#help-17
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that's what i told shah, but apparently this whole server is just about maths, maths and maths
1 test never ruins anything bro; but either way dont discourage him @karmic imp check yt and khan acad bro; doesnt seem like we can help
Please leave this channel
its my dream school please i want to ensure i can get in
my math scores in the past hasnt been the best already
Schools look at your gpa, if you have a 4.0 and you fail one test in one subject, you won't get knocked down to a 1.0
wow bro u just wasting ur time here at this point; just going to sleep or messaging a classmate is better bro
damn bro i'm actually studying and wanted to help this guy because he was nice
...
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but can't understand everything because i don't know all the math terms in english
Pls tell me what you don’t understand I’ll try to translate or even simplify it
Well I don’t want to be wasting my time, I actually want to get a good grade at this test
am i supposed to somehow care abt ur little story? @slim lily i think u've already helped quite a lot this past 20min
Please it’s already really late for me and I have to wake up early
if u not gonna trust me its fine but bro this place isnt guaranteed help; msg a friend; wake up early and study
oh really, then help him einstein
Mehdi can you help me out
I think these questions are the most important
If I can at least get them right I might be able to achieve a decent grade
This is question 8 graph
By my friend
Hello?
<@&286206848099549185>
Please
I can help I guess
What is the question though? I don't understand what you've written down
Right but I don't understand those questions lol
Can you type them here in complete sentences
Is this homework that you're getting from a friend?
@azure zinc Has your question been resolved?
Ping me whenever you reply
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how can I prove easily that sqrt(5) + sqrt(7) is irrational. I tried to assume for the sake of the contradiction that it was, in fact, a rational number, so that:
sqrt(5) + sqrt(7) = a/b, where a and b have no common factors. got some weird result in regards to a being equal something with b, but it also contained a on the other side.
Try squaring
Did you prove that sqrt(5) is irrational?
nope
Consider showing it to understand why this proof has actually sense
alright, thanks, I'll give it a shot
Suppose sqrt(5) + sqrt(7) is a rational number, what do you get when you multiply it by sqrt(5) - sqrt(7)?
Exactly, now take sqrt(5) + sqrt(7) to the other side of the equation
No, we won't use that definition for this proof, since it will be hard to find a contradiction
(√5 + √7)(√5 - √7) = -2
(√5 - √7) = -2/(√5 + √7)
This shows that (√5 - √7) is also a rational (Since division by 2 rationals is a rational, and we supposed that √5 + √7 is a rational)
oh, I see
2sqrt(5)
Yep, sum of two rationals is also a rational
and if you divide that result by 2, it should also be a rational (since 2∈ℚ)
but we know that √5∉ℚ, which is a contradiction
right, but from the get-go, we also knew that sqrt(7) is also not a part of Q.
and don't we also know that the sum of two irrational numbers is also irrational?
so from that point, it's easy to prove
expect for like sqrt(2) + (-sqrt(2)), or something like that
Not always
what if we write sqrt(5) + sqrt(7) as an x, where we assume that x is a rational number
and do the calculations
Consider (1 - pi) + pi
oh, that's a fair point, not gonna lie
technically, we could get to a form where we have x on one side and an irrational number on the other side, which would prove that it is not true to say that sqrt(5) + sqrt(7) is a rational number
right?
Yep, but you would have to manage to get that result in the first place
[Proving that sqrt(7) + sqrt(5) is irrational]
Certainly, let's use a different approach to prove that (\sqrt{5} + \sqrt{7}) is irrational without explicitly introducing the ratio (a/b).
Proof:
Assume for the sake of contradiction that (\sqrt{5} + \sqrt{7}) is rational.
Square both sides of the equation (\sqrt{5} + \sqrt{7} = x), where (x) is a rational number:
[(\sqrt{5} + \sqrt{7})^2 = x^2]
Expand the left side of the equation using the FOIL method:
[5 + 2\sqrt{35} + 7 = x^2]
Combine the constants on the left side:
[12 + 2\sqrt{35} = x^2]
Now, we have:
[2\sqrt{35} = x^2 - 12]
Rearrange the equation:
[\sqrt{35} = \frac{x^2 - 12}{2}]
Now this, we do know that is irrational. [\sqrt{35}] is not part of Q, we know that already.
