I know a bit of French but I don't think we say "cancels" in english
#help-26
rose sierra
short iris
in less on time
rose sierra
i'm not sure what that means
short iris
I know a bit of French but I don't think we say "cancels" in english
ok sorry
i search how to translate in english
thorny flameBOT
kebesque
short iris
Let f : [0,1] → R be of class C1 such that f (0) = f (1) = 0. Show that for all a ∈ R, the function f′ + a f is equal 0 at least once on ]0,1[. I need some help. With rolle f is egqle 0 at least once but for f'
topaz sinewBOT
Let f : [0,1] → R be of class C1 such that f (0) = f (1) = 0. Show that for all ...
@short iris Has your question been resolved?
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visual moon
this might be a dumb question but do sinx=1 and sinx=-1 have any triangle on the unit circle? Im learning trig right now and i was solving an equation.
topaz sinewBOT
this might be a dumb question but do sinx=1 and sinx=-1 have any triangle on the...
meager dawn
they have lines, which are “degenerate” triangles in a way
one side will have length 0
it’s a good question
visual moon
so I can't really show it on the unit circle in that case unless i just draw the line between the circle showing the lines of sinx=1 and sinx=-1
bitter hemlock
why do they need triangles?
meager dawn
well if you’re on the unit circle you don’t really have to draw triangles anyways, you’re just trying to find points on the unit circle whose y coordinates are 1 (or -1 respectively)
bitter hemlock
visual moon
why do they need triangles?
they dont im just doing it for my sake because im curious lol
visual moon
okok i see
bitter hemlock
so I can't really show it on the unit circle in that case unless i just draw the...
what are you saying you can't show on the unit circle for these particular positions on the unit circle?
if you are starting out with the equations sin t = 1 or sin t = -1 and want to think of it visually using the unit circle:
well that means, we're looking for points on the unit circle whose y coordinates are 1 and -1 respectively, so we can draw the lines y = 1 and y = -1 and see where they intersect the circle:
visual moon
see here for example you can see the triangle for pi/6, in this case would it even be possible to have a triangle like that on the unit circle for sinx = 1 and sinx=-1?
bitter hemlock
bitter hemlock
see here for example you can see the triangle for pi/6, in this case would it ev...
?
do you mean for cos x = 1, -1 ?
those ones are degenerate ones i guess
visual moon
https://youtu.be/Dsf6ADwJ66E?si=a7cNIfQkT6dgPwt_
I’ll check it out thanks
bitter hemlock
well for certain points on the circle, you'll simply have an x or y coordinate of 0 so you can't draw triangles with the components along x axis or y axis
but just think of the cos as the x coordinate and sin as the y coordinate, you don't need triangles for visualizing stuff necessarily, just helps to see the relationship cos^2 + sin^2 = 1, but that still holds if cos or sin is 0 and the other is 1 right?
topaz sinewBOT
@visual moon Has your question been resolved?
neon iron
The equation 63x + 70y + 15z= 2010 has an integral solution. [True/False]
terse seal
=0??
Oh, I was thinking another topic I'm not great with calculus. Hope someone else can help!
Friend just sent me this :To determine whether the equation (63x + 70y + 15z = 2010) has an integral solution, we can utilize the properties of linear Diophantine equations, which are of the form (ax + by + cz = d) with integer coefficients.
The general solution exists if and only if the greatest common divisor (GCD) of the coefficients (a), (b), and (c) divides the constant term (d). In this case, we need to check whether (\gcd(63, 70, 15)) divides 2010.
-
Calculate (\gcd(63, 70, 15)):
-
(\gcd(63, 70)):
- Prime factorization of 63: (63 = 3^2 \cdot 7)
- Prime factorization of 70: (70 = 2 \cdot 5 \cdot 7)
- Common factor: (7)
- (\gcd(63, 70) = 7)
-
(\gcd(7, 15)):
- Prime factorization of 15: (15 = 3 \cdot 5)
- Common factor: None
- (\gcd(7, 15) = 1)
-
Therefore, (\gcd(63, 70, 15) = 1)
-
-
Check whether 1 divides 2010:
- Since 1 divides any integer, 1 divides 2010.
Since the GCD of the coefficients is 1 and it divides the constant term 2010, the equation (63x + 70y + 15z = 2010) has an integral solution.
