Answer:
or
Step-by-step explanation:
yz-plane has the equation 
where 
is a real constant. The normal vector of this plane (vector perpendicular to the plane) is
If the line is perpendicular to this plane, then it is parallel to the normal vector, so the equation of this line is
where 
are coordinates of the point the line is passing through.
In your case, the line is passing through the point (0,0,8), so the canonical equation of the line is
Write a vector parametrization for this line
The answer is (2z-3)(z+3)
Answer:
x=0 x=9
Step-by-step explanation:
f(x) = x^3 - 9x^2
Set equal to zero to find the zeros
0= x^3 - 9x^2
Factor out x^2
0=x^2 (x-9)
Using the zero product property
0=x^2 0=x-9
x=0 x=9
Answer:
Step-by-step explanation:
The identities you need here are:
and 
You also need to know that
x = rcosθ and
y = rsinθ
to get this done.
We have
r = 6 sin θ
Let's first multiply both sides by r (you'll always begin these this way; you'll see why in a second):
r² = 6r sin θ
Now let's replace r² with what it's equal to:
x² + y² = 6r sin θ
Now let's replace r sin θ with what it's equal to:
x² + y² = 6y
That looks like the beginnings of a circle. Let's get everything on one side because I have a feeling we will be completing the square on this:
Complete the square on the y-terms by taking half its linear term, squaring it and adding it to both sides.
The y linear term is 6. Half of 6 is 3, and 3 squared is 9, so we add 9 in on both sides:
In the process of completing the square, we created within that set of parenthesis a perfect square binomial:
And there's your circle! Third choice down is the one you want.
Fun, huh?
First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.
Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.
![\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20cot%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B6%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B-7%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B6%5E2%2B%28-7%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B36%2B49%7D%5Cimplies%20c%3D%5Csqrt%7B85%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