This is what I got from my good old pal, GPT
here is a better picture
calculations seem legit
Yep, I supposed it would be easier to get a good proof with √5 rather than √35, since it should have been proven before
hi
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What is the difference between the slope of a tangent line and secant line and how do you identify when to use each?
@limber tiger Has your question been resolved?
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Tangent lines are lines
Lines only have one slope
But a curve may have different tangent lines at different points on the curve
And a secant line isn't a tangent line
It's a line that's formed by two points on the curve
Whereas tangent lines are formed only by one point on the curve
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What's the original question?
create 5 vectors that will create a state of equilibrium with the given equillbrant (13N N27degreesW)
@vast shale Has your question been resolved?
Does the 13N mean 13 newtons?
yeah 13 newtons and then north 27 degrees west
Ok so we need 5 vectors that when added up with the 13N N 27° W vector we have 0?
Ok that makes sense
i have a copy of the assignment that i can send in a bit
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Slope of Secant line = f(2) - f(0) / 2 - 0
(0 , f(0)) (2,f(2))
f(x) = distanced travelled in km , x = 0 at start of the race
how can I find the slope without an equation, I have no idea what the graph should look like
I guess f(0) = 0 since at the beginning of the race he travelled nothing
I know the average velocity at the first 2 hours mark and the average velocity
so would I just do 45 - 39 / 2?
6/2 = 3
but that doesn't sound right
Lets start here: whats f(2)?
39
Not quite
x = 0 at start of the race? so I'm guessing 2 is 2 hours after
which gives my average velocity to 39 no?
Reread the problem
Two hours after the start ... a cyclist passes the 78-km mark
correct
So what is f(2)
Indeed
I thought it was the velocity
Can you figure out the slope of the secant line from there?
like I found it, it's 39
assuming f(0) is the start of the race which should be 0 right?
if f(0) is just 0, it would just be 78 / 2 = 39
... I'm kind of upset at myself for not seeing it earlier
It happens, glad you could get it fixed
Also because it basically asked the same thing twice that I didn't think it could be the same answer
but thanks a lot for the help
.close
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Youre welcome
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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
@dusty void Has your question been resolved?
@dusty void Has your question been resolved?
So to know the speed the plane goes you can use $s=\frac dt$ where $d$ is the distance and $t$ the time he took to do the distance.
Joseph.P
okay
And s is the speed
Did you find it ?
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i'm not really sure what to do here
do I distribute the t, then set one vector equal to the other to find that t value? then what?
i assume youre aware of the dot product?
you only need to focus on the direction vectors and find the angle between them
thats equivalent to finding the angle between the lines
are the direction vectors the ones with a t next to them?
t and s, yes
so I do the dot product of <6,2,8> and <-3,-6,-1>?
like the one with the cos theta in it
cos, but yeah
solve for theta
check its the smallest angle though, it can help to draw a diagram for that
youll either get the b or the a
note they will sum to 180 so you can subtract one from 180 to get the other
ah
okay let me try part a...
r(t) dot r(s) = -18+-12+8 = -22
-22 = |r(t)| |r(s)| cos(theta)
-22 = sqrt(104)*sqrt(46)*cos(theta)
arccos(-22/(sqrt104*sqrt46)) = 108.55
180-108.55 = 71.75
check your last line
hmm, am I not supposed to subtract from 180? It is the larger angle...
oh wait
71.45
heh
so how come I use the direction vector here?
What do the first set of vectors (without the t/s) indicate?
the first set of vectors are a position vector for some point that lies on the line
the direction vector is a vector that determines the direction of the line, you can get any point on each line by changing the value of s or t respectively
so for any two lines, the angle between the direction vectors is the angle between the lines when extended infinitely
i should probably note you can only find the angle between lines that will actually intersect at some point, but that may be obvious
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What's another way to write out 42 times 57? At first I tried (40x50) + (2x7) and realized it's wrong. What is an easy way to think it out like this? Thx
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Not sure what steps to take from here to get the final answer
@blazing trail Has your question been resolved?
