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2 years ago

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Chris and Becky are comparing mortgages. The mortgage is for $183,500. If they choose 30 years at 5%, their monthly payment will

be $985.07 with a total payback of $354,625.20. If they choose 20 years at 4.5%, their monthly payment will be $1160.91 with a total payback of $278,618.40. How much interest will be paid on each loan?

1 answer:

Anestetic [448]2 years ago

4 0

Answer:

Total payback for Loan type 1 = $354,625.20

Total payback for Loan type 2 = $278,618.40

Explanation

Total payback of a loan consists of the summation of the amount of mortgage to be paid and total interest to be paid after the given number of years.

Total payback of a loan can also be calculated by multiplying your monthly payment of the loan by the total number of months given for payment.

a. For Loan type 1

In the question we have been given or monthly payment = $985.07

The formula for Total payback for Loan type 1 = Monthly payment × number of months

Numbers of years for payment of loan = 30 years

1 years = 12 months

Hence,

Number of months = 30 years × 12 months

Number of months = 360 months

Total payback of loan = $985.07 × 360 months

= $354,625.20

b. For Loan type 2

In the question we have been given or monthly payment = $1160.91

The formula for Total payback for Loan type 1 = Monthly payment × number of months

Numbers of years for payment of loan = 30 years

1 years = 12 months

Hence,

Number of months = 20 years × 12 months

Number of months = 240 months

Total payback of loan = $1160.91 × 240 months

= $278,618.4

Step-by-step explanation:

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Answer:

27.

<em>Equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

28.

<u>Part 1:</u> 400 feet

<u>Part 2:</u> Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

<u>Part 3:</u> acceleration at any time t is -32 feet/second squared

<u>Part 4:</u> t = 10 seconds

29.

<u>Part 1:</u> Average Rate of Change = -15

<u>Part 2:</u> The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

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Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

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Where m is the slope, or the derivative of the function

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If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

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$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

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Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

<em>So equation of 2 tangent lines at the given curve going through point P is:</em>

<em>y = -7x + 1</em>

<em>y = x + 1</em>

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28.

<u>Part 1:</u>

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

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max height = 400 feet

<u>Part 2:</u>

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

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