#help-33
1 messages ¡ Page 53 of 1
Yeah, I might be missing something that eliminates extraneous solutions
What answers were you getting
This got 60 and 180
Which does seem to be the correct answer
I got w = 1/2 and -1
Which is indeed x = 60 and 180
Thanks for the help
Yeah I messed something up but I don't know where

Oh
Wow, dumb
Messed up arithmetic in the discriminant
Otherwise, a perfect solution đ
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On surface integral questions how do you know the direction of the normal vector you get after performing the cross product to double check the correct direction for the oriented surface?
@dense pollen Has your question been resolved?
Help pls
can you give an example?
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I have the acceleration rate of 980cm/s². How would I change this to a graph with algebra?
x-axis is seconds and y-axis is cm
Just graph it
Is y measuring displacement?
You currently have the form $\frac{d^2y}{dx^2}=980$
(...is that x at the end supposed to be there?)
Oh dear
Frosst
ie. your y (position) valueâs 2nd derivative wrt x (time) is 980, as in 980cm/s² acceleration
Crepas, why do you keep asking calculus questions?
yes
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xD
...what
he hasn't taken any calc/precalc before
supposedly these are "challenge" questions from diffferent math server he is trying to do
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Pick any number $x+y+z=0$. Explain why $\frac{\mathbf{v}\cdot\mathbf{w}}{|\mathbf{v}||\mathbf{w}|}=-\frac{1}{2}$, where $\mathbf{v}=\langle x,y,z\rangle$ and $\mathbf{w}=\langle z,x,y\rangle$
Math Is Fun
!status
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
Have you tried subbing in the vector forms of v and w into the equation
It makes a bigger mess
If you have, show us
well yeah
Ok wait a second
First convince yourself this is true
I got $\frac{xz+yx+zy}{x^2+y^2+z^2}$
Math Is Fun
Please read #âhow-to-get-help
think about what the bottom looks like
for example
what does x^2 + y^2 resemble?
Math Is Fun
You just have to play around with it!
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an airplane is flying at an airspeed of 345km/h on a heading of 040 degrees. The wind is blown at 18km at a bearing 087 degrees. Determine the ground velocity of the airplane and include a diagram in your solution.
@loud field Has your question been resolved?
!status
What step are you on?
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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
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I am confused with properties of matrices
trace of AB= traceA*traceB
should be
so first need is to sequare?
suppose it is defined
this property will hold?
yep
it's pretty special since you know that matrix multiplication is not commutative
i.e. AB=/=BA
but tr(AB)=tr(BA)
assuming A and B are both nxn square matrices ofc
yes
did that answer your question?
you asked a different question though
that 4th property in the 1st screenshot is different from the 3rd
Afaik Trace is a linear functional that's only defined for elements in M_n(F), that is, nxn matrices with entries from F
$tr: M_n(F) \longrightarrow F
tr(A) = \sum_{i = 1}^{n} a_{ii}$
Riku
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
Wait uh why does it say compile error
Try to see what are individual elements in the sum for trace(AB), that is, basically how the diagonal elements of AB look like
Then you can compare
@tender mantle Has your question been resolved?
So what do you think about it?
Hint: If $C = AB$, then
$c_{ii} = \sum_{j = 1}^{n} a_{ij}b_{ji}$
Riku
C ii are the diagonal elements of the new matrix you get multiplying A and B
No it shouldn't
It will be equal 
How does trace A look like
$a_{11} + a_{22} + ...... + a_{nn}$
Riku
All the diagonal elements sum
You get that these signify the diagonal elements?
Similarly calculate trace of B
$b_{11} + b_{22} + ...... + b_{nn}$
Riku
Oh wait hm
Actually i might be wrong
Let's see if I am wrong
That's also going to get the answer
how will the diagonal elements of the matrix be which you get from multiplying A and B
Get those, and then sum and equate
If not equal then yes this is not true always
I tried this
Sorry had to go somewhere, I'm back
Lemme see
Yeah no
They won't be equal
@cold fox
Yes. Thanks riku
Have a good day
Good morning
Ahh wrong tagged
I am arjuuuunnn
I'm sorry if you see this that was a complete misping
Good morning
You can edit it
I mean it still shows the ping to them so anyways
They will be nore confused if I edit it
But yeah, tr(AB) = tr(BA), this is 100% true
You apply the same method
In the end everything will equate
oh shit i gotta cancel my nutro trial or i'll be even more broke
đ
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something wrong with the calculation here?
maybe I should try radian? or it's just wrong altogether
radian doesn't work either
,w 77.25 deg to radians
are you allowed to input trig expressions into your answer thing?