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help
Because 3/2 is a constant
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would anyone know how to do this porblem (linalg)
Find the nullspace (kernel or whatever you call it) of the matrix formed by [v1 v2 v3] and then just scale it so that the first entry is 1
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I found average velocity in A by substituting 1 and 4 into x^2 + x
which gave me 20 and 2
so the change would be 20 - 2 / 4 - 1 = 18 / 3 = 6
but for B, f(1) = 2 and f(1+h) = h^2 + 2h + 1
so the velocity would be h^2 + 2h + 1 - 2/1+ h - 1 which really just h
h^2 + 2h -1 / h is not factorable
so how can I proceed?
Isn't f(1) = 2 and f(1+h) = h^2 + 3h + 2? It's y = f(x) = x^2 + x
But your answer seems to be based off just f(x) = x^2
let me check
ah
I see
yes you are right
so it should be h^2+3h/h
which I can factor out an h
yup!
but that doesn't give me the average velocity right?
at least entering 3 gives me the wrong answer
but h approaches 0
ok so I can't just plug that in
and is instantaneous velocity just the derivative?
well yes but also that's exactly when you let h approach 0
even thought it says x = 1?
the average velocity between t = 1 and t = 1+h is h+3, so instaneous is let h = 0
I guess I mean x, not t, but yes we have from x = 1 to x = 1+h right
so you just set h = 0 for instaneous velocity at x = 1
which would be 3
ok thank you! I always try to plug in 0 at h when I see limit approaching 0
Haha no limits in averages! only instantaneous
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Many income tax systems are calculated using a tiered method. Under
a certain tax law, the first $100 000 of earnings are subject to a 35%
tax; earnings greater than $100 000 and up to $500 000 are subject to
a 45% tax. Any earnings greater than $500 000 are taxed at 55%.
Write a piecewise function that models this situation.
Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.
Find the value of k that makes the following function continuous.
Graph the function.
f (x)= { x^2-k, if x <-1
2x -1, if x >= -1
Statufi
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
@vast shale Has your question been resolved?
no
Statufi
thats not what the questons asking lol
Statufi
@vast shale Has your question been resolved?
Limit
nah im not doing limits
Sorry for that, i don't know your math background
That's ok
You just have to find a k that makes the two expression of your function coincides at -1, so replace x by -1 and solve
can u explain a lil. more
You want to have k such that $x^2-k=2x-1$ for $x=-1$
Statufi
Because you know how f behaves to the left of -1 and to the right, and you want these two pieces to be connected at -1 so there values must be equal
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anyone here has a video that can teach me in about 2-3 days Analytic Trigonometry II (SummDiff, Mult Angle, Power Red, ProdSum, SumProd) and fundational trigonometric identities 😄 we got exam in a week
by teach i mean master that shit 😄
Khan academy
anymore?
Youtube
no i mean either a channel name or a website..
You're not gonna watch all of Khan videos in 3 days
Help channels are for specific problems. Come back when you start doing them
Google them
Like look up that topic in youtube, watch the videos
aye thx
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Im in calculus 3 and I just want to know if my answers are correct to my questions?
I tried to graph both of these and i can show my sketches so im unsure 😓
pulling up my phone to show sketches
sorry for the bad sketched
this is byfar my worst math
💀
ignore #3, i'm working on that one rn
@high abyss
<@&286206848099549185>
uh
😭
Ig while I wait I'l lzoom the questiosn in
^ I updated to a Family of Circles
<@&286206848099549185>
that's right
@spiral inlet the first question looks like 2 parabolic graphs on the xy plane right??
parabolas? or hyperbolas?
hyperbolas*
yeah, it's a set of hyperbolas. the two different directions come from if f(x,y) is positive or negative
so x^2 - y^2 would also be correct
but x^2-y^2 = z gives hyperbolas with a vertical or horizontal axis of symmetry
no, x^2 - y^2 gives a hyperbola, but one with a horizontal or vertical line of symmetry
not the diagonal ones like we see in that graph
a diagonal hyperbola is like xy = 1
or like, you're probably more familiar with it in the form y = 1/x
yep that is more recognaizable
so that means
f(x,y) = xy
would be the correct answer as it is not horizontal line of symmetry nor vertical
this one is diagonal
@spiral inlet may i check my work with one more question?
