bye its my chem time now
nocturne sundial
I should also mention that in a Venn Diagram, putting an element in the overlapping section also indicates that it is in each circle independently as well.
nocturne sundial
ok bro i will come back bye its my chem time now
Good luck!
round dagger
thank you very much
obtuse pebbleBOT
@round dagger Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
lucid pendant
hello
gleaming nebula
Im not sure if I am being incredibly silly but I have no idea how to get the answer for this numerical reasoning example question for a job application, see images below:
obtuse pebbleBOT
Im not sure if I am being incredibly silly but I have no idea how to get the ans...
gleaming nebula
It says the correct answer is 68.2% which boggles my mind because I cant get 68.2% no matter what I do
Wait
Nevermind I just got it
thin star
Ok good.
gleaming nebula
I guess I just spent 10 minutes plugging the wrong numbers in
thin star
It's just (new - old) /old
gleaming nebula
its like 30c today maybe my brain is just getting cooked 😅
I am actually a little confused on the follow up question
nvm I think I got that too
I think its the heat and stressing a bit about wanting to get the questions in the actual quiz right
I did 7637,000,000(3063,000,000/1000) = 2493.31
I knew I wasnt this mathematically challenged
obtuse pebbleBOT
Available help channel!
stone ermine
I can't do math
obtuse pebbleBOT
I can't do math
thin star
Keep going
thin star
you want to re-isolate y on one side
stone ermine
How do I do that?
stone ermine
Im just doing the inverse
stone ermine
give me a sec
sinful nova
okay
stone ermine
stone ermine
sinful nova
why the handwriting so bad tho
stone ermine
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
plain nebula
help
plain nebula
idk how q^2 is equal to that
shouldnt it be the same since 1^2 and 0^2 are both the same
like they stay the same
eternal thistle
$\lim_{x\to \infty} \left(1+\dfrac{1}{x}\right)^{\sqrt{x}}$
obtuse pebbleBOT
$\lim_{x\to \infty} \left(1+\dfrac{1}{x}\right)^{\sqrt{x}}$
warm shaleBOT
tales
eternal thistle
Without l'hopital
Any tip?
I know $\lim_{x\to \infty} \left(1+\dfrac{1}{x}\right)^{x} = e$
warm shaleBOT
tales
royal basin
your function is that but raised to the power of 1/sqrt(x)
1/sqrt(x) approaches 0, and (1 + 1/x)^x approaches e
eternal thistle
$\lim_{x\to \infty} \left(1+\dfrac{1}{x}\right)^{\sqrt{x}} =\lim_{x\to \infty}\left[ \left(1+\dfrac{1}{x}\right)^{x}\right]^{\frac{1}{\sqrt{x}}$
warm shaleBOT
tales
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
eternal thistle
Can i take the limit inside tho?
It would be 1
then
But I don't think it's legal to say $\lim{x\to \infty} \left(1+\dfrac{1}{x}\right)^{\sqrt{x}} =\lim{x\to \infty}\left[ \left(1+\dfrac{1}{x}\right)^{x}\right]^{\frac{1}{\sqrt{x}}} = e^0 =1$
warm shaleBOT
tales
obtuse pebbleBOT
@eternal thistle Has your question been resolved?
eternal thistle
.close
royal shard
which step?
granite sparrow
1-(1/3+1/3x2/3+1/3(2/3)^2
viscid gull
Is it even right?
viscid gull
yeah
royal shard
ahhh
ok then P(x>2)=1-P(x<=2)
in that case, the calculation should be correct, but the first line is wrong
royal shard
P(x>2)=1-P(x<=2)
do you understand this?
granite sparrow
yes
granite sparrow
mhm
royal shard
what they did, was they wrote down P(x<=1) instead of P(x<=2)
but then they did include the x=2 term as if they were using P(x<=2)
try to calculate P(1) and P(2)
royal shard
$P(1)=\frac{1/3}{(3/2)^1}=\frac{1}{3}\cdot\frac{2}{3}$
obtuse pebbleBOT
Available help channel!
warm shaleBOT
Martin
timid silo
Using mathematical induction prove that n e N
-3+3+9+...+(6n-9)=3n^2-6n
obtuse pebbleBOT
Using mathematical induction prove that n e N
-3+3+9+...+(6n-9)=3n^2-6n
timid silo
@loud mist
timid silo
nothing
loud mist
just take a picture of the problem
loud mist
okay
btw, 0 is not in N right? because there are 2 different conventions
N starts at 1 right?
do you know
@timid silo I will help you solve it dont run away
timid silo
i dont even know what you mean im clueless
loud mist
do you know what N is?
