Please show work!

1 answer:

3 0

Okay. So the formula for the amount of time is I / Pr, meaning you multiply the principal and the rate in decimal form together, and you divide the amount of interest by that number. The formula for finding the interest rate is I / Pt, meaning you multiply the principal and time, and the you divide the amount of interest by that number. With a decimal, be sure to multiply by 100 to get the whole percentage. With that being said, if you solve this correctly, here are your answers.

Principal: $6,631
Interest Rate: 8.05%
Time: 33 months
Simple Interest: $1,467.94

Principal: $884,706
Interest Rate: 5.42%
Time: 40 months
Simple Interest: $159,931

Note: For times in months, you'll have to do some conversion into decimals, but you know that 12 months equal 1 year, so you can divide the time period by 12. For the second part, I happened to get 5.43% when I worked it out, but a calculator happened to round off to 5.42%, so it's a bit confusing, but those are the answers.

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$