A company estimates that the revenue (in dollars) from the sale of x doghouses is given by R(x)=14,000ln(0.01x+1). Use the differential to approximate the change in revenue fro the sale of one more doghouse if 110 doghouses have already been sold. (Round to nearest cent as needed)

The change in revenue from the sale of one more doghouse if 110 doghouses have already been sold is dR/dx=66.67 $/doghouse.

Step-by-step explanation:

We have a revenue function that is:

We have to approximate the change in revenue from the sale of one more doghouse, if 110 doghouses have already been sold.

That is the marginal revenue at x=110.

The marginal revenue is expressed as the first derivative of the revenue.

Then, we calculate the derivative of R:

We then evaluate this marginal revenue at point x=110:

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions