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IgorLugansk [536]
3 years ago
12

What is the equation of the line that is parallel to 8x-5y=2 and goes through the point (-5,-2)

Mathematics
1 answer:
Stella [2.4K]3 years ago
3 0

Answer:

8x - 5y = - 30

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange 8x - 5y = 2 into this form

Subtract 8x from both sides

- 5y = - 8x + 2 ( divide all terms by - 5 )

y = \frac{8}{5} x - \frac{2}{5} ← in slope- intercept form

with slope m = \frac{8}{5}

• Parallel lines have equal slopes, hence

y = \frac{8}{5} x + c ← is the partial equation of the parallel line

To find c substitute (- 5, - 2) into the partial equation

- 2 = - 8 + c ⇒ c = - 2 + 8 = 6

y = \frac{8}{5} x + 6 ← in slope- intercept form

Multiply through by 5

5y = 8x + 30 ( subtract 5y from both sides )

0 = 8x - 5y + 30 ( subtract 30 from both sides )

8x - 5y = - 30 ← in standard form

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Step-by-step explanation:

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Hence, the principle after one month is $1010.83

b. The principal after 6 months:

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P_{6m}=P(1+i_m)^n, \ n=6m, P=1000, i_m=0.1380\\\\P_{6m}=1000(1+0.1380)^{6/12}\\\\=1066.77

Hence,  the principal after 6 months is $1066.77

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P_{1y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{1y}=1000(1+0.1380)^{12}\\\\P_{1y}=1138

Hence,  the principal after 1 year is $1138.00

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P_{20y}=P(1+I_m)^n, n=1/12, i_m=0.1380, P=1000\\\\P_{20y}=1000(1+0.1380)^{12}\\\\P_{20y}=13269.22

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