#help-26
226100 messages · Page 233 of 227
I thought to apply identity sin(2x) = 2sin(x) cos(x) to simply my product.
makes sense
Some terms from numerator and denominator will be cancelled out.
That's why I tried to find the product upto n, and I'll apply limit of n tends to infinity.
But I stuck.
Just looking at an approximation using desmos, it looks close to $\frac{\sin\left(2x\right)}{2x}$, but I have no proof of it other than just how the equation looks.
Chixen
Ohh
If you needed it for some program, sin(2x)/(2x) is a good if not perfect approximation.
I need to solve mathematical problem (calculator is not allowed).
oh oh ok
so you would need a proof of sin(2x)/(2x) as well as how you would come to that conclusion.
other than "the equations look the same"
Hmm
well the product of two cosines is the average of the cosines of the angle's sum and difference.
$\cos\left(\alpha\right)\cos\left(\beta\right)=\frac{\cos\left(\alpha+\beta\right)+\cos\left(\alpha-\beta\right)}{2}$
Chixen
Can you check this?
This is so long.
This seems right
Oh okay, thanks for checking...
Now, how to compute this result if n tends to infinity?
instead of stopping at 2^(n-1), stop at some other variable 2^m.
see what happens then
then take a limit as m->∞
I thought to make n terms for easily calculations.
That tends to break whenever limits are involved
Ohh
$$\lim_{m\to \infty} \frac{\sin(2x)}{2^{m+1} \sin(x/2^{m-1})}$$
rewriting cos(x) as sin(2x)/2sin(x) for all the terms we get the partial sum formula here which you can work out using l'hopital's rule
the infinite product is sin(2x)/2x
Ohh
Thank you!
yeah no prob
i can send work if you need it
I'm unsure if I'm wrong here.
the product of the first n terms is
$$\lim_{n\to \infty} \frac{\sin(2x)}{2^{n+1} \sin(x/2^{n})}$$
dansman1090
using l'hopitals rule you get the limit of 2^(n+1)*sin(x/2^n) is 2x
But how can we apply Lhospital rule only for denominator?
gimmie a min ill send my work
Ok
@neon iron
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Hi guys, im having trouble with my homework, i need to find the two equation of straights passing through point P which are tangent to the circumference
Have you tried anything?
I tried to put the circumference equation and the general straight formule with P in it in a system
y - 1 = m(x -6)
x² + y² + 2x - 4 = 0
What class is this for. Might help us to know
im from italy im in 3rd year of high school
you should 1) complete the square so you can find the center of the circle
- you should use the fact that the normal slope to a circle would just be (y coordinate - circle y coordinate)/(x coordinate - circle center x coordinate)
Oh no man i know what u are talking bout
then 3) you should use the normal slope equation (opposite recripocal) to find the tangent slope
and then you are done
Our teacher wants us to solve in another way
or is it precalc
sry man i dont know
It's the same method.
No, chad was talking bout the second one
Umm... Can you translate if possible?
Yea
First method
We systematize the equation of the generic line for P (1; 5) with the equation of the given circumference:
numbers
By eliminating y between the two equations we obtain the solving equation of the system, in the unknown x
numbers
we impose the condition of tangency Delta = 0
We will have so
numbers
Like i understood the process but i cant get the calc right
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hI
please tell me how to solve this?
<@&286206848099549185>
@neon iron start another chanel, this one is occupied
@rain lotus Has your question been resolved?
umm what do you mean by guessing and checking
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so i'm trying to solve part b to this question, i got the answer to part a as 5cos2x - 4sin2x + 2
for part b i tried to set this equal to 4 and then rearrange for tan but the +2 always got in the way, anyone got any ideas?
what...?
so for part a of the question
i got 5cos2x - 4sin2x + 2, which i checked is correct
and then for part b i used this same identity and set it equal to 4
so 5cos2x - 4sin2x + 2 = 4
and then i had no idea what to do, i tried to rearrange this to get it all under one trig function but i couldn't manage it
<@&286206848099549185>
Are you allowed calculator?
@wild tapir
considering how the question asks for answer upto 2 decimal places
yeah
yeah
do i need to use that though? wouldn't that just get me to where i started
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I did 300 * cos(25) and got 271.892336
is it suppose to be in Kilo joules?
or jouoles
@cosmic cedar Has your question been resolved?
Joules
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i cant get the cdf
i integrated it but
i dont know how to get the result shown
nvm got it
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I have a function f. So f: R → (0, ∞), defined by f(x) = e^x. I have to show it's injective or not. And surjective or not. I've showed that this is injective, so now I'm confused about how to show that it's either surjective or not surjective
@tall halo Has your question been resolved?
Not so much
He didn't go over it much and the textbook is confusing me
@tall halo You show for any number in (0, infinity), e^x can equal that number for some x.
Perhaps you can just use the log function to show this
So I can choose any number (basically positive) to plug into x?
I know that there's, let's say an a and b. f(a) = b. If this is true, then the function is surjective
Maybe you can just use that exp is continuous and unbounded above
the definition of surjectivity is: for every b in the codomain (in this case your codomain is (0,\infty)), you can find an a in the domain (in this case R) such that f(a) = b
so the idea for any b in (0,\infty), you find some a in terms of b such that e^a = b
What will be an acceptable way to show it really depends on what facts about the exponential function you can use
Like using ln x?
If you can assume ln as you know it exists, then yea that’s probably the easiest way
There is some theory that goes into why ln is a mathematically valid function and this exercise would seem circular to me if you were allowed to use the fact that it exists
If you can use the fact that ln exists then f is invertible so it’s injective and surjective
Seems too cheaty
Idk though, could use some more info on what has been covered in your course
In my course
Hmm
Honestly
I was just given what a function is when injective, surjective or bijective
And then he just gave us pictures of each
Is this an analysis class or something else?
Oh ok you can probably use ln then
Given any y in (0, infinity), ln y is in R.
Therefore the function can take on any value in (0, infinity) so it is surjective.
That's it.
Really?
Yea, assuming a and b are as Camilleone said
And when I plug it into the function, I end up with b, which makes the function surjective
You can expand on this a bit and do as above
In this way, it's more of like the representation of it. Like if I wanted to picture it kind?
making a = ln(b)
Ah
Given any y in (0, infinity), ln y is in R. Let x = ln y. Then y = e^x.
Therefore the function can take on any value in (0, infinity) so it is surjective.
Thank you both very much
I guess I just needed to see an example from the homework to understand subjectivity better
You guys are the best🥺
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I dont know how to solve this task
Draw the parabola p:y=2x+1
a) Draw a line connecting this parabola
intersects at points A(1/3) and B(1.5/7).
@stable totem Has your question been resolved?
@stable totem Has your question been resolved?
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what have you tried
does $\int{\frac{\sin(x)}{x}}$ look familiar at all?
a disappointing son
Absolutely not
perhaps just $\frac{\sin(x)}{x}$?
a disappointing son
possibly learned something in class about this and its integral?
this problem can be hard for first time seeing integrals
im clueless hmm
probably because it doesn't have an elementary answer
And why is that
i thought it was graph sketch smh
none of our normal integration techniques work
Rude teacher
honestly lmao
Sometimes
what class is this for
There's a complicated multivariate way
And a trivial complex analysis way
yes
...........
I'm so confused rn lol
But I don't know a single variable calculus way
you can express it as a power series lmao
whats your class
agreed
what year uni
I'm 17
bro what
Names of courses are meaningless. What topics are on the syllabus
Jesus christ
Power series and term-by-term integration works. But is the answer obvious from the series?
Lol
Si?