i already finished it and can send neat outlined work
sure
I found the domain limits but I'm more unsure abt the shading I think
for the ln, you should have 2-y > 0, not just 2-y not equal to 0
oh 💀
ln(x) is not defined in the reals for negative x
y < 2
yeah
Your graph looks mostly right, you just shouldn't have the bigger area on the top right
just the bit by the origin
yes, like this
that's where y > x^2 (above the parabola) but y < 2 (below the line)
and yeah x being positive or negative doesn't make a difference
👍
if you dont mind as well soryr to hold oyu
could you explain the 2nd question furtheR?
i just completely wildly guessed with the x^2 + y^2 and the 2x
Repost as to not scroll @spiral inlet
oh yeah sure
seems like you are pretty familiar with the general equations for conic sections?
you recognized x^2 - y^2 as a hyperbola
iffy but yeah that's the unit we're working on
are you pretty comfortable with like the equation of a circle, ellipse, etc
yes
so the equation of the curve given is $z = \frac{2x}{x^2+y^2}$ where $z$ is the height of the function where we're taking the level curve
tatpoj
$$z = \frac{2x}{x^2+y^2}$$
$$z(x^2+y^2) = 2x$$
$$zx^2 + zy^2 = 2x$$
tatpoj
if you gather the x terms, and gather the y terms (there's only one in this case), and complete the square on each, you will find an equation for a circle
so it would be
(zx^2 + 2x + 1) + zy^2 = 2x + 1?
not quite, I haven't actually done it out on paper yet, one sec
are you comfortable with completing the square?
yes
i understand how to do it and the lesson itself, i think here just the z is throwing me off 😵💫
well we just have to treat z as a constant
since, for any single level curve, it is a constant
ok so treating z as a constant i like ot think of any constant as just "1" in the scenario
x^2 + y^2 = 2x
(x^2 - 2x + _ ) + (y^2 + 0y + 0) = 0
(x^2 - 2x + 1) + (y^2) = 1
(x-1)^2 + (y)^2 = 1
add back in z i believe
(zx-1)^2 + (zy)^2 = 1?
well, you can interpret z as 1z, but you can't just interpret it as 1
$$zx^2 + zy^2 = 2x$$
$$(zx^2 - 2x) + zy^2 = 0$$
$$z(x^2 - \frac{2}{z}x) + zy^2 = 0$$
tatpoj
this is how I would start
yes 👍 awesome
and to make it look like the circle equation fomr of (x-h)^2 + (y-k)^2 = r^2
i need to simplify the first part of the circle equation
well just finish completing the square, yeah
would you re-distribute the z back in to help with this part?
$x^2 - \frac{2}{z}x + \frac{1}{z^2} = (x-\frac{1}{z})^2$
zx^2 - 2x + 1/z
(zx - 1/z)^2?
tatpoj
ok
yeah that makes sense
i would double check the answer with foil to mkae sure too
so with level curves, z would be treated as a constant
then try to get it to fit within a certain type of equation shape
whether it's a sphere, ellipse, circle, cone, etc
oh hence the 2d answers
oof that's never good
so right away we oculd've elimianted cones and cylinders
yeah and
even as far back as zx^2 + zy^2 = 2x
you can already kind of see that it's going to be a circle
since the x^2 and y^2 terms both have the same coefficient: z
they are scaled by the same amount
just didnt know the math to actually get to the formula like in case it was an FRQ and not MCQ like the last one
oh yeah I gotcha
so a scenario to look out for hyperbola
completing the square like we did here should always work, but you might have to do it on both the x and y parts
is to get it to be 1/x and 1/y
and parabola would be different coefficients of x and y?
hm
I think you'll recognize hyperbolas like that if you can make the x's and y's multiply with each other
like y = 1/x
xy=1
no other conic section is going to have a term like xy
okok
and what abt parabolas?
