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
Answer:
A -60i - 14j
Step-by-step explanation:
u = -9i + 8j
v = 7i + 5j
2u = 2(-9i + 8j) = -18i + 16j
6v = 6(7i + 5j) = 42i + 30j
2u - 6v
(-18i + 16j) - (42i + 30j)
(-18i - 42i) + (16j-30j)
-60i - 14j
For this case we have the following fraction:
(1-cos ^ 2 (θ)) / (sin ^ 2 (θ))
We must take into account the following trigonometric identity:
cos ^ 2 (θ) + sin ^ 2 (θ) = 1
Therefore rewriting we have:
sin ^ 2 (θ) = 1 - cos ^ 2 (θ)
Substituting in the given fraction we have:
(1-cos ^ 2 (θ)) / (sin ^ 2 (θ))
= (sin ^ 2 (θ)) / (sin ^ 2 (θ))
= 1
Answer:
1
Answer:
Step-by-step explanation:
consider downloading photomath