Yes
Non-elementary function
not surprising lol
Nope
what integration techniques do you know
not that they help here but i'm curious
Partial integration, substitution and some trigonometric identities
damn lmao
yeah your teacher either goofed or is borderline insane
lol
,w integrate from 0 to pi ((sin x) / (2x))
@normal citrus Here
Yea fuck your teacher
This reminds me of the time my linear algebra instructor, who was a grad student, told us to disprove Fermat's Last Theorem
Well thanks guys
Fuck that guy
lmfao
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Is it possible for upper bound of integral to be less than lower bound?
No
Like I was solving an integral.
$$\int_0^{\frac{\pi}{3}} \sin(x) \ln(\cos(x)) dx$$
Is it possible?
yes
Umm... But how?
Mosh
Could have googled that btw.
I thought I was doing some mistakes.
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Can someone help with this question? Im not sure where to even begin
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Borre
I do know how to use partial integration but now it's a function in the exponent and I'm lost
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Borre
Here we go again I guess
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Im having struggle finding the right u and du
I thought it was x^2 for u
Then 2x for du, but how would i get to
Oh
it is
3?
yes
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Stuck on how to calculate the partial sum. Calculator is not working.
@valid rampart Has your question been resolved?
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'
im doing calc limits
it says to take the reciprocal to simplify but I'm not sure how they are getting the answer they get
from my understanding reciprocal meant just flipping a fraction around
a disappointing son
tell me im not crazy
am i wrong or did they pull that shit out of thin air
bro i am so dumbn
close
idk how to close
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please explain it to me i will be back in about 5 - 10 minutes
<@&268886789983436800> academic dishonesty
@neon iron is this a test?
im back, and no this is homework
i got it just needed a little bit of research thanks guys
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Use the cos ratio to find the hypotnuse of the big triangle
^
Ok
that isn't needed
then use sin ratio to find the angle
I’m confused
there is much better trig function that can be used
Okay let me try
it's a little insidious that the point with the right angle is given no name
I didn’t get it
fixed

So what do I do
Y
can you name some right triangles in this diagram?
there are two of them here.
i would like you to name them.
There’s a scalene and right-angled triangle
no, that's not what i'm asking you to do.
i'm asking you to name the two right triangles present in this picture,
not just describe them by type
if you were asked to name a person you would rather say "Frank" than "a tall man" wouldn't you?
Yes.
There is one ‘orthogonal triangle’.
Y
okay, tell me how to name a triangle.
The right angled triangle is CDB
...that is one of the right triangles in the picture, yes.
you'vev given a name finally
And the other is ACB
CAB
that's the same triangle as ACB...
in this context, putting the vertices in a different order makes no difference
BAC
are you being dense on purpose here
?
immediately after i say
in this context, putting the vertices in a different order makes no difference
you put the vertices A, B and C in yet another order
hoping, apparently, that this time you'll get it right
Oh right
you have named CDB as one of the two right-angled triangles in the picture.
what's the other one?
BDA
Okay
can you write out what data is known for each triangle?
They are both right- angled triangles
thank you captain obvious
if you're going to say something as trivial as that, then at least say which angle in each of them is the right angle.
Oh yeah triangle BDA has one angle as 39 degrees
which angle?
Angle A
No
It’s third angle is 51 degrees( angle b)
angle B, not b. math is case-sensitive.
Ok
and that's not given, but sure.
what else do we know about triangle ABD?
again, look at the diagram
all i'm asking you to do here is read off what's given.
no calculations.
just reading.
The base length is 15.9m
Side ad
math is case-sensitive
Side AD
please read and understand everything i say.
preferably, please do so the first time round.
Ok
okay
now
what part of triangle ABD are we looking for?
i.e. what is asked for in part (a) of the problem?
again, no calculations, only reading.
What is the width w of triangle ADB
yes, w is what's asked for.
so
to bring that all back together
we have
a right triangle ADB, with D as the right angle
in which:
the angle A = 39° is known
the side AD = 19.5 is known
the side DB = w is sought
can you write down a trigonometric ratio statement relating these three quantities?