just one of them is x^2
and the other is to a degree of 1
right
yes
well, do you have sort of a picture in your head of what a level curve is?
it's like
you have a 3d curve on the plane
the 3d curve's projection onto the xy axis
is the level curve
i think
well, i wouldn't quite describe it as a projection
like imagine you have this function f(x,y) = whatever
every point has a (x,y) input that determines the output z
mhm
right, so the height of the surface represents the output z
z = height of surface
imagine chopping this with a horizontal plane, like at a height of 0.4
do you see that the cross section would be a circle?
yeah
that is the level curve for z=0.4
we chose a "level" of 0.4
and that circle is specifically defined to have a z of 0.4
so z is constant: 0.4
so the level curve at 0 would be kind of like 2 circles?
yeah u get what i meant
so what's the purpose and stuff of finding domain for level curves??
well essentially it's the set of values that satisfies an equation like f(x,y) = 0.4
if you want your function to output a certain value, the level curve is the set of all inputs (x,y) that give that value
so what abt stuff like this
Does finding the domain only tell us like
how the graph would look like on a 3d plane?
what you were finding in that problem was essentially the "footprint" of the graph
yeah, wherever z is defined
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I don’t really know how to start on this either
if a+b is a factor of the equation, then setting a=-b in the polynomial must yield zero
How do I factorise it tho
divide it by (a+b)
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Help please
Math 124 it’s algebra
could you take a screenshot instead of a picture of your screen
No I can’t it’s a practice test
It’s locked
Sorry
I could take a better photo tho
As good as it gets
?
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Why here when i add 60 degrees it counts up to 330 but the correct answer is 300
Here i add 30 and its righr but why is. the 60 wrong
<@&286206848099549185>
sorry, I can't read your handwriting. What is the problem to solve
let me log into pc its easier to thpe there one sec
np
so basically, in second picture, i have complex number -sqrt(3)-i, and then i get that the angle is -30 and -30, and Re<0 Im>0 so i draw the angle 30 in the third sector anticlokcwise, and i get 180+30 = 210 degrees and thats right
but in first, i have complex number 1/2 - sqrt(3)/2 * i , and i get that sin angle = -60 and cos angle = 60, and as re < 0 , and im > 0 its fourh sector so i draw the angle in fourth sector and i get angle 270+60=330 but thats not what i should get, so i want to know did i draw it wrong or what
you mean 270 + 60?
idk should i even draw it there, someone explained to me yesterday and i did understand it but on this one i get trick
okay thank u
maybe i should have drawn the angle different way, so then i get 270+30 , but i think it should be drawn anticlockwise right?
ok im getting it now
the values of cos and sin tell u its Q4, that's definitely right. So you can be safe to say that the angle of theta is a form of pi/3 in the 4th quadrant
it's not really a matter of matching a pattern of where to rotate the angles, but you're basically using the given info to deduce where and what angle theta is
So i should just use -Pi/3 and not add anything to that?
yeah, there's no pattern to the problems or anything if that's what you were thinking before. You just need to figure out the quadrant and angle based on the given formula. If its in quadrant 2 and sin and cos tell you it's a pi/4 angle, you would deduce that it was 3pi/4. Get it?
Always keep a unit circle handy if you don't have it memorized already too 😉
So thats 45 + Pi/2 = 135 right?
we have Pi/4 in second, and then we add value of the one behind it ( first only ) and we get Pi/4+ Pi/2
I understood that, but i didnt know why i dont add the 270 to 60
so this is this circle, and as Re>0 and Im<0 , i can just say its -60 as it belongs in the right quadrant right?
That's what I would think. What does the exact question ask?
I just need to convert to complex trigonometric form
Like r(cos angle + i sin cos angle)
yeah that should be fine then. do you know if the answer needs to be in radians or degrees
Also here, would it be right if i just used the value of -30 degrees ( -Pi/6) instead of 210 ( 7Pi/6) ?
you should use the actual angle
what would the actual angle mean
well you can say it's pi/6 but that doesn't tell you what quadrant it's in really
if you say 7pi/6 you know exactly where that is
So its important to mention the quadrant?
no because the angle tells you what quadrant its in
pi/3 is in the first quad
7pi/6 is in the 3rd quad i believe
Y
Yes
Does it stand for like
-Pi/3 is in fourth?
