Answer:
Angle< PQO and Angle < OQP
Step-by-step explanation:
The angle you are naming would be in the middle.
Answer:
x = 0, 7
Step-by-step explanation:
Step 1: Write out equation
x² - 7x = 0
Step 2: Factor out <em>x</em>
x(x - 7) = 0
Step 3: Find roots
x = 0
x - 7 = 0
x = 7
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
</span>
Answer:
y=5(x-3)^2+2
Step-by-step explanation:
You plug the point into f(x)=a(x-3)^2+2... that already has already been solved for the vertex that you want. Then you swap it out for the solution you have solved for.
Answer:
not proportional
Step-by-step explanation:
because they arent apart of any of the same multiple