Therefore, the statement is True.
thorny flameBOT
Nicole
terse seal
I don't get it but hope it helps 🙂
neon iron
Thanks
earnest solstice
**Nicole**
thats kinda odd, cuz every linear eqn will have an integral soln that ways, cuz 1 divides everything
earnest solstice
oh its like that?
neon iron
In that case it will depend on whether gcd(a,b,c) | d or not
earnest solstice
then yh i got it
distant totem
x/2-((x+3)/6) = x-2((x-4)/4)
topaz sinewBOT
x/2-((x+3)/6) = x-2((x-4)/4)
distant totem
How could I make the fractions to whole numbers
gilded edge
Multiply everything by 6
distant totem
Multiply everything by 6
So just curious, why?
gilded edge
Do it and you will see wht
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still goblet
Can someone explain the center right please
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Can someone explain the center right please
@still goblet Has your question been resolved?
still goblet
Can someone explain the center right please
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still goblet
Can someone explain the center right please
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pearl fog
ABC is a right angle triangle with hypotenuse BC, there is an incircle with center O with radius r being an integer, line l is parallel to BC and passes trough center O, it touches AB and AC at P and Q respectively, and passes the incircle at x and y such that x is closer to AB than y is to AB, if PX=1.5 and QY=4, find the area of the circle
topaz sinewBOT
ABC is a right angle triangle with hypotenuse BC, there is an incircle with cent...
silk crest
still got 5m in the purse?
pearl fog
i dont have the question on me
pearl fog
still got 5m in the purse?
what
pearl fog
oh yea i think
pearl fog
wait wtf you have a 3 letter username??
cursive patrol
no it's mez3141
pearl fog
no his mc username
cursive patrol
oh ok
topaz sinewBOT
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cursive patrol
setting up a relation to solve for r is probably how i'd start this
refer to line QP and make some right triangles you think might help
topaz sinewBOT
@pearl fog Has your question been resolved?
pearl fog
waiy
topaz sinewBOT
@pearl fog Has your question been resolved?
topaz sinewBOT
@pearl fog Has your question been resolved?
topaz sinewBOT
@pearl fog Has your question been resolved?
topaz sinewBOT
@pearl fog Has your question been resolved?
molten raven
ABC is a right angle triangle with hypotenuse BC, there is an incircle with cent...
use the similar triangles and pythag
B and C are irrelevant, just focus on APQ
and use the similar triangles formed by the tangencies of the semicircle with AQ and AP
pearl fog
yeah i think so
oh
tangent AP to semi is N, and AP to semi is M
r^2+MQ^2=(r+4)^2
r^2+NP^2=(r+1.5)^2
(NP+r)^2+(MQ+r)^2=(2r+5.5)^2
like this?
molten raven
yeah that works
you can also write MQ or NP in terms of r
because NPO similar to MOQ so youll have a system of 2 variables instead of 3
molten raven
NO/PO=MQ/OQ
you know NO, PO, MO
then MQ^2+MO^2=OQ^2 and everything is in terms of r
pearl fog
uh ok
molten raven
does that make sense?
pearl fog
ye
pearl fog
okay
pearl fog
NO/PO=MQ/OQ
would this be r/r+1.5=MQ/r+4
molten raven
OQ=r+4
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fading zenith
Hello! I just want to ask you guys for second opinion. Is the instruction asking us to find the summation formula for i or 2i+1 (considering that it was emphasized by: "given that" and so on)
topaz sinewBOT
Hello! I just want to ask you guys for second opinion. Is the instruction asking...
gilded edge
For i
fading zenith
so the equation after the "given that" is just a distractor? (since I can get the formula for the summation of i even without that)
gilded edge
It wants you to derive it from that
fading zenith
now that makes it complicated...
thanks though! let me think this thoroughly
last question, so I could have some directory. Is the Sn still n(n+1) all over 2?
old crystal
Yess
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frail reef
I would assume that n is an integer if you are being asked to prove this
thorny flameBOT
Meolve
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bronze bough
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fast jay
yep
inner wren
!noping
topaz sinewBOT
Please do not ping individual helpers unprompted.
inner wren
,w 14*193
bronze bough
i did it
thorny flameBOT
bronze bough
then?
bronze bough
i have 1 doubt
topaz sinewBOT
No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/
bronze bough
see if a^2+b^2 can be equal to (a+b)^2
inner wren
?
inner wren
?
fast jay
it's faster
inner wren
you mean +2ab?
fast jay
but (a+b)^2=a^2+b^2+2ab
from this you get a^2+b^2=(a+b)^2-2ab
which is faster to compute
inner wren
yeah
first expand (a+b)^3 = a^3+b^3+3ab(a+b)
a^3+b^3 = (a+b)^3-3ab(a+b)
= (a+b)(a^2+b^2-ab)
fast jay
and you already have the a^2+b^2 aswell
inner wren
and a+b and ab
fast jay
same as before
bronze bough
should i rationalize the denominator or directly add it?
