Answer:
5(1+x)
Step-by-step explanation:
Answer: 131.37 and the volume is 341.25
Step-by-step explanation:
Answer: OPTION C.
Step-by-step explanation:
Observe the triangle ABC attached.
Notice that the angle of depression is represented with
.
Knowing that the top of a lighthouse is 260 feet above water and the ship is 270 feet offshore, you can find the value of
by using arctangent:

In this case you can identify that:

Therefore, substuting values into
, you get that the angle of depression is:

Answer:

Step-by-step explanation:
We are asked to find the tangent line approximation for
near
.
We will use linear approximation formula for a tangent line
of a function
at
to solve our given problem.

Let us find value of function at
as:

Now, we will find derivative of given function as:




Let us find derivative at 

Upon substituting our given values in linear approximation formula, we will get:


Therefore, our required tangent line for approximation would be
.