All categories

4 years ago

Arnold borrowed $7890 at 11.5 percent for five years. How much did Arnold pay in interest?

Mathematics

2 answers:

Rudiy274 years ago

5 0

937.5 is what I got.

Anna71 [15]4 years ago

4 0

Answer:

The total amount Arnold pay in interest is $ 2521.30

Step-by-step explanation:

Given :Arnold borrowed $7890 at 11.5 percent for five years.

We have to find the amount Arnold pay in interest

Using formula for monthly payment ,

$Monthly\ payment = \frac{rate \times Principal}{1-(1+rate)^{-time\ period}}$

Given : Principal value = $7890

Rate of interest = 11.5% anually

So monthly interest rate is $\frac{11.5}{1200}$

Time = 5 years

In months 5 years = 5 × 12 = 60 month

Substitute, above we have,

$Monthly\ payment=7890\cdot\frac{\frac{11.5}{1200}}{\left(1-\left(1+\frac{11.5}{1200}\right)^{-60}\right)}$

Simplify, we get,

Monthly payment = $ 173.52

Total payment that Arnold pay = No. of months × Monthly Payment

⇒ Total Payment = 60 × 173.52

⇒ Total Payment = $ 10411.30

Since, Money Arnold borrowed = $7890

Thus, Amount of interest = Total payment - Amount borrowed

⇒ Interest = 10411.30 - 7890

⇒ Interest = $2521.30

Thus, The total amount Arnold pay in interest is $ 2521.30

You might be interested in

Find f(a), f(a+h), and 71. f(x) = 7x - 3 f(a+h)-f(a) h if h = 0. 72. f(x) = 5x²

Leni [432]

Answer:

$71. \ \ \ f(a) \ = \ 7a \ - \ 3; \ f(a+h) \ = \ 7a \ + \ 7h \ - \ 3; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 7$

$72. \ \ \ f(a) \ = \ 5a^{2}; \ f(a+h) \ = \ {5a}^{2} \ + \ 10ah \ + \ {5h}^{2}; \ \displaystyle\frac{f(a+h) \ - \ f(a)}{h} \ = \ 10a \ + \ 5h$

Step-by-step explanation:

In single-variable calculus, the difference quotient is the expression

$\displaystyle\frac{f(x+h) \ - \ f(x)}{h}$ ,

which its name comes from the fact that it is the quotient of the difference of the evaluated values of the function by the difference of its corresponding input values (as shown in the figure below).

This expression looks similar to the method of evaluating the slope of a line. Indeed, the difference quotient provides the slope of a secant line (in blue) that passes through two coordinate points on a curve.

$m \ \ = \ \ \displaystyle\frac{\Delta y}{\Delta x} \ \ = \ \ \displaystyle\frac{rise}{run}$ .

Similarly, the difference quotient is a measure of the average rate of change of the function over an interval. When the limit of the difference quotient is taken as h approaches 0 gives the instantaneous rate of change (rate of change in an instant) or the derivative of the function.

Therefore,

$71. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{(7a \ + \ 7h \ - \ 3) \ - \ (7a \ - \ 3)}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{7h}{h} \\ \\ \-\hspace{4.25cm} = \ \ 7$

$72. \ \ \ \ \ \displaystyle\frac{f(a \ + \ h) \ - \ f(a)}{h} \ \ = \ \ \displaystyle\frac{{5(a \ + \ h)}^{2} \ - \ {5(a)}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{{5a}^{2} \ + \ 10ah \ + \ {5h}^{2} \ - \ {5a}^{2}}{h} \\ \\ \-\hspace{4.25cm} = \ \ \displaystyle\frac{h(10a \ + \ 5h)}{h} \\ \\ \-\hspace{4.25cm} = \ \ 10a \ + \ 5h$