thorny flameBOT
inner wren
,w 1/(root5+root2)
thorny flameBOT
inner wren
rationalize or take lcm
fast jay
again ratio
bronze bough
but the denominators are same
bronze bough
cant i do 7-4root3 - 7+4root3?
inner wren
?
modern rock
thorny flameBOT
Meolve
bronze bough
can u pls tell me the answer?
inner wren
$(5+root3)(7+4 * root3)-(7-4 * root3)(5-root3)$
thorny flameBOT
Meolve
inner wren
yw
bronze bough
now this is coordinate geometry
can u help me?
pls
@inner wren
hey
are you there?
inner wren
hi
ok
find the height of the triangle
it's an equilateral triangle so side lenght = QR
bronze bough
can i use pythagoras theorem?
inner wren
yes, to find the height
inner wren
yeah, 4
bronze bough
-4 or +4?
inner wren
yeah
bronze bough
therefore co-ordinates of p=0,4?
inner wren
noooooooooooo
bronze bough
oh
inner wren
the height will be the coordinate
bronze bough
ok
inner wren
(0, height)
it is asking of the point P.
so find height that will give the distance of P from 0.
inner wren
that is just the origin
bronze bough
so px is the height
inner wren
yeah
inner wren
yeah
draw the perpendicular bisector of QR. Use congruency to prove that QX = XR
if you want to
since ang PQX = 60 = ang PRX
Use AAS.
inner wren
ydea
bronze bough
xr=2?
inner wren
yeahy
inner wren
xr^2+xp^2=rp^2
yeah
inner wren
yeah
inner wren
or 2*root3
bronze bough
so the coordinates are 0,2root3
inner wren
correct
bronze bough
thank you for your help
you have cleared my doubts in number system and co-ordinate geometry
thank very much
bye
topaz sinewBOT
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potent blade
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potent blade
Don’t know how to solve this
topaz sinewBOT
@potent blade Has your question been resolved?
tulip basin
Is the answer (a)
neat quarry
Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
topaz sinewBOT
Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\t...
thorny flameBOT
_bensbeans
topaz sinewBOT
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heavy timber
Limit of (2x^(-1) -x^(-2)) (4x^2 +1)^1/2 as x approaches infinity
heavy timber
So what i did was multiply by x^2/x^2
The equation became( (2x-1)(4x^2+1)^1/2 )/x^2
That can be broken down into two limits
(2x-1)/x and (4x^2 +1)^1/2 /x
The first limit is zero
Why is the answer not zero
coarse tusk
$\lim_{x\to\infty}(2x^{-1}-x^{-2})\sqrt{4x^2+1}$
thorny flameBOT
kheerii
coarse tusk
Is it this?
heavy timber
Yes
heavy timber
Yes.
coarse tusk
Okay but both of those limits are non-zero
heavy timber
Okay but both of those limits are non-zero
They are undeterminate
Both exists
Right?
thorny flameBOT
kheerii
heavy timber
**kheerii**
How
coarse tusk
divide the numerator and denominator by x
or just separate the 1/x lol
like separate the fractions
$\lim_{x\to\infty}\left(2-\frac1{x}\right)$
thorny flameBOT
kheerii
heavy timber
2-1/x
coarse tusk
it is
heavy timber
Yeah and the other limit two and the answer is indeed 4
coarse tusk
yeah
heavy timber
Wait wait lemme comprehend
coarse tusk
but splitting limits like this isn't a good idea
you should evaluate the whole limit together, unless you know that one of the limits you removed exists and is non-zero, finite
heavy timber
**kheerii**
I did the same thing by dividing by x and somehow i got 0/1
coarse tusk
that means you went wrong somewhere
$\lim_{x\to\infty}\frac{(2x-1)\sqrt{4x^2+1}}{x^2}$
heavy timber
that means you went wrong somewhere
I said 2x/x is 0
thorny flameBOT
kheerii
coarse tusk
solve this as one limit
heavy timber
Nooo why
coarse tusk
but splitting limits like this isn't a good idea
heavy timber
you should evaluate the whole limit together, unless you know that one of the li...