Or i need to write it as 5Pi/3 ( 300 )
no you need to write it as the whole 5pi/3 thing
measure angles in radians anticlockwise from the positive x-axis
if that's the case then you need to find what that angle is if you rotate it the opposite direction
like forwards vs backwards
yh, im stating this as a convention.
the usual one.
So that means i should write -60 like this?
oh omg im so dumb i totally misinterpreted that
that -60 is good
and then it makes sense, 270 + 30 = 300
yes -60 will work lol
i havent seen the original question
ye
did i draw the angle right here ?
for that one u either write 210 degrees in the counterclockwise direction or 150 degrees in clockwise direction
so i did 30 and then +90+90 its 210
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- Find the probability that x < 30.
X ~ N(54, 8)
z score is so confusing
Never seen this notation. But is this what it means
μ=54
σ=8
yea
it's like
(x-54)/8
but it's confusing finding Z score because it's like
the same formula
but u come out w a decimal place number
and then you randomly find the Z from that?
idk
. X ~ N(4, 5)
Find the maximum of x in the bottom quartile
this one also confuses me
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quick q, can I use ln properties to move the ^2 so that it becomes 2xlnx?
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Hello! Could you help me with this fraction division? The result should be 68.
But even though I double-check and redo the exercise, it still gives me 17/3 divided by 27/16.
i think the problem must be here
(5/10) * (2/9) should be like (10/90) = (1/9)
@vocal lodge
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can someone please help me understand the solutions of this exercise?
One of my main questions is about the boundsof the first integral which I dont understnad
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Hi! I'm still a bit confused on this delta stuff for limits. Could someone help me approach this problem?
The distance between (x,y) and (5,1) should be less then delta
(x - 5, y - 1) lies inside a circle centered around the origin with radius delta
right
But for all points in this circle the x coordinate is between -delta and delta, right?
I should say disk btw
yes
so that's basically the problem?
Yeah
I think you can prove this more rigorously with the triangle inequality, haven't figured out the exact details though
That's a good idea I'll try that. Thanks!
how would I go about that?
You need to prove that if a^2 + b^2 < delta^2 then a^2 < delta^2 => |a| < delta
Ok
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You have a road that is 1 mile and a car that only has 30 meters before it hits the travelling limit, how much of the could the car cross before hitting the limit? (the limit makes the car stop even at full fuel)
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helppp

since it's pr then you know it's what kind of function?
nevermind
bro what
what is the name of the function
you've already drawn it, right
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point of origin?
direct proportional function
the equation is y=kx
I don't know if you graph it the right way
so when I said function and that you've drawn it I meant that it's a line. the relationship of $ to amount is a line with non-zero slope
just tell me what to wright please
and it' a + slope meaning it point up and to the right like you've drawn
2 pounds of candies=5 dollar
5=2k
It's kinda weird that the slope aren't the same
my bad
dude
it is
it is
5/2
the function expression will be y=5/2x
how
you just graph it bro
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What does this mean? conv is for convex hull, I don't know what is the bar above conv or what cl is (I think its closure)
yes, the closure of the convex hull
so the conv bar is just notation for the closure of convex hull? but what does that mean (closure)?
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I don't get how closure is different from convex hull
That's true! Thank you very much, have a good afternoon.