Angle A might be useful here
Ok I think I just got the answer
I got it thank you
For your help
I understood how to do it
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hi just a quick question on twiddle in relation statements
is twiddle always an equivalence
~
i guess i should say is twiddle always an equivalence with a condition
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✅
<@&286206848099549185>
question
if you're reading it from a keyboard-typed text, then it may also mean negation, but it's most likely an equivalence relation
$$\verb|~| A \leftarrow \text{may be negation}$$
$$A \verb|~| B \leftarrow \text{most likely equivalence relation}$$
Khazali
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Dose anyone know what is Bases in Japanese? <@&286206848099549185>
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is there a formula for this summation?
5^0 + 5^1 + 5^2 + 5^3 + 5^4 + 5^(k-1) + 5^k
yes
what is it
you can view this as a geometric series
but what is the formula
try looking up formula for geometric series
@livid scarab Has your question been resolved?
i can't find the one for mine
were you able to find a general formula when looking up geometric series
ya
what did you find
try and identify the three things listed at the bottom
what if it was something like this:
5^0 * 7^k + 5^1 * 7^(k-1) + 5^2 * 7^(k-2) + 5^3 * 7^(k-3) + 5^4 * 7^(k-4) + 5^(k-1) * 7^(k-k+1) + 5^k * 7^(k-k)
same idea
not sure what common ratio is for r
i got 5/7 as the common ratio
Ye
$\frac{a_1\left(1-r^n\right)}{1-r}=\frac{5^0\cdot 7^k\left(1-\frac{5}{7}^k\right)}{1-\frac{5}{7}}$
The-Elite
is this right ? @keen venture @restive inlet
not quite
note that the exponents go from 0→k (and k→0)
so you actually have k+1 terms
and you should also have parentheses when raising your r to a certain power
oh
$\frac{a_1\left(1-r^n\right)}{1-r}=\frac{5^0\cdot 7^{\left(k+1\right)}\left(1-\frac{5}{7}^{\left(k+1\right)}\right)}{1-\frac{5}{7}}$
The-Elite
that look good ?
no
but i added k+1
the first term is still 5^0*7^k, that part was fine
oh
oh
$\frac{a_1\left(1-r^n\right)}{1-r}=\frac{5^0\cdot 7^k\left(1-\left(\frac{5}{7}\right)^{k+1}\right)}{1-\frac{5}{7}}$
The-Elite
yes
thanks
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Bro what the hell is this
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latex
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<@&286206848099549185>
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$\frac{\left(\left(7n-5^{\left(k+1\right)}\right)\right)}{2}$
The-Elite
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I have an assumption that 7^k = n
I want to make the base 5 in 5^(k+1) into a base 7
but not sure how to
7^[log7(5)(k+1)]
u could use logs
Where log7(5) is the logarithm, base 7, of 5
$\log _7\left(5\right)^{k+1}$
The-Elite
The-Elite
sorry this
$7^{\log_7(5)(k+1)}$
Kaynex
oh
Note the k+1 can be put as an exponent into the log, then the logs cancel
Giving 5^(k+1) back
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how did you know this ? Is there a formula
or rule
@livid scarab Has your question been resolved?
hi im new
$5 = 7^{\log_7{5}} \implies 5^{k + 1} = 7^{\log_7{5}(k + 1)}$
Beans
do you know if there another formula for this ?
$\frac{\left(\left(7n-5^k\cdot ::5\right)\right)}{2}$
The-Elite
anyway to remove the denominator ?
@livid scarab Has your question been resolved?
$\frac{\left(\left(7n-5^k\cdot ::5\right)\right)}{2}$
The-Elite
anyway to remove the denominator ?
multiply by 2?
where ?