Ah- too many requirments
coarse tusk
yes.
coarse tusk
**kheerii**
it's the same thing though
just divide the x^2
$\lim_{x\to\infty}\frac{2x-1}{x}\cdot\frac{\sqrt{4x^2+1}}{x}$
thorny flameBOT
kheerii
coarse tusk
$\lim_{x\to\infty}\left(\left(2-\frac1{x}\right)\cdot\sqrt{4+\frac1{x^2}}\right)$
thorny flameBOT
kheerii
coarse tusk
now just replace all the 1/x terms with 0
heavy timber
2*2
heavy timber
**kheerii**
Wait uhm
heavy timber
**kheerii**
Oooh okay
heavy timber
**kheerii**
But this is lowkey the same thing
Ig i was lucky when i seperated
Ok ty kheeri
Ill keep the channel open
Im solving the advanced questions section in my workbook
And theres only 1 ledt
Soo
Ehm ehm
I failed
Or idk i dont wanna continue bc its getting out of hand
Lemme send the question on pAper
Ignore the stuff in arabic
Uhm
Yeah
Ifk where to start tbh
I divided by x^3
By x^2
By x
And by sqrt(x)
I faced a problem in each attempt
And i dont think lhopitals would work here
Maybe we can use it to simplify the expression but idk
coarse tusk
you actually have to divide by x^(3/2)
heavy timber
you actually have to divide by x^(3/2)
Oh wait
coarse tusk
whenever you have a rational function like this, with "polynomial-like" bits (square roots etc) with x tending to infinity just check the highest degree of the numerator and denominator
heavy timber
whenever you have a rational function like this, with "polynomial-like" bits (sq...
X*sqrtx
X^3/2
coarse tusk
yeah, that's the highest degree
you can actually ignore all the other terms
if you don't want to you can divide by x^(3/2) and then see that the other terms in the numerator will go to 0
coarse tusk
the cube term inside the square root in the denominator is also of degree 3/2!!
heavy timber
Yeah i see i see
heavy timber
the cube term inside the square root in the denominator is also of degree 3/2!!
Yes
I saw
So
coarse tusk
so, you can ignore all other terms
heavy timber
yes indeed
I dont think ill check the answer since you said its 5/2 
tysm sir kheeri
coarse tusk
nw
heavy timber
.close
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neon iron
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neon iron
is this the correct way of doing it
i got the answer, just wanted to know if there's some other easier and less time taking method/way
(ping when reply pls)
topaz sinewBOT
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covert slate
@neon iron i guess this is basically the same thing you did, but recall X' = AdjX/|X|, so AdjX = |X|.X'. (using apostrophe for inverse)
so Adj(Q'BP') = |B|.(Q'BP')' = |B|PB'Q (since |P|=1=|Q|), which is then just PAQ
covert slate
the determinant is just a number, you can sneak it into anywhere
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heavy timber
Ehm
topaz bolt
yea
cedar wagon
the coordinates of A(x;y) where x is the number on the vertical of A and y on the horizontal
topaz bolt
just need b)
cedar wagon
bold nebula
it'll be the same rectangle in fact
bold nebula
you've tried A = (7,4), right?
bold nebula
here
it'll be same rectangle, but the A'' is mirrored about the line y = 5
and after the first transformation it was (7,6)
topaz bolt
ty I get it now
topaz sinewBOT
@heavy timber Has your question been resolved?
civic schooner
this is a proof from my book,
topaz sinewBOT
this is a proof from my book,
civic schooner
why exactly did we need the prime subfield step?
ie, why does $|(K^{\mathcal{B}})_0|=|\mathcal{B}|<2^{|\mathcal{B}|}\le |K^{\mathcal{B}}|$ not work along the same lines?
thorny flameBOT
esca (@ with reply)
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wintry plinth
i'm currently learning about probability and i'm not sure how to tell whether an event is independent or dependent
topaz sinewBOT
i'm currently learning about probability and i'm not sure how to tell whether an...
wintry plinth
like would it be P(both freshmen) = (P(being freshman))^2
or P(both freshmen) = P(being freshman) * P(being freshman | freshman already selected)
broken niche
wintry plinth
i see
i'm not sure when the whole replacement thing applies cuz there was another problem where it was like 60% of the us population owns a smart tv, whats the probability that you select 2 americans with smart tv
and it was just (0.6)^2
even though you're not replacing anyone
ig at that scale it doesn't really make a difference
sudden spade
Yea when the set is "large", we assume no replacement would hardly effect the probability so it's a valid approximation
In the case of 20 students, we have a high dependency on whatever the first student is. The answer is quite simple though:
20Choose2=20C2=number of ways of picking unique pairs
Likewise
7Choose 2=7C2=number of ways of picking unique pairs of freshman
I'll let you get to the final conclusion given that info
wintry plinth
Yea when the set is "large", we assume no replacement would hardly effect the pr...
how large does the set have to be for us to ignore replacement
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sour egret
Am I right?
topaz sinewBOT
Am I right?
sour egret
I drew the blue lines
Angle angle side
Nope I’m wrong ahaha
Why why whyyyy
So sad
deft granite
Am I right?
