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

Answer:h=1 df/dx=-15h=0.1 df/dx=-10.5h=0.01 df/dx=-10.05h=0.001 df/dx=-10.005h=0.0001 df/dx=-10.0005Step-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^2\\\\f(a+h)=5(a+h)^2=5a^2+10ah+5h^2Then, we have:\frac{\Delta f}{\Delta x} =\frac{f(a+h)-f(a)}{h}=\frac{5a^2-(5a^2+10ah+5h^2)}{h}=-10a+5hThe value for a is a=1For h=1\Delta f/\Delta x=-10a-5h=-10-5=-15For h=0.1\Delta f/\Delta x=-10-5(0.1)=-10-0.5=-10.5For h=0.01\Delta f/\Delta x=-10-5(0.01)=-10-0.05=-10.05For h=0.001\Delta f/\Delta x=-10-5(0.001)=-10-0.005=-10.005For h=0.0001\Delta f/\Delta x=-10-5(0.0001)=-10-0.0005=-10.0005