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anyone see whats wrong here
@neon iron Has your question been resolved?
i'll be honest i have no idea the answer but seeing as nobody else is responding, going off the pattern i see, the first one seems incorrect
why
wait sorry meant the second one lol
essentially no basis i'm just looking at the arrows
still wrong
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x^3-2x^2-x+2 / x-1
What exactly are you stuck on?
got to x^2 - x then dont know what to do to get -2 at the end
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trying to find general formula for projectile motion given an initial vector but this looks wrong and I can't figure out the right way to do it
a would be x velocity and b would be y velocity
seems wrong because 0 should be a zero of the parabola
wait I might be dumb zero might be a zero and I might just be wrong gimmie a minute
wait wait no it's not
okay so I actually do need help, I feel like that b/100 should be squared or something, because otherwise f(0) != 0
and if for some reason I change it to b^2 it miraculously works???
???????
Like, I'm elated that it works but I'm really confused
like, this is beautiful to finally see but why is it b^2???
OOOOOOHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
I found it
.close
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,tex $\int_\frac{-\sqrt{3}}{2}^0 \frac{1}{2}x - x\sqrt{1-x^2}+ \int_0^\frac{\sqrt{3}}{2} x\sqrt{1-x^2} - \frac{1}{2}x$
life improver 912839
Okay so I know how to find areas from these sorts of equations (I think) but I am stuck trying to simplify $\frac{1}{2}x - x\sqrt{1-x^2}$
life improver 912839
I found the boundaries of the shaded region from a graph as well as set up the equation, but I am not sure how to simplify from here so that I can find the antiderivative of each side in order to plug in my boundaries and solve
Here is the problem as well because it is possible I've already made a miscalculation
Why do you need to simplify it
I'm unsure how to take the antiderivative of a square root that has an expression or exponent in it
Also the fact that the square root is multiplied by x throws me off
the fact it's xsqrt(1-x^2) makes it easier.
Well xsqrt(x) is simply x if im not wrong
so would xsqrt(x^2) be x^2
Oh crap I am wrong
$\int x\sqrt{1-x^2}\dd{x}$
Mosh
With the power rule the antiderivative of $x = \frac{x^2}{2}$
life improver 912839
And $\sqrt{1-x^2}$ can be rewritten as $(1-x^2)^\frac{1}{2}$
yes.
life improver 912839
So with the power rule $(1-x^2)^\frac{1}{2}$ the antiderivative is $3\frac{(1-x^2)^\frac{3}{2}}{2}$
life improver 912839
You need to use substitution first.
Ah
Try taking the derivative of your expression with chain rule to see why it's wrong. You're close, just fix minor details
$u = 1-x^2$, so the antiderivative of $x(1-x^2)^\frac{1}{2}$ is $x\frac{(3u)^\frac{3}{2}}{2} + C$
The x has not been antiderived yet
life improver 912839
I think you need to go back and review your substitution for integrals
Okay
@golden falcon Has your question been resolved?
This is what I've got now
Sorry if it's confusing I have limited space probably should rewrite it on a different page
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if you are given two collinear points can the third point be placed anywhere in the plane?
wdym
a line doesn't define a plane
yeah ok
what even is the question
but the line is in a plane lmao
what?
ok, that's still infinitely many planes
ask a proper question
there's one plane
There isnt
the xy plane
wait im so confused wtf
Ok yeah
You guys are making no sense at all
and i give you two points on that line
Yep, with you
now the question is can a third point be placed anywhere not on the line itself
to create a triangle
yes
it can be placed anywhere not on the line and you'll have a triangle
anywhere on the line, it'll be 3 collinear points
what is this called
geometry
but this theorem
ok so ignoring transformations and rotations let's let the two collinear points be (a,0), (b,0)
and then let C = (x,y)
sure
then the sides of the triangle are:
b-a
sqrt((x-a)^2 + y^2)
sqrt((x-b)^2+y^2)
side lengths, but yes
now the task is to prove that this satisfies the inequalities
you mean the triangle inequality?
yeah
have fun
lol ok
ok those inequalities should hold if i square the lengths
hmm this is tricky
(b-a)^2
(x-a)^2 + y^2
(x-b)^2 + y^2
<@&286206848099549185>
How do I prove this makes a triangle
what is this?
the bottom two look like circles and the top seems like a point related to the centers of the two circles
Right thats the equilateral case
@narrow torrent Has your question been resolved?