Why do you think the blue angles are equal?
sour egret
Bisector?
deft granite
What is the bisector here?
sour egret
The line in the middle
deft granite
Why do you think it's the bisector?
sour egret
It’s in the middle
deft granite
What is the definition of a bisector?
sour egret
What is the definition of a bisector?
Divide into two equal parts or congruent parts
deft granite
This is kind of circular, don't you see?
The angles at the tip are not necessary equal
sour egret
Yeah
deft granite
Yeah
Can you imagine two triangles that share a side and have the opposing angle equal, but are not congruent?
sour egret
Yes
deft granite
Well, that's all the information you're given about the two triangles in this task
sour egret
So?
deft granite
Answer 4 is the right one, though it's not worded correctly (they're not necessarily congruent, but there's nothing in the picture that proves they are)
gilded edge
Yes
deft granite
Yes. The triangles are intentionally drawn this way, so that you can't immediately discard the 4th answer
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At
hexed wyvern
help please
topaz sinewBOT
help please
hexed wyvern
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@hexed wyvern Has your question been resolved?
whole geode
!status
topaz sinewBOT
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
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topaz sinewBOT
Show your work, and if possible, explain where you are stuck.
whole geode
@hexed wyvern ^
hexed wyvern
i am stuck on part b and c
i got A doing 2500/30(0.3+.08) since d(x)=.08
2
.close
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whole geode
@hexed wyvern did you figure it out or give up?
Sorry I was helping someone else out
hexed wyvern
oh i joined another channel, would i still be able to get help?
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gleaming badger
Hi
gleaming badger
Im trying to figure out if 3 points are in the same line
My idea was seeing if the slopes of the segments sharing a common point matched
But I dont know how to calculate the slope in R3
coarse tusk
you need to check the direction ratios of the line joining two points
that's how we talk about slope in 3D
gleaming badger
So they need to have the same unitary vector?
coarse tusk
and see if the direction ratios of the line joining A and B are proportional to those of the line joining B and C
coarse tusk
So they need to have the same unitary vector?
sure, you can check the unit vector as well
same thing
coarse tusk
and see if the direction ratios of the line joining A and B are proportional to ...
or just this
gleaming badger
So the slope can be calculated directly by formula only in R2
coarse tusk
the word "slope" only has a meaning in 2 dimensions.
for 3 dimensions the word "slope" gets replaced by either direction ratios or direction cosines
glossy musk
Yo guys! I need some help. I'm going into honors math 7th grade and I'm nervous. Can yall give me some advice?
gleaming badger
for 3 dimensions the word "slope" gets replaced by either direction ratios or di...
Oh i understand
thanks bro
fiery saddle
topaz sinewBOT
rigid ivy
Let u be one of those polynomials
thorny flameBOT
fiery saddle
what ever is at 4
restive inlet
pretty much the same as what they had
restive inlet
and just complete the sub
sub your (x^4 - 3x^2 + 6) for u
and the (4x^3-6x) dx for du
fiery saddle
sub your (x^4 - 3x^2 + 6) for u
and the (4x^3-6x) dx for du
what do u mean dx for du
restive inlet
wdym by first
fiery saddle
nothing
restive inlet
$\int \underbrace{(x^4-3x^2 + 6)}{u}\underbrace{(4x^3-6x) \dd{x]}{\dd{u}}$
restive inlet
yes
fiery saddle
i just don’t u derstand how they got the second last thing
restive inlet
$\int \underbrace{(x^4-3x^2 + 6)}{u}\underbrace{(4x^3-6x) \dd{x}}{\dd{u}}$
fiery saddle
u^1/2
thorny flameBOT
ℝαμΩℕωⅤ
fiery saddle
yeah
restive inlet
there is no u^1/2 there
fiery saddle
i mean u^2
restive inlet
integrating u, applying (reverse) power rule
fiery saddle
but never have i done that in the other examples
is this different cause there’s no obvious compisites
restive inlet
no
no compositions makes this easier
after the sum, you have one of the simplest types of integration problems
fiery saddle
so do i find the anti derticotve of u?
restive inlet
yes
restive inlet
no
fiery saddle
where i go wrong
fiery saddle
like the anti derovotuve of u
restive inlet
why are you setting that equal to x^4
fiery saddle
like it’s not proper way
restive inlet
and setting that equal to x^3/3
and setting that equal to x^3
fiery saddle
im just find ing the anti derivotve
my bad it’s messy
but isn’t the anti derisive of x^4 : x^3
restive inlet
no
fiery saddle
what’s the right aenser
restive inlet
note that you want to integrate u wrt u
you shouldn't be trying to bring x back this early either
the whole point of the sub was to get something simple to integrate
$\int \red{u} \dd{\red{u}}$
thorny flameBOT
ℝαμΩℕωⅤ
restive inlet
can you state the (reverse) power rule (for integration)?
