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Dafna1 [17]
3 years ago
6

Solve the system of equations. y = x + 6 y = x^2 + 8x + 16

Mathematics
1 answer:
rewona [7]3 years ago
7 0

Answer:

(-5,1)

(-2,4)

Step-by-step explanation:

y = x + 6

y = x^2 + 8x + 16

Since both equations are equal to y, we can set them equal to each other

x + 6  = x^2 + 8x + 16

Subtract x from each side

x -x + 6

= x^2 + 8x-x + 16

6 = x^2 +7x +16

Subtract 6 from each side

6-6 = x^2 +7x +16-6

0 = x^2 +7x +10

Factor

What 2 numbers multiply together to give us 10 and add together to give us 7?

5*2 = 10

5+2 =7

0 = (x+5) (x+2)

Using the zero product property

x+5 = 0  x+2 = 0

x = -5   x=-2

Now we need to find the y's that go with the x's

x=-5

y= x+6

y =-5+6

y=-1

(-5,1)

x=-2

y = -2+6

y=4

(-2,4)

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Becca's house is assessed at $157,000 and her property tax rate is 3.3%. How
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Answer:

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Clark and Lana take a 30-year home mortgage of $128,000 at 7.8%, compounded monthly. They make their regular monthly payments fo
svp [43]

Answer:

Step-by-step explanation:

From the given information:

The present value of the house = 128000

interest rate compounded monthly r = 7.8% = 0.078

number of months in a year n= 12

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To find their regular monthly payment, we have:

PV = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{r}{n})^{-nt}}{\dfrac{r}{n}}    \end {bmatrix}

128000 = P \begin {bmatrix}  \dfrac{1 - (1 + \dfrac{0.078}{12})^{- 12*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

128000 = 138.914 P

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P = $921.433

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To find the unpaid balance when they begin paying the $1400.

when they begin the payment ,

t = 30 year - 5years

t= 25 years

PV= 921.433 \begin {bmatrix}  \dfrac{1 - (1 - \dfrac{0.078}{12})^{25*30}}{\dfrac{0.078}{12}}    \end {bmatrix}

PV = $121718.2714

C) In order to estimate how many payments of $1400  it will take to pay off the loan, we have:

121718.2714 =  \begin {bmatrix}  \dfrac{1300  (1 - \dfrac{12.078}{12}))^{-nt}}{\dfrac{0.078}{12}}    \end {bmatrix}

121718.2714 = 200000  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

\dfrac{121718.2714}{200000 } =  \begin {bmatrix}  (1 - \dfrac{12.078}{12}))^{-nt}   \end {bmatrix}

0.60859 =  \begin {bmatrix}  (1 - \dfrac{12}{12.078}))^{nt}   \end {bmatrix}

0.60859 = (0.006458)^{nt}

nt = \dfrac{0.60859}{0.006458}

nt = 94.238 payments is required to pay off the loan.

How much interest will they save by paying the loan using the number of payments from part (c)?

The total amount of interest payed on $921.433 = 921.433 × 30(12) years

= 331715.88

The total amount paid using 921.433 and 1300 = (921.433 × 60 )+( 1300 + 94.238)

= 177795.38

The amount of interest saved = 331715.88  - 177795.38

The amount of interest saved = $153920.5

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