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KiRa [710]
4 years ago
11

A sequence of transformations maps triangle ABC to triangle A’B’C’. The sequence of transformations that maps triangle ABC to tr

iangle A’B’C’ is a reflection across the
- y-axis
- x-axis
- line y=x
OR - line y=-x

...followed by a translation

- 4 units to the right and 10 units up
- 8 units to the right and 10 units up
- 10 units to the right and 2 units up
OR - 10 units to the right and 4 units up.

Mathematics
1 answer:
Zolol [24]4 years ago
6 0

Answer:

Reflection across the line y=x followed by translation 10 units to the right and 4 units up.

Step-by-step explanation:

Triangle ABC has vertices at points A(-6,2), B(-2,6) and C(-4,2).

1. The reflection across the line y=x has the rule

(x,y)→(y,x).

Thus,

  • A(-6,2)→A''(2,-6);
  • B(-2,6)→B''(6,-2);
  • C(-4,2)→C''(2,-4).

2. The translation 10 units to the right and 4 units up has the rule

(x,y)→(x+10,y+4).

Thus,

  • A''(2,-6)→A'(12,-2);
  • B''(6,-2)→B'(16,2);
  • C''(2,-4)→C'(12,0).

Points A'B'C' are exactly the vertices of the triangle A'B'C'.

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Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

The price of the home = $275,000

The mode of purchase of the home = Mortgage

The percentage of the loan amount payed as down payment = 20%

The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n -  \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }

The formula for monthly payment on a mortgage, 'M', is given as follows;

M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1}  \approx 1,519.28

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1}  \approx 1,576.15

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

The monthly payment, M = $1,576.15

C. When t = 30 year and r = 5%, we have;

n = 12 × 30 = 360

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1}  \approx 1,181.01

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

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