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that's a Diophantine Equation, do you know how to solve them? :)
It's a very basic one. I saw in the other chat you outlined your strategy and it's good
.close
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T
The top problem with C
ok so I plug in 6 for the original function but then I got a weird decimal number. but I continue that and plugged that answer into the inverse function and it not work. so I try the inverse of 2 first, got that answer and plug in for the original equation and that one makes sense
Little fact about inverses, which is what this exercise was trying to show you $f^{-1}(f(x))=f(f^{-1}(x))=x$
MrFancy
albeit they could've chosen a better function to introduct such a concept
indeed, one of my favorites!
what were they trying to show me
to constraint a domain / range whatever ?
or that maybe one of the function isn’t a function
MrFancy
take some value and put it into f(x)
ok
ok
MrFancy
plug this into $f^{-1}(x)$
MrFancy
,rotate
dang so u gotta do it like that
it should almost work if you decimal anyway
yes I was just deadly weirded out by the 6.000000000001
like it might give you 6.00000000 something but it should be really close still
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for "which of the following could be the graph of y=-2(x)^2" the graph would open downward right?
I think that the one opening downward would be right
I would ask you professor becuase the website may be wrong
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how can i figure this out, im pretty clueless rn
was any more details given about ABCD?
nevermind I was going to ask whether it was a parallelogram but the answers don't line up with that
abcd is a paralelogram
oh it is
but im so confused i tried everything ik i just dont know how to do it
ok so a parallelograms area is base times height
and we have the base of 30 already so we need to find out its height
yea thats where im stuck
yeah me to
and i have no idea
let me think about this lol
any ideas?
ok i think i may have it
give me a second
ok i got it
so
we have the base but we need the height
so notice that angle NDC equals angle DAB
(i'm assuming that they are equal but I think my assumption was right)
but do you see why finding angle NDC/ angle DAB would be useful?
@regal stirrup do you still need help?
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,rotate 270
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I’m trying to complete the square but idk where to go from here.
the LHS is a perfect sqaure trinomial :)
$(a+b)^2=a^2+2ab+b^2$
MrFancy
I feel like I've helped you like 5 times today 
Sounds abt right, srry for troubling u
Teacher wasn’t here today and I got so confused during class
I’ll try that
no worries man just pointing it out, nothing wrong with asking questions :)
its how we learn
do I need to factor it first in order to do the perfect square trinomial equation? bc if I do, idk how to do that
well take a look at the x^2
that's going to be our a^2, right?
well is x=a
ooo
I could see where this could be confusing
ok
so bear with will me for a sec 
ok lol
so imma do a new binomial expansion (d+e)^2=d^2+2ed+e^2 right?
same thing just different letters you'll see why in a sec
ok
Yep
do that middle term 2ed just becomes 2xe, right?
Yes
now compare that with b/a*x and you'll get e :)
Hint: look at the square you drew. What is x^2 + b/a*x + b^2/(4a^2) (the area of the square) equal to?
Like this? 
oh actually that's also an idea
Whats the side length of the square?
Wait that’s the wrong picture
close
My brains tryna process this
Look at what you drew for #4 again
Wouldn’t it be equal to what’s written on the right side of the equation ?
The area of that is what you are trying to factor
yes
So take a look at the side lengths of that square
b/2a * x ?
sorry I’ve never done perfect squares b4 until rn, I’m rlly confused and them
You need to add not multiply
So just b/2a + x ?
Yep
Thats the side length
And area = (side length)^2
We know the area is x^2 + b/a*x + b^2/(4a^2) and we can rewrite that as the side length squared (since we know the side length!)
Would the right hand side of the equation still be -c/a + b^2/4a^2 ?
can you please check if I did this right
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<@&286206848099549185>
yeah and prob 5
product rule on the left
d/dx [xe^y] = d/dx[x] * e^y + d/dx[e^y] * x
that's chain rule
see since y is a function of x
you have to apply the chain rule
so
d/dx [e^y] = e^y * dy/dx
you have to differentiate both sides when doing implicite differentiation
oh you did it later
fine
the derivative of y with respect to x wont be y dy/dx
it's just dy/dx
(on the RHS)
for the LHS, use chain rule, and for the RHS use quotient rule
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log 8 =0.90
, find the value of log√32
What is the common base between 8 and √32
its 10
I am talking about the numbers
yea
Try to do the same thing to √32
thts it 2^5
Well there is a root
so 2^5*1/2
So you know that log 2^3 =0.9
Try to figure out what log(2^(5/2)) will look like
@craggy granite to solve this problem you need to know this
i know this so 2/5 log 2
No its 5/2
Yup
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