fiery saddle
it’s just minusing the exponent by 1
restive inlet
no
fiery saddle
oh
i’m so confused
like i got it wrong
can u tell me how u would do it
so i can understand
the issue i’m having it’s just a lack of steps @restive inlet
restive inlet
look up reverse power rule integration
notation is lacking a bit,
but from what it looks like you had ZERO issue applying said rule for the question above this
fiery saddle
oh wait
i i was just finding the rule for actual u 💀💀💀
like what it equaled to
i didn’t know it was just u
okay i got it tho tysm
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grave hedge
Can someone help me with this part like how did it get to square root X? This is the given working.
topaz sinewBOT
Can someone help me with this part like how did it get to square root X? This is...
coarse tusk
$\frac1{2\sqrt{x}}\cdot 2x=\sqrt{x}$
thorny flameBOT
kheerii
grave hedge
But wouldn’t it only cancel out the 2? So like X/square root X?
grave hedge
Oh yeah right
grave hedge
The x/sqrt(x) becomes sqrt(x)
Thanks
half field
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violet topaz
2H1. Suppose there are two species of panda bear. Both are equally common in the wild and live in the same places. They look exactly alike and eat the same food, and there is yet no genetic assay capable of telling them apart. They differ however in their family sizes. Species A gives birth to twins 10% of the time, otherwise birthing a single infant. Species B births twins 20% of the time, otherwise birthing singleton infants. Assume these numbers are known with certainty, from many years of field research.
Now suppose you are managing a captive panda breeding program. You have a new female panda of unknown species, and she has just given birth to twins. What is the probability that her next birth will also be twins?
<@&286206848099549185>
long stirrup
suppose another birth happened while you were thinking, what is the probability that it was a species a birth?
violet topaz
suppose another birth happened while you were thinking, what is the probability ...
1/2
violet topaz
what is the probability that it was a twin birth AND a species A birth
1/20
long stirrup
yeah
so 1/20 of the births are that way, and 1/10 of the births is twins and species b
so 2/3 of the twin births are species b
violet topaz
so 2/3 of the twin births are species b
Yes 🙂
long stirrup
so that's the answer, (2/3)(1/5) + (1/3)(1/10)
violet topaz
Makes sense!
Thank you for not just giving me an answer and helping me to solve it.
😀
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sly lotus
The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The longer of the two bases measures 22 feet long. The sum of the base angles is 140° part a use the law of cosine to find the length of the diagonal part B use the law of signs to find the length of the shorter base round your answers to the nearest hundredth.
topaz sinewBOT
The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The l...
sly lotus
I am stuck with part b the diagonal is 20.67 feet I just can’t find the shorter base
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sly lotus
No
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patent silo
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patent silo
I'm going to upload a photo of my work so far
fossil crow
Ello mate.
patent silo
Hello there
The same question is in the minima and maxima section of the textbook as well, and you're supposed to solve it using both lagrange multipliers and partial derivative stuff, with the discriminant like this
patent silo
Ello mate.
I did get the correct answer after a LONG, LONG time with the first section, but I got a negative discriminant so there's that. Kinda just wrote a little note next to my answer and asked for the profs mercy
patent silo
So basically you want to simplify it
Sorry its pretty unclear so I'm going to try to write out what I was trying to do
fossil crow
Yea
patent silo
So the A = (xcostheta + y)(xsintheta) is the area of the cross section. I tried to use a lagrange multiplier with the variables being x, y, and theta, but of course I didn't get the nice, clean stuff like this
I got what kind of looked like an answer, as I was able to get both x and y in terms of theta, but when I put it into symbolab I didn't get a value for theta (it was a very messy answer)
The issue is that it seems like the theta must be a variable for both the area and total length = 2 function, which just makes everything so incredibly messy once I start taking derivatives
I ran into that issue a few times when I was working on the problem by finding all the extrema first
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crystal wing
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eager timber
$u = 8x^2 + 2dx \$
$du = 16x dx \$
$dx = \frac{du}{16x} \$
$when x = 1, new limit = 10 \$
$when x = 0, new limit = 2 \$
topaz sinewBOT
$u = 8x^2 + 2dx \\$
$du = 16x dx \\$
$dx = \frac{du}{16x} \\$
$when x = 1, new l...
