Answer:


Step-by-step explanation:
Given

Required
Select equivalent expression





Start by solving the given expression (using a calculator)

Then we solve the given options (using a calculator)










The equivalent expressions are (a) and (e)
The answer for this will be
-1/3x+3
Answer:
The rate of change is
ft^(2)/min
Step-by-step explanation:
The area of a circle is given by the following equation:

To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So

We have that:
.
We want to find 
So


Since the area is in square feet, the rate of change is in ft^(2)/min.
So the rate of change is
ft^(2)/min

keep in mind that, a negative coefficient to "x", will make the graph reflect over the y-axis.
Answer:
Step-by-step explanation:
Given the angle ∠AOB
It is stated that CO is the angle bisector of ∠AOB.
Given that ∠AOB = 30°
As we know that the angle bisector bisects the angle into two equal angles.
Thus, the angle bisector CO bisects the angle ∠AOB into two equal angles, which are:
as
∠AOB = 30°
Thus, the two formed angles i.e m∠AOC and m∠BOC by the angle bisector would be half of the angle bisector as the angle bisector bisects the angle ∠AOB into two equal angles.
Therefore,