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Svet_ta [14]
4 years ago
10

1) Draw the arrow diagram and the matrix representation for the relation: R={(1, 2), (3, 4), (2, 3), (3, 2), (2, 1), (3, 1), (4,

3)} with domain {1, 2, 3, 4)

Mathematics
1 answer:
tensa zangetsu [6.8K]4 years ago
6 0

Answer:

The arrow diagram and the matrix representation for the relation is shown below.

Step-by-step explanation:

The given relation is

R={(1, 2), (3, 4), (2, 3), (3, 2), (2, 1), (3, 1), (4, 3)}

If a relation is defined as

R=\{(x,y)|x\in R,y\in R\}

Then the set of x values is domain and set of y values is range.

The domain of the function is

Domain={1, 2, 3, 4)

The range of the function is

Range={1, 2, 3, 4)

In arrow diagram, we have two sets first set represents the domain and second set represents the range. The arrow connecting the element represent the relation.

In matrix representation,

M_{ij}=\begin{cases}1 & \text{ if } (x_i,y_j)\in R \\ 0 & \text{ if } (x_i,y_j)\notin R\end{cases}

The arrow diagram and the matrix representation for the relation is shown below.

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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
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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

duration of time t = 30 years

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

P = 128000/138.914

P = $921.433

∴ Their regular monthly payment P = $921.433

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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