thorny flameBOT
Tomi
eager timber
whas the problem
the solution is ln 5
now if I evaluate it between the 2 limits, I get something else
I will end up with ln |u |
coarse tusk
yeah
coarse tusk
no, you plug in t0 for log|u|
eager timber
no, you plug in t0 for log|u|
I changed the bounds of the integration, no?
coarse tusk
$\int_{x=0}^{x=1}\frac{16x}{8x^2+2}\dd{x}=\int_{u=2}^{u=10}\frac{\dd{u}}{u}=\ln |u|=\ln |8x^2+2|\eval_{x=0}^{x=1}$
eager timber
log? it is going to be one over u with respect to du, sure you can combine it like that but it should be still ln, no?
coarse tusk
yeah log means ln
eager timber
no you plug back the stuff that youve taken out as u
coarse tusk
you don't need to do that if you're performing a substitution
thorny flameBOT
kheerii
eager timber
you don't need to do that if you're performing a substitution
ah yes
coarse tusk
if you substitute the value of u back in the limits that you input also change
eager timber
I forgot about this totally, but it still wont be ln 5
coarse tusk
I forgot about this totally, but it still wont be ln 5
it will
$\ln|u|\eval_{u=2}^{u=10}=\ln|8x^2+2|\eval_{x=0}^{x=1}$
thorny flameBOT
kheerii
eager timber
**kheerii**
ln 10 - ln 2 is not ln 5
verbal crater
ln 10 - ln 2 is not ln 5
two logs subtracted divides their argument
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amber star
Let A be not tautology. Let L be Hilbert’s system and we gonna add A as an axioms so we get the following axioms:
1. $A_1 :A\to (B\to A)$
\ $2. A_2:(A\to (B\to C))\to ((A\to B)\to (A\to C))$\
3. $A_3:(\lnot B\to \lnot A)\to ((\lnot B\to A)\to B)$\
4. A \
Prove that in the new system there exists some B such that B and not B are provable.
topaz sinewBOT
Let A be not tautology. Let L be Hilbert’s system and we gonna add A as an axiom...
thorny flameBOT
mtr123
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
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lucid junco
how do i start?
i feel like i don have enough info
topaz sinewBOT
how do i start?
i feel like i don have enough info
pearl fog
you can start by writing it algebraicly
lucid junco
like?
pearl fog
likke lemme write the first
$$ar=\frac{2}{3}$$
thorny flameBOT
Skill_Issue
pearl fog
do the same with the 4th them
lucid junco
yuh i have that step down
so ar^4 = 32/405
wait
oops
wrong question wait
acctually
doesnt matter its the same ques with different numbers
sudden spade
It'll be $ar^3$
lucid junco
ok we do that
thorny flameBOT
Max
lucid junco
yeah
sudden spade
$u_n=ar^{n-1}$
thorny flameBOT
Max
glass maple
try to eliminate a term so that u are only solving for one
lucid junco
is there another equation i can use or something
glass maple
This is sufficient information to solve the question
glass maple
is there another equation i can use or something
think of how u usually solve simultaneous equations
pearl fog
ar = 2/3
ar^3 = 8/27
divide second equation by the first
lucid junco
i thought i could only do that with terms like directly consecutive from eachother like
a, ar, +...+
ar/a so common ratio is r
glass maple
what is the difference between 2 terms
glass maple
its $r^{2}$ right?
thorny flameBOT
Dootud
lucid junco
yeah
glass maple
so then u can find the value of r
if u know what r^2 is, u can determine the value of r quite straightforward from this point
pearl fog
$$ar=\frac{2}{3}$$
$$ar^3=\frac{8}{27}$$
$$\frac{ar}{ar^3}=\frac{\frac{2}{3}}{\frac{8}{27}}$$
thorny flameBOT
Skill_Issue
lucid junco
$\frac{ar^{3}}{ar}=\frac{9}{4}$
thorny flameBOT
tox was here
lucid junco
**Skill\_Issue**
shouldnt it be ar^3/ar
pearl fog
shouldnt it be ar^3/ar
alr, now cross out ar from numerator and denomerator
lucid junco
$r=\frac{3}{2}$
thorny flameBOT
tox was here
glass maple
then u can find a
lucid junco
do i ignore the + or -
glass maple
since it has a limiting sum u can safely ignore the negative solution
wait
ignore that
that was a stupid comment from me
lucid junco
but it doesnt matter if r < 1 right cuz |r|
pearl fog
i dont think you can know weather its ratio is posiyive or negative
lucid junco
do i test both?