timid silo
natural numbers
timid silo
its very confusing because the thing has -3 and other negativ numbers (not natural)
timid silo
or does it start at 1
starts at 1
loud mist
okay
timid silo
btw let me translate it a bit better
loud mist
so remember with the marbles, we had to know that the zero marble is red, in this case it is 1 instead of 0, so you have to fill in n with 1 and see if the equation is correct that is step 1
timid silo
"prove by mathematical induction that for all n e N is valid:" idk if this matters
timid silo
so remember with the marbles, we had to know that the zero marble is red, in thi...
um
so
idk
wait my stupid brain is having issue processing
or i am
im the brain
im stupid
timid silo
yes but why is it 3 -6
loud mist
so this is -3 = -3
loud mist
so step 1 is finished
timid silo
i see alrght
loud mist
so we have proven that the first marble is red
loud mist
now we need to prove that if the equation is correct for n, that then it is also correct for n+1
and then the proof is finished
we have proven that all marbles are red
loud mist
no
timid silo
because n was 1
so its not 1 anymore?
im sorry im very stupid
im really trying tho
loud mist
so we have to prove that if A = 3n^2 - 6 that then A+(6(n+1)-9) = 3(n+1)^2 - 6(n+1)
timid silo
give me like a min to process this
loud mist
okay
timid silo
what is A
loud mist
it is just a name i gave to 3n^2-6
timid silo
ah okay
loud mist
so you can also prove 3n^2 - 6+(6(n+1)-9) = 3(n+1)^2 - 6(n+1)
timid silo
how did you get all of these other numbers
"A+(6(n+1)-9)"
where did (6(n+1)-9) come from
loud mist
so A is basically the right hand side of -3+3+9+...+(6n-9)=3n^2-6n this equation
timid silo
i see
loud mist
and by the hypothesis we can assume that this equation is true
so -3+3+9+...+(6n-9)=3n^2-6n = A
so both sides are equal to A
and now we have to see what happens when we do n+1 instead of n
loud mist
well gave -3+3+9+...+(6n-9) the name A, and we know that -3+3+9+...+(6n-9) = 3n^2-6n
its kinda like
if a = b
and b = c
then a = c
timid silo
alrighti get that so we assume that -3+3+9+...+(6n-9) = 3n^2-6n is correct
loud mist
yes
timid silo
what then
loud mist
we want to know if that equation is still true if we replace n with n+1
loud mist
well the right hand side of the eqation if we use n+1 instead of n is A + (6(n+1)-9)
do you understand that?
loud mist
okay let me give you a simpler equation so you understant what im doing
so imagine A = 1 + 2 + ...+ n
timid silo
alright
timid silo
hm
loud mist
do you understand
timid silo
ah so i just add n + 1 every time
timid silo
you need to prove that if n is true then that n+1 is true
I understand that i just dont know how
timid silo
nope 😦
loud mist
dont worry
timid silo
yeah i get that
loud mist
but now if we do it for n+1 then B = -3+3+9+...+(6(n+1)-9)
and this is the same as B = A + (6(n+1)-9)
timid silo
oh you just added 1
loud mist
now we need to prove that B = 3(n+1)^2-6(n+1)
timid silo
so -3+3+9+...+(6n-9) + (6(n+1)-9)
loud mist
yes
loud mist
well yes but
we can replace -3+3+9+...+(6n-9) with 3n^2-6n
becauce we know -3+3+9+...+(6n-9) = 3n^2-6n
is true
thats the hypothesis
so in the end we need to prove that 3n^2-6n + (6(n+1)-9) = 3(n+1)^2-6(n+1)
do you get it
loud mist
do you know how to prove it?
timid silo
but i think i will not be able to repeat this to solve anything by myself
timid silo
do you know how to prove it?
no
timid silo
alrght well thank you for your time
loud mist
do you know how to prove it?
to prove this you need to do some algebra
timid silo
im sorry you wasted it on my worthless self but im sure ayou get some good karma for it
loud mist
you need to show that the right hand side is the same as the left hand side
with some algebra
loud mist
okay
narrow jackal
obtuse pebbleBOT
narrow jackal
I am supposed to draw the graph of the function. How the hell should I even start?
does the graph "tighten" because of the bigger number (the one we multiply with x)?
stray shoal
for the first question the graph is stretched in the x direction by a scale factor 2
I like to think of it as though I had some values I could plug in and what would happen to them. For the first one:
original is y = f(x). then we get y = f(2x). To get the same value we have had to multiply the x value by 2
narrow jackal
I think I understand, but could you give me a moment to post the results?
to make sure I am right, I guess
stray shoal
sure
stray shoal
I think that's correct
narrow jackal
Thank you very much!
stray shoal
np
narrow jackal
.close
eager latch
Is the standard deviation given in this problem considered the population standard deviation?
The fresh egg weight of a randomly selected egg of a Mallard Duck is a random variable with a mean of 50.1 grams and standard deviation of 3.5 grams. Calculate the 90th percentile of the sample mean of a random sample 34 fresh Mallard Duck eggs. Use one decimal place accuracy.
So I would still need to calculate the sample standard deviation using this formula?