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online math moment
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@frank orbit Has your question been resolved?
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how do i solve physics problems like this in calc 3?
i dont know what to google/lookup to find problems similar like this with solutions
I'd start by drawing it*
yeah but it's gonna end up coming down to a formula i dont know
and will have to memorize
You will know which formula(s) to use, if needed, once you have first visualized the situation
i dont know any of the formulas related to physics types of problems
and i know this isnt an initial value problem
because we dont start out with acceleration
and have to integrate back to position
we're given an angle which means theta/alpha/beta will be present in the formula i have to use
we start with the velocity
we can integrate that to position
and there is an initial value
at time 0, it is in the javelin thrower's hand, at the point (0, 7)
how do i parameterize velocity
into x,y
oh wait
80cos(45) 80sin(45)
i think i see now
a = dv/dt = (dv/dt)(dx/dx) = (dv/dx)(dx/dt) = (dv/dx)v
So a= v(dv/dx)
adx = vdv
And a = -g (y direction) and a=0 (x direction)
đ
For x tho since there is no acceleration, thereâll be a constant velocity
So you donât have to integrate
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I am in Geometry Honors and was wondering what the forumla for this is? I remember something about calculating chords using a tangent line but I can't find the formula on google. Thanks!
Please ping me if you know the formula
The formula is known as tangent secant theorem
@hasty elbow
lol I literally found the exact website while searching
But thank you!
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I donât really understand this and could use some help
Hint: 70 + 52 + 58 = 180
Do you understand what "congruent triangles" means?
Yes
You know that the sum of the three angles in a triangle sum up to 180 degrees. Right?
Yes
Well, knowing that: what are the angles not shown in the ABC triangle and in the DEF triangle?
E58 and C52
Well. If we look at both sets of angles:
ABC angles = {70, 58, 52}
DEF angles = {70, 52, 58}
They're the same.
If two triangles have the same angles, so their sides are proportional. Right?
Ya I think I get it
Is there some non obvious part in my explanation until now?
Ok.
Their sides are proportional, so what is that proportion?
What is the ratio between the sides of ABC to DEF?
That i donât know
Like: proportional means that one is a multiple of another.
You take one, multiply by a number and get the other.
Multiply what
I will give a example.
Let say candy bars cost 3 dollars per unit. 10 candy bars costs 30 dollars. 21 candy bars costs 63 dollars.
(Amount payed)/(number of candy bars) = constant.
The amount payed is proportional to the number of candy bars. You can multiply one quantity by a number (the constant) and get the other.
So is 122 and 128 the constant
How did you get 122 and 128?
70 plus 58 and 70 plus 52
That is just the sum of two angles.
In that case: the sides of one triangle are multiple of the other.
You can take the ABC triangle, multiply every side by a number and get the DEF triangle.
Are you following?
Ye donât how do I find out the sides
That is the neat part: you don't have to find the sides.
What the problem gives us about the sides?
What information it telled about the sides of the triangles?
Are you asking me a question
Yes.
It only told me about the angles
Look at the sides AB and DE.
Oh there equal
If the sides are equal, so what number should i multiply one side (AB) to get the other side (DE)?
Yes
Multiply what
The side AB.
The triangles are proportional.
I can multiply the sides of one and get the sides of the other.
Multiply what numbers
That is what i am asking you.
The sides donât have numbers just the angles
I know the sides don't have numbers in them, but they do have a length. The question just didn't give it.
Nope
You think the sides don't have length?
No i donât but I donât know them
Oh. Ok.
We don't have to know the exact length of sides.
The question is just asking for us to prove that the triangles ABC and DEF are congruent (both has the same sides and angles).
We already know that the angles are equal.
Half of the job js done.
How can we show that the sides are also equal?
I donât know
Am i even helping you?
I don't know what to say.
I am asking to know if i have to change how i am explaning.
Maybe i am beeing too complicated and wanted to know.
Probably
Ok. Give me a minute.
Ok
Right.
We just have to prove that both the triangles ABC and DEF have the same sides (we already did it with angles).
The question telled us that AB and DE are equal.
If the triangles ABC and DEF have the same angles, so the sides of ABC are proportional to the sides of DEF. Right?
Ya
Well, what is the definition of "proportional"?
You, what you think when reads "proportional"?
Crosponding sizes
Almost.
Let say you running such that you traverse 10 meters each second.
If you ran for 6 seconds, how much meters you traversed?
60
What if you ran for 4.5 seconds?
45
You may note that the distance traversed is a multiple of the time running. Right?
Yes
That is what proportional means.
Ok
"A is proportional to B" means that you can multiply B by some number and get A.
Ok
Now. We know the sides of the triangle ABC are proportional to the sides of triangle DEF. Right?
Yea
What does it means their sides to be proportional?
One side is equivalent to the other side
Not quite.
Remember what i said.
I do remember that but I donât get it in this context
I see.
You can multiply all the sides of the ABC triangle by some number and get all the sides of the DEF triangle.
I donât quite get it
I can give a example.
You can note that all the sides of the blue triangle are double the sides of the red triangle.
What number should i multiply the sides of the red triangle to get the sides of the blue triangle?
2
What number should i multiply the sides of the blue triangle to get the sides of the red triangle?
Wouldnât you divide
Yes, exactly. By what number would i divide the sides of the blue triangle to get the sides of the red triangle?
2
Great.
Just a note: divide by 2 is the same as multiplying by 1/2 and that was what i was asking but you are right the same way.
Do you understand now "multiplying the sides of a triangle".
Ya
So, there's very little left to do.
Their sides are propotional.
So they are all equal to ab side
Oh, sorry.
You mean that every side in the ABC triangle is the same?
I did but now I donât
Ok.
So i can multiply the sides of the ABC triangle and get the sides of the DEF triangle.
Right?
Yes
We have to multiply every side of the triangle by the same number, right?
Yes
Like: i cannot multioly one side by 3 and other by 2 and the other one by 6. I have to multiply all the sides by 2.
Uh huh
This means you understand or you didn't?
I did
Ok. I asked because you normally say "right" but ok.
My bad
You can note that if i only gave you one side of each triangle: it would be suficient.
Yes
If i only gave you the 10 side of the blue triangle and the 5 side of the red triangle, so you could discover that you have to multiply by two.
Yea
I will try a example.
Consider a triangle with sides 13, 5, 17 which i will call A and other triangle i will call B.
Assume the sides of A and B are proportional. Assume one of the sides of B is three times one of the sides of A. What are the sides of B?
39 15 51
Ok
We know that one of the sides of ABC (AB) is equal to one of the sides of DEF (DE) and we know they are proportional.
Ye
Hou much do i have to multiply 4 to get 4?
1
How much do i have to multiply 13 to get 13?
1
Why 1?
You said how much do i have to multiply 13 to get 13 and 13 x 1 is 13
Can you elaborate on that one
Sorry.
1
Do you notice something?
If i have some number, what should i multiply it by to get itself?
What?
Of the ab side is equal and I only need one side to find out if itâs proportional and it equal that means there all equal right
Not quite.
Think about this.
It would be 1
1
If i have to multiply AB by 1, so i have to multiply every other side of ABC by one.
Yes
If i have something and then multiply by one: i will get the same thing. Right?
Yes
If the sides of ABC are 12, 24, 17: what are the sides of DEF?
12 24 17
We have proven in the start that the angles of both triangles are equal.
I think I get it
Now we just proved that both triangles have the same sides.
So I did it
Yes.
Thanks dawg
Did you understand my explanation?
Yes
What do you mean by "dwag"?
Did some thing not feel obvious?
No
I like friend kinda like homie
Thank you for beeing patient with me.
Same to you
Thank you (even if i didn't have to be patient).
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,rotate
what have you tried?
Let x=oldest
what are the 3 sister's ages then
I dont understand how to do this
In terms of x
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I found a solution online but it was complicated. Is there any easy way to solve it?
What if I use modular arithmetic?
(10^n)mod6 is always 4 where n is an integer and n>0
Well, old biscuit is right too
@tender mantle
What do you mean by old biscuit?
He deleted his message
The given number is product of three consecutive numbers
Yes
Think about divisibility of product of three consecutive numbers by 6
@tender mantle Has your question been resolved?
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What property is this?
ĺăăăžăă
How would i simplify these so they don't have log 
wolfgram doesn't help
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Would ask for someone to check over my work and tell me if itâs right or wrong and the answer yes this is just a check over for extra credit (letting you know bc somone who helped me noticed the blue background meaning extra credit)
I wouldâve just turned it in for my credit but knowing my grade is low
I need it
hmm
Hello!
can you remember soh cah toa?
right
Which one are we working on
you can try memorizing soh cah toa since it'll make it a lot easier for you
number 1
do you remember what adjacent, opposite, and hypotenuse mean?