lucid junco
huh
pearl fog
$$\frac{ar}{ar^3}=\frac{\frac{2}{3}}{\frac{8}{27}}$$
$$\frac{ar}{ar^3}=\frac{2}{3}\cdot\frac{27}{8}$$
$$\frac{1}{r^2}=\frac{1}{1}\cdot\frac{9}{4}$$
$$r^2=\frac{4}{9}$$
$$r=\frac{2}{3}$$
thorny flameBOT
Skill_Issue
pearl fog
its just flipped
pearl fog
ar is 2/3
ar^3=8/27
pearl fog
cuz u get a different ratio
no
if you simplify youll get 1/r^2=9/4
just flip them so 4/9
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lucid junco
im stuck on part C), i solved the equation and got a = 2,6 but i dont know what to do with these values of a afterwards
topaz sinewBOT
im stuck on part C), i solved the equation and got a = 2,6 but i dont know what ...
grim sable
im stuck on part C), i solved the equation and got a = 2,6 but i dont know what ...
Find the coordinates of b as well
lucid junco
Find the coordinates of b as well
how
grim sable
how did you find a?
lucid junco
i solved the equation
grim sable
when you solved, you should have gotten two solutions for u, right?
also can you send a picture of part A?
lucid junco
grim sable
so when you solve for u, you get both 2^(a/2), and 2^(b/2) as your two solutions
so just use the other solution to solve for b
or, just use the fact that a+b=8
lucid junco
but i get two values of a which one do i use in a + b =8?
grim sable
they should add to 8, right?
lucid junco
yeah
grim sable
what are your two solutions for a?
lucid junco
a = 2
a = 6
grim sable
So a = 2, b = 6
grim sable
looking at the diagram we can say that a < b
lucid junco
right
grim sable
so now you can just plug your x-values for a and b into the equation y = 2^(x/2) to fid their y-values
and then you can just solve for the equation of the line
lucid junco
$P\left(2,2^{\frac{2}{2}}\right)?$
thorny flameBOT
tox was here
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spare smelt
John has 4 apples in one hand and -1 apples in the other. Now suppose he puts the 4 apples through a squaring machine and the -1 apple through a square rooting machine. How many apples does he have now?
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John has 4 apples in one hand and -1 apples in the other. Now suppose he puts th...
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smoky owl
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smoky owl
I factored out the 1/2 and I see that it's integration by parts
not sure next step though
ionic oar
IBP how
ionic oar
[ \f 12 \int e^x \left { \f 1{1 + x} - \f 1{(1 + x)^2} \right } \dd{x} ]
ionic oar
unless there's another way to do it?
there is
smoky owl
how?
smoky owl
wait how did you split the integral like that?
thorny flameBOT
argon
ionic oar
wait how did you split the integral like that?
- 1 - 1 in the numerator
smoky owl
hmm. interesting.... but then how to proceed from there? now I'm even more confused lol
haughty wren
Of course neon finds the one guy asking for integral help before me
smoky owl
with e^x as numerator for both?
smoky owl
and then? I feel like that's even more weird to solve
smoky owl
Of course neon finds the one guy asking for integral help before me
how would you have done it?
ionic oar
and then? I feel like that's even more weird to solve
trust me
craggy haven
"it's not IBP"
proceeds to do IBP
ionic oar
it's gonna cancel
haughty wren
Yeah trust argon, he's an integral connoisseur
ionic oar
or don't, but you'll realize what I meant when you figure it out on your own
haughty wren
"it's not IBP"
*proceeds to do IBP*
it's not IBP tho? Just add and subtract one from the numerator like (x+1-1)e^x and split
thorny flameBOT
argon
lucid junco
ionic oar
does this look familiar
smoky owl
ok so I tried to do integral of e^x/1+x
ionic oar
I hate doing this but it's just that
smoky owl
Idk how to solve that
haughty wren
Bro brought out a whole cheat code
ionic oar
good
smoky owl
I'm so confused but thanks ig
haughty wren
Yeah you can also just do by parts right from the get go.
if you want more hint: || u = xe^x, dv = 1/(1+x)^2 dx ||
haughty wren
I wanted to avoid Ei stuff, assuming they haven't seen special functions 
haughty wren
Ah alr
ionic oar
foresight
smoky owl
if you want more hint: || u = xe^x, dv = 1/(1+x)^2 dx ||
so du is (e^x)(e^x)(x) dx and v would be -1/(1+x)^3?
haughty wren
Uh, no. I think you may have gotten confused and differentiated what you were supposed to integrate and vice versa
Differentiate u and integrate dv
smoky owl
yeah, I thought that's what I did?
differentiate u with product rule
oh wait I forgot the plus sign
so du is (e^x) + (e^x)(x) dx and v would be -1/(1+x)?