5/1 or 5 as slope is the rise or y, over the run or x
Unlike the previous problem, this one requires application of the Law of Cosines. You want to find angle Q when you know the lengths of all 3 sides of the triangle.
Law of Cosines: a^2 = b^2 + c^2 - 2bc cos A
Applying that here:
40^2 = 32^2 + 64^2 - 2(32)(64)cos Q
Do the math. Solve for cos Q, and then find Q in degrees and Q in radians.
We have
-9 + 7 = -2
You can also rewrite this as
7 - 9 = -2
If it makes you more comfortable.
Hope this helps.
Y = -2(x-5)^2 + 4
Note: vertex form is 0, f(x), y, or nothing= a(x-h)^2 + k
The opposite of the x value of the vertex is h
The y value value of the vertex is k.
Find a. Plug in (6,-10)
-10 = a(6-5)^2 + 4
-10 = 1a + 4
-10 = 5a
a = -2
