Find the ordinary interest on a loan of $11,000.00 at 7.0% annually for 92 days.

h=1 df/dx=-15

h=0.1 df/dx=-10.5

h=0.01 df/dx=-10.05

h=0.001 df/dx=-10.005

h=0.0001 df/dx=-10.0005

Step-by-step explanation: The function should be 5x^2.

If the function is linear, the answer is very simple: it is 5 for every value of h.

The rate of change can be defined as:

\frac{\Delta f}{\Delta x} =\frac{f(a+h)-f(a)}{h}

For this function f=5x we have:

f(a)=5a^2f(a+h)=5(a+h)^2=5a^2+10ah+5h^2

Then, we have:

\frac{\Delta f}{\Delta x} =\frac{f(a+h)-f(a)}{h}=\frac{5a^2-(5a^2+10ah+5h^2)}{h}=-10a+5h

The value for a is a=1

For h=1

\Delta f/\Delta x=-10a-5h=-10-5=-15

For h=0.1

\Delta f/\Delta x=-10-5(0.1)=-10-0.5=-10.5

For h=0.01

\Delta f/\Delta x=-10-5(0.01)=-10-0.05=-10.05

For h=0.001

\Delta f/\Delta x=-10-5(0.001)=-10-0.005=-10.005

For h=0.0001

\Delta f/\Delta x=-10-5(0.0001)=-10-0.0005=-10.0005

practice. practice. practice.