A bit not to much Im a bit fuzzy i sit here and do math all day for credit
ok
so hypotenuse is the longest side in the right triangle
its the one that doesnt touch the 90 degree angle
yes
what do you mean ac?
in number 1?
it would be ba
I mean line
remember, its the longest side
Oh yes your right
So that means
these are what you call the isdes of the triangle that arent the hypotenuse
Sin(a)= 4/13
right!
Okokok
do you remmeber what opposite and adjacent mean?
And then
Not as much
ok
Remind me please
so this is relative to the angle that you're using the trig function on
opposite would be the leg that isn't touching the angle youre using it on
adjacent would be the leg that is touching
hmm
remember which two sides are you gonna use?
like adjacent, opposite, hypotenuse...
it's actually adjacent/hypotenuse
this is where its really useful to memorize soh cah toa!
did you learn about how to use it yet?
ok ill give you an explanation since i found it reallt helpful when i was learning it
But it is still 4/12
actually 4/13
yeah
but soh cah toa is this acronym
the beginning letters are the trig functions, like sine, cosine, tangent,
Wait so son a and cos b is the same answer
.
yep
do you want to learn about soh cah toa?
i found it really helpful but if you would prefer to work on the questions we can do that too
I would like to if I had a lot of time but I do need this credit to be in
Yea
It does seem helpful
sure, i'd highly recommend learning it later just since you'll struggle on tests if you don't have your trig functions down
And I would like to learn it from you 100% but still my credit is low and it has to be in by Wednesday
Yea on my test I got like a 2.3
Should we do this one now
actually tan(b) is wrong too...
Oh
so remember which sides do you need to take for tan?
adjacent,hypotenuse
no, that woukd be cosine
remember soh cah toa?
ah (adjacent hypotenuse) would be cosine
but tangent would be oa or opposite over adjacent
yep!
Yess
see this is where the acronym is really useful
This
the photo is a little blurry, is it asking to find angle H?
Itâs asking to
Not that
hat is the correct trig ratio below in terms of Cosine of angle H? *
COSH=15/17
yep
remember, cah is adjacent over hypotenuse
C
Alrigjt ili note it down
Note
I resent a few
ok for the first one
look at the angle they give you can the sides they give you
relative to that angle, what are the sides?
like are they adjacent, oppposite, hypotenuse,
yep
its the hypotenuse
remember, the hypotenuse is the longest side
its also the one that's kinda slanty
ok so if you have opposite and hypotenuse, which trig function are you gonna use?
remember soh cah toa
Uhh
remember how soh cah toa works?
can you try memorizing it? itll REALLY help you on tests
so what two letters do opposite and hypotenuse turn into?
Wait a sec
i really think we should go over soh cah toa
i know you just want the answers to extra credit, but i really don't think you're learning anything currently
if you can't remember what the trig functions do, you're really gonna struggle as time goes on
i'm sorry, youve been away for some time now and i have other things to take care of
if you come back hopefully someone else can take over
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Hi, i need help with part b) I have forgotten how to do when X is not equal to x (2 in this case)
you could calculate the complement instead
N is normal distribution right?
Yeah
do you know about pdfs?
hmm which is short for?
probability density function
No I haven'f done that yet
P(Xâ 2) = P(X<2) + P(X>2)
Oh ok
(in general, the probability at a single point is 0 for continuous distributions)
so do i make P(X>2) 1-P(X<2)? or do i just make 2 the lower bound?
if you know that result, you can
alright
P(X>2) + P(X<2) = 1 - P(X<2) + P(X<2) = 1
(and using this fact, P(Xâ 2) = 1 - P(X=2) = 1 - 0 =1)
if you use exact results, you would get exactly 1
but you dont need to calculate anything ^
Ok so i just know it i 1
yes
Thanks mate!
yw :)
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What is theta intercept?
I don't understand 2 c
@soft wigeon Has your question been resolved?
you need to find the points that cos(theta) = 0
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Whatâs the difference between doing sin(x), referring to opposite/hypotenuse, and sin(x) with Taylorâs series
The kinda x - (x^3/3!) + (x^5/5!)
It's the same function. In right triangle with one angle being x, opposite/hypotenuse = sin(x) = x - (x^3/3!) + (x^5/5!)...
Iâm not sure I understandâŚ
sin(x) is exactly equal to the Taylor series; that's actually how it's calculated on your computer probably
it's also exactly equal to opp/hyp, but if you don't have those values then you won't be able to calculate it that way
ofc the Taylor series is infinite so if you don't have infinite time you won't get an exact answer that way
3/10 = x - x^3/3! + x^5/5! etc
I think there's more sophisticated algorithm on computers, taylor series is rather useful to make computation in head
,w solve 3/10 = x - x^3/3! + x^5/5!
,w sin(0.304693)
close enough
Here, the angle x was computed and in the second calculation, ramonov tried to make sin(x) and as you can see, it's aproximately 3/10. Which is your opp/hypo ratio
so if you wanted to compute e.g. sin(0.5), you could instead compute 0.5 - 0.5^3/3! + 0.5^5/5! ...
which is easier to do by hand
So the taylor series is to like calculate the o/h of a triangle of which we donât know o or h?
yes, if you know the angle x, the normal way you would calculate o/h is by doing sin(x) on your calculator. But instead, you can do it by hand by evaluating the taylor series
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I've tried to think of a way to solve this but have yet to find how to solve this
so you have two lines that each go through the point P and are tangent to f(x)
so the first line goes through some (a, f(a)) and is tangent to f(x) at that point
no! it takes some thought dw
ok so you know what the slope of a tangent line is right?
Yea f'(a) of the function f(x)= x^2 -x +2 is equal to f'(x) = 2x-1, and the point where tangent t_1 and f(x)=x^2 -x+2 intersect will have the same slope which is equal to the slope of the tangent line with a function rule in the form of y =mx+c, so the m in that function rule will be the derrivative of the function F(x)= x^2 -x +2 where the x values are the same
(sorry if it's unclear English isn't my first language and the maths terms have to be translated)
right okay so we have that it goes through (a, a^2 - a + 2) and (3, -1) and that it has slope (a, 2a - 1)
if I gave you two points how would you compute the slope of the line?
Do I use the formula Y - F(a) = F'(a) x (x-a)
please don't use x as a multiplication symbol, use * instead
sure that'll work
I'm still lost
have you seen the formula $m = \frac{y_1 - y_0}{x_1 - x_0}$?
Hayley
Yeah differntial quotient
But I'm a bit lost I know one point would be (3,-1) but the other
the interception point with the tangent line and the function F(x)=x^2 -a +2
How do i find that x coordinate
the other point would be (a, f(a))
which i know sounds like a non-answer
but we also have that the slope matches the derivative at that point, so $m = f'(a)$
Hayley
I get that m = f'(a) but how would the point be equal to (a, f(a))
the tangent line to f at a point a is defined to be the line with the slope m = f'(a) and that goes through the point (a, f(a))
is that like a general law? I have never heard of that before let's say the second point is (a,f(a)) and I fill it into my differential Quotient formula of (Y1- Y2) / (x1-x2) then I get ( F(a) +1) / (a-3)
Alright
a law sorta implies that there's some mathematical argument behind why it's true
but this is one of those things that just describes
what it means to be "tangent" in this particular case
Alright
but I filled my differential equation in correctly right? How do I get the formula of the slope from that? I'm sorry I'm very lost right now
honestly maybe Hayley has a better way of doing it, because I'm way too lazy to scroll up and read. I'm just thinking of the brute-force way of giving variables to everything
and then stacking up the equations and solving
you've got $m = \frac{f(a) + 1}{a - 3} = \frac{a^2 - a + 3}{a - 3}$
Hayley
you also have $m = f'(a) = 2a - 1$
Hayley
so $2a - 1 = \frac{a^2 - a + 3}{a - 3}$
Hayley
go forth and algebra
yeah ;) it suddenly becomes clear once you solve it
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\frac{p-1}{1+\frac{2}{p}+\frac{3}{p^3}}
\frac{p-1}{1+\frac{2}{p}+\frac{3}{p^3}}
$\frac{p-1}{1+\frac{2}{p}+\frac{3}{p^3}}$
Arctic
!status
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
explain what is troubling you
^
bro i dont know where the begin
so its 1 not 6
- don't call me bro
- why say 6 and not 1?
the question is simplified this form
Because im blank
id start by getting rid of fractions
i mean you could at least read what we're saying & what we're asking you
anyway like
ok
so you have the expression $\frac{p-1}{1 + \frac{2}{p} + \frac{3}{p^3}}$, and you need to simplify it, and you don't know where to start.