@narrow torrent Has your question been resolved?
read #❓how-to-get-help
@narrow torrent can you restate your question again
and/or whether it's still relevant
sure
given two collinear points in the xy plane
prove that a triangle can be formed with any other point not on the line
that... sounds like theres nothing to prove?
also "two collinear points" what?
any two points by defn have a line passing through them (and exactly one line in fact)
and like, any three points not on the same line make a triangle
why wouldnt they
who's asking you this question
by defn?
ok i guess it's as simple as saying the lines must have different slopes so they mutually intersect
what's unclear?
you're refusing to tell me who's asking you this question
relevance?
From the very start, your question made little sense.
Mosh somehow read your mind, but the rest of us can't
i'm not sure i can explain it more clearly
Can you show us the original question.
none
A question sheet, you having an idea, or what?
stop asking questions about the question and answer the damn question haha
I cant answer something that dont make sense
there's a big difference between shit wording handed down from above and shit wording that came out of your own mind
I give you two points on a line. Will any point not on that line form a triangle with those other two points?
don't really know how many ways i can rephrase that
and now the challenge is to prove it
ok sure you have two points A and B and a point C not on AB
the lines AC and BC will not coincide with AB
AB and AC will intersect at A and AB and BC will intersect at B
are you happy now or will you insist something is missing here
ok then that's all
Any 3 points form a triangle regardless. Any 3rd point not collinear with the first two will form a non-degenerate triangle.
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Can I ask again now?
Ask what
Do I have to send my math problem again?
Before someone was solving another math problem
I think I interrupted it sorry
Idk where your problem is, they do disappear if you don't reply I think
I'd just send it again now:)
The graphing image
found it
,rotate
what are ya havin trouble on m'friend
gathered that, but what specifically
I'll show you what I got
i see you labeled H.A and V.A, so good start
oop sorry
didn't see this
all looks good except for f(0) being 1
seems you drew it defined at two points
make sure you make a hole in your graph at x=0
looks good 👍
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for the question(first pic)
how did the circled equations come in the solution(second pic)
@void summit Has your question been resolved?
@void summit Has your question been resolved?
this is a very weird solution
but those red circled equations simply state that the normal of your plane is perpendicular to the normals of the two given planes
@void summit Has your question been resolved?
Let $\hat{a}$ be the normal to the plane $2x-2y+z=0$ and $\hat{b}$ be the normal to the plane $x-y+2z=4$, then you can determine that $\hat{a} = \qty(2, -2, 1)$ and similarly the $\hat{b}$ from how the equation of a plane is determined, i.e., $(x-x_1,y-y_1,z-z_1)\cdot \hat{n} = 0$
Ansh
The red circled equation is tantamount to saying $\hat{a}\cdot \hat{n} = 0$ and $\hat{b} \cdot \hat{n}=0$ as the question clearly says that the plane is perpendicular to these two planes and hence the dot product of their normals must be 0
Ansh
that's how you got the red circled equation
from that, you clearly get a = b, c = 0 and therefore, the equation of your plane
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or is it just wrong
@tribal zephyr Has your question been resolved?
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thanks!
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Basically the question is equivalent to finding the values of k such that the equation has real roots
Try to simplify the equation into the form ax^2+bx+c = 0 first
@copper lotus Has your question been resolved?
this function should model with some maxima and minima
find the derivative's values at x = 0
you'll get two points
plug in the x values of those points into the original equation
you'll likely have minima -> maxima as the range from the y component
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Hi there! needing a bit of confirmation to my solution in regards to the problem on the picture. Is the reason the proof is wrong because M has to be greater or EQUAL to 1? Belle says she doesn't like A because x is lower than 1 but actually its equal to it. Many thanks!
no, the proof is wrong for a different reason
the proof asserts "M=1 is a non-example of a strict upper bound for [0,1], therefore Belle cannot like it!"
conveniently sweeping under the rug that Belle may well like it, but with a different strict upper bound
Is there missing info for the Q or is that it?
