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kirill115 [55]
3 years ago
11

Determine the monthly payment of a loan for $3,000 at 7.5% interest compounded monthly for 36 months.

Mathematics
2 answers:
kolbaska11 [484]3 years ago
7 0

Answer:

The monthly payment is $93.32 ⇒ answer a

Step-by-step explanation:

* Lets explain how to solve the problem

- The monthly payment is \frac{P(r)}{1-(1+r)^{-n} }

  where:

# P is the loan amount

# r is the rate per period in decimal

# n is the number of periods

- The loan is $3000

- We need to find the monthly payment

∴ P = $3000

- The compound monthly interest is 7.5% for 36 months

∵ The period is 12 (1 year = 12 months)

- Divide the rate as a decimal by 12

∴ r=\frac{7.5}{100(12)}=0.00625

∴ r = 0.00625

∴ n = 36

* <u><em>Lets calculate the monthly payment using the rule above</em></u>

∵ Monthly payment = \frac{(3000)(0.00625)}{1-(1+0.00625)^{-36}}=93.32

∴ The monthly payment is $93.32

Makovka662 [10]3 years ago
3 0

Answer:

$93.32

Step-by-step explanation:

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4 years ago
Integrate the following problem:
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Answer:

\displaystyle \frac{2 \cdot sin2x-cos2x}{5e^x} + C

Step-by-step explanation:

The integration by parts formula is: \displaystyle \int udv = uv - \int vdu

Let's find u, du, dv, and v for \displaystyle \int e^-^x \cdot cos2x \ dx .

  • u=e^-^x
  • du=-e^-^x dx
  • dv=cos2x \ dx
  • v= \frac{sin2x}{2}

Plug these values into the IBP formula:

  • \displaystyle \int e^-^x \cdot cos2x \ dx = e^-^x \cdot \frac{sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx
  • \displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \int \frac{sin2x}{2} \cdot -e^-^x dx

Now let's evaluate the integral \displaystyle \int \frac{sin2x}{2} \cdot -e^-^x dx.

Let's find u, du, dv, and v for this integral:

  • u=-e^-^x
  • du=e^-^x dx
  • dv=\frac{sin2x}{2} dx
  • v=\frac{-cos2x}{4}  

Plug these values into the IBP formula:

  • \displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} - \int \frac{-cos2x}{4}\cdot e^-^x dx

Factor 1/4 out of the integral and we are left with the exact same integral from the question.

  • \displaystyle \int -e^-^x \cdot \frac{sin2x}{x}dx = -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx

Let's substitute this back into the first IBP equation.

  • \displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ -e^-^x \cdot \frac{-cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]  

Simplify inside the brackets.

  • \displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} - \Big [ \frac{e^-^x \cdot cos2x}{4} + \frac{1}{4} \int cos2x \cdot e^-^x dx \Big ]

Distribute the negative sign into the parentheses.

  • \displaystyle \int e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} -  \frac{e^-^x \cdot cos2x}{4} - \frac{1}{4} \int cos2x \cdot e^-^x dx

Add the like term to the left side.

  • \displaystyle \int e^-^x \cdot cos2x \ dx  + \frac{1}{4} \int cos2x \cdot e^-^x dx= \frac{e^-^x sin2x}{2} -  \frac{e^-^x \cdot cos2x}{4}  
  • \displaystyle \frac{5}{4} \int   e^-^x \cdot cos2x \ dx = \frac{e^-^x sin2x}{2} -  \frac{e^-^x \cdot cos2x}{4}  

Make the fractions have common denominators.

  • \displaystyle \frac{5}{4} \int   e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x}{4} -  \frac{e^-^x \cdot cos2x}{4}

Simplify this equation.

  • \displaystyle \frac{5}{4} \int   e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4}

Multiply the right side by the reciprocal of 5/4.

  • \displaystyle \int   e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{4} \cdot \frac{4}{5}

The 4's cancel out and we are left with:

  • \displaystyle \int   e^-^x \cdot cos2x \ dx = \frac{2e^-^x sin2x - e^-^x cos2x}{5}

Factor e^-^x out of the numerator.

  • \displaystyle \int   e^-^x \cdot cos2x \ dx = \frac{e^-^x(2 \cdot sin2x-cos2x)}{5}

Simplify this by using exponential properties.

  • \displaystyle \int   e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x}

The final answer is \displaystyle \int   e^-^x \cdot cos2x \ dx = \frac{2 \cdot sin2x-cos2x}{5e^x} + C.

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