Ann
yes
multiply top and bottom by p^3 to get rid of the nested fractions.
okey
$\frac{p^3(p-1)}{p^3+2p^2+3}$
!nosols and also dont say blatantly wrong shit
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
Arctic
youre the one who told them exactly how to simplify it
i gave the instruction
it was up to him to execute it
and either achieve his goal or screw up and complain
miss, what's the next stepp?
is there anything you see in the fraction right now that needs extra simplification?
if you don't see anything in the fraction that screams "this needs to be simplified", there is probably nothing.
and in this case there really is nothing
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Just need a math check for all of these to ensure no mistakes.
@placid mountain Has your question been resolved?
<@&286206848099549185>
4 to 8 are ok
1 c) is wrong
slope is already expressed in the function
it's -3/2
Ah ok
everything else is right
12 is also wrong
remember that perpendicular graphs have a slope that is not only opposite but also inverse
.
you forgot to invert the slope
purrthagorascat
that the slope is -3/2
you really don't have to calculate anything, it's already expressed in the equation of the function
in 12, it says it's the same as the b of the other function
which is 4
But thereâs no b in 11 or 12
Ah ok for 11?
both
yup
So whatâs wrong with it?
on ex 12, you wrote b right, but the slope is wrong
Right?
sorry, wym by right?
There two different questions Iâm confused by what you mean?
i know they are two different question
and on exercise 12, the function has to be perpendicular, which means that the slope is inverted and opposite, you only did opposite
you also have to invert the slope in order for the other function to be perpendicular
bruh
Iâm sorry Iâm very confused. How is the question b=4 ??
11
Not question answer
Excuse me? No Iâm not Spanish not that that was any of your business in the first place. Secondly your not making any sense your saying question 11 is b=4 I am asking you how itâs that answer and not what I put down. If your not able to explain that to me Iâll leave and find help elsewhere.
okey, i am going to rephrase. I only asked if you speaked spanish because i do, and since you were writing confusing sentences, maybe that way we could understand each other
i will try my best to make my point clear
Iâm not trying to be rude but I made two typos and you kind of lost it for no reason. Alright then letâs start over.
Hm okay. I see what your saying, can I ask where you got the +4 from?
the -2/3 is just the same as writing -4/6 (i just simplified the fraction)
and the y=-2 and x=6 are just info given by the instructions (that's the point the function goes through)
Right okay, wouldnât -2/3 be simplified?
Ah okay.
was that more helpful?
Kind of. However I still donât see where you got +4 from. I see the simplifying however wouldnât-2/3 be the simplified equation?
you mean the simplified fraction?
Yes.
yes it is, but it is the same thing
purrthagorascat
i just rewrited the fraction -4/6 as -2/3, but it is the same really, nvm that
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hi
can someone explain how a limit approaching from the left and a limit approaching from the right are different in the context of this problem
the first one says 0 remains positive while the second one says 0 remains negative
Approaching from the left would be like if we considered the numbers 4.9, then 4.99, then 4.999, etc as we approached 5. If we do this, then 5-x is positive, since x is always a tiny bit less than 5, but we still approach 5, so 5-x still approaches 0, it just stays positive
From the right is similar: we can consider x being, like, 5.1, then 5.01, then 5.001, etc, and then 5-x is still approaching 0, but since x is always slightly larger than 5, 5-x is negative
ofc!
.close
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I dont understnad how they went got the laststep
$\log_2(x) = y \iff 2^y = x$
Ann
do you understand this Y/N
@mystic terrace Has your question been resolved?
how do i solve the last step
i made a writing error
it should be
5/2=log2(x)
i did it wrong
how do i solve
@mystic terrace Has your question been resolved?
Which one do you want to solve you posted 3 different
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how do i get this started
Are you aware of any property of pure polynomial functions with regards to their continuity?
i dont think so
Alright, do you know the definition of continuity in limit form?
And what about the definition in graphical form
doesnt that mean that you can write it down without like putting ur pencil up
Yes graphically it means that
Now the question says the function shouldn't be continuous in the CLOSED interval [-1,3]
Now option C is a quadratic polynomial
And if you're aware of this, their graph is a parabola
Which can be drawn without lifting the pen at all everywhere
In fact if you had known the definition in limit form I'd have proven it to you for any polynomial
Polynomial functions are always continuous