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3 years ago

14

A couple can afford to make a monthly mortgage payment of $700. If the mortgage rate is 2% and the couple intends to secure a 30

-year mortgage, how much can they borrow?

1 answer:

Sever21 [200]3 years ago

5 0

<h3>Answer: $189,383.96</h3>

========================================================

Explanation:

r = 0.02 = decimal form of 2% annual interest
i = monthly interest rate
i = r/12 = 0.02/12 = (2/100)/12 = 1/600.
n = number of months = 12*30 = 360
P = monthly payment = 700

We'll plug those items into the formula below.

L = loan amount

L = P*( (1+i)^n - 1)/( i*(1+i)^n )

L = 700*( (1+1/600)^360 - 1 )/( 1/600*(1+1/600)^360 )

L = 189,383.961539639

L = 189,383.96

The couple can afford to borrow at most $189,383.96

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3 years ago

How would the average rate of change over years 1 to 5 and years 6 to 10 be affected if the population increased at a rate of 8%

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(A): As year increases the number of pikas reduces.

(B): As year increases the number of pikas increases as opposed to when the rate reduces.

Step-by-step explanation:

<em>See comment for complete question</em>

<u>Given</u>

$a = 144$ --- Initial Population

$r = 8\%$ --- rate

<u>(A) WHEN THE RATE DECREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1-8\%)^x$

Convert % to decimal

$f(x) = 144(1-0.08)^x$

$f(x) = 144(0.92)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(0.92)^x$

$f(1) = 144*0.92^1 =144*0.92=132.48$

$f(5) = 144*0.92^5 =144*0.66=95.04$

$Rate = \frac{95.04 - 132.48 }{4}$

$Rate = \frac{-37.44}{4}$

$Rate = -9.36$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(0.92)^x$

$f(6) = 144*0.92^6 =144*0.61=87.84$

$f(10) = 144*0.92^{10} =144*0.43=61.92$

$Rate = \frac{61.92-87.84}{4}$

$Rate = \frac{-25.92}{4}$

$Rate = -6.48$

So, we have:

$Rate = -9.36$ for year 1 to 5

This means that the number of pikas reduces by 9.36 yearly

$Rate = -6.48$ for year 6 to 10

This means that the number of pikas reduces by 6.48 yearly

So, we can say that, as year increases the number of pikas reduces.

<u>(B) WHEN THE RATE INCREASES</u>

First, we need to write out the function when the population decreases.

This is given as:

$f(x) = a(1-r)^x$

Substitute values for a and r

$f(x) = 144(1+8\%)^x$

Convert % to decimal

$f(x) = 144(1+0.08)^x$

$f(x) = 144(1.08)^x$

Next, we calculate the average rate of change for both intervals using:

$Rate = \frac{f(b) - f(a)}{b-a}$

For 1 to 5:

$Rate = \frac{f(5) - f(1)}{5-1}$

$Rate = \frac{f(5) - f(1)}{4}$

Calculate f(5) and f(1)

$f(x) = 144(1.08)^x$

$f(1) = 144(1.08)^1 = 144*1.08= 155.52$

$f(5) = 144(1.08)^5 = 144*1.47= 211.68$

$Rate = \frac{211.68 - 155.52}{4}$

$Rate = \frac{56.16}{4}$

$Rate = 14.04$

For 6 to 10:

$Rate = \frac{f(10) - f(6)}{10-6}$

$Rate = \frac{f(10) - f(6)}{4}$

Calculate f(6) and f(10)

$f(x) = 144(1.08)^x$

$f(6) = 144(1.08)^6 = 228.52$

$f(10) = 144(1.08)^{10} = 310.89$

$Rate = \frac{310.89-228.52}{4}$

$Rate = \frac{82.37}{4}$

$Rate = 20.59$

So, we have:

$Rate = 14.04$ for year 1 to 5

This means that the number of pikas increases by 14.04 yearly

$Rate = 20.59$ for year 6 to 10

This means that the number of pikas increases by 20.59 yearly

So, we can say that, as year increases the number of pikas increases as opposed to when the rate reduces.

5 0

3 years ago

If h(x)= 3x+5 and h(a) =27, then what is the value of a ?

Usimov [2.4K]

h(x) = 3x + 5

h(a) = 27

We can write this as 3a + 5 = 27

Now, we just need to solve for a.

<em><u>Subtract 5 from both sides</u></em>

3a = 22

<em><u>Divide 3 from both sides.</u></em>

a = 7.33

Now, let's plug 7.33 back into the original equation.

3 * 7.33 + 5 = 27

22 + 5 = 27

27 = 27 √ this is correct

6 0

4 years ago