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leonid [27]
3 years ago
6

Aline passes through (-3,-2) and is perpendicular to 3x - 2y = 7.

Mathematics
2 answers:
Natali5045456 [20]3 years ago
8 0

Answer:

y=-\frac{2}{3}x-4

Step-by-step explanation:

3x-2y=7

-3x      -3x

-2y=-3x+7

/-2      /-2

y=\frac{3}{2}x-\frac{7}{2}

perpendicular lines always have opposite reciprocal slopes

3/2--> -2/3

y=-2/3x+b

-2=-2/3(-3)+b

-2=2+b

-2    -2

b=-4

y=-2/3x-4

frez [133]3 years ago
6 0

Answer:

y = - \frac{2}{3} x - 4

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 3x - 2y = 7 into this form by subtracting 3x from both sides

- 2y = - 3x + 7 ( divide all terms by - 2 )

y = \frac{3}{2} x - \frac{7}{2} ← in slope- intercept form

with slope m = \frac{3}{2}

Given a line with slope m then the slope of a line perpendicular to it is

m_{perpendicular} = - \frac{1}{m} = - \frac{1}{\frac{3}{2} } = - \frac{2}{3}, thus

y = - \frac{2}{3} x + c ← is the partial equation

To find c substitute (- 3, - 2) into the partial equation

- 2 = 2 + c ⇒ c = - 2 - 2 = - 4

y = - \frac{2}{3} x - 4 ← equation of perpendicular line

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3 years ago
The compound interest on a sum of money in
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Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

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The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

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x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

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5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

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The compound interest in 3 years is therefore;

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The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

The difference between the principal and the compound interest in three years, d = Rs 17,994.

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