Cinderella likes sets with a nonstrict upper bound and Belle likes sets with a strict upper bound
to put it briefly
@chilly plaza M=1 is not a witness for Belle liking A, but that alone does not mean Belle doesn't like A
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can someone explain what am I supposed to do after factoring?
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I need help with my math test and multiple questions
If you can possibly help that would be much appreciated
You can't get help on tests
You just said you needed help with your math test, is that not a test?
Yea the pic I sent is one of the questions on the test
So that is a test?
It’s a performance task before the test to make sure you know the material before the test happens
But it has similar questions to the test
Well, if you know, getting help on tests is academic dishonesty, so if that was a test, that's a bannable offense from the server
It’s not a test man look what I just sent you
Obviously its not a test he just want to know how to solve this
It was a reply to that person
Ok are u able to help with me so I’m able to be ready for the test?
Do you know how to apply SOH CAH TOA?
No, this is a culminating activity and we did SOH CAH TOA a while ago so at this point I forget
Thank you for trying to be helpful man
Here is my suggestion, since you already know SOH CAH TOA, you should watch this as a refresher
https://www.khanacademy.org/math/geometry-home/right-triangles-topic/intro-to-the-trig-ratios-geo/v/basic-trigonometry
Then if you are still stuck, someone can help you through the process
If both sides have the same value for example both are 5.7 meters than you have 45 degree angle, but 3.8 meters are way smaller than 5.7 so the degree must be smaller than 45
In this case its only 34
Don't give out answers, and if you read, it said to use SOH CAH TOA
Bruh you could just explain him sog cah and toa, you gave him a whole video😭😭
Basically explained him a way efficient and logical way to solve this problem😁
He said he knew it but forgot what it was, that video is to refresh he memory
Can you tell me which part of the triangle is opp and adj and the hyp??
Solving it the logical way may not be the right way the teacher wanted. If it asks to use SOH CAH TOA, you should use it
Bro but I don’t remember how I used it at the start of the course
its multiple choice so efficiency is important
It says tan B, what side is opp to that point?
3.8 m?
Even if it's multiple choice, the teacher would ask for full work to get full credit
Yes
Opp?
Now, what side is adj to B?
hahahahah
Opp?
So? Even online, they can still ask for work, if they wanted
My teacher doesn’t, he’s easy-going, plus I’m in a college level course not in a university one when it comes to math
You determined opp as 3.8, now what about the adj side to B?
@stuck linden are you allowed to use a calculator on the test?
Yea
Idk Mann
What does adjacent mean?
I’ll just wing it I have a lot of questions on this to ask so it’ll take too long to slowly go through each one because it’s due by two, which is in an hour
Idk
you know the formulars for soh cah and toa?
in this case the opposite and the adjacent are given, so you can only solve this problem with toa
If you don't want to put the effort in understanding what I'm asking, or using the resource I sent to refresh your memory, then tbh you're out of luck, especially in university level classes. Higher level classes, you're going to need to use resources, outside of the book, to understand material, and you have to understand the concepts and terms used
I said college level, in Canada college level is a lower level than university courses
Don’t be L mans bro cmon
yea i think i can help this guy you can help someone else dldh
you see that
if you dont have the hypotenuse you cannot use soh or cah
Same applies to any level of classes
what are you studying dldh or what have you studied?
Are you able to help me with this type of stuff?
obv yes
Haven't you noticed that I've been trying to help you but you're not putting any effort into it
Computer Science
Nice man, I have a computer science course for the first time this semester and I’m doing pretty good in it, I think I got an 80 in it right now
That's fake stuff, i isn't real