Mathematics, 02.09.2020 05:01 Sjanaya853

11. Through (-3,-5), perpendicular to -2x - 5y = -19 A. y=2/5x+2/5 B. y=3/5x-19/5 C. y=-5/2x+5/2 D. y=5/2x+5/2

12. Through (-8,1), parallel to -8x + 5y = 89

A. y=-8/5x-69/5 B. y=8/5x+69/5 C. y=8/5x-89/5 B. 5/8x-1/8

14. Through (-8,1), perpendicular to -5x + 9y = 49

A. y=-9/5x-67/5 B. y=-9/5x C. y=-5/9x-67 D. y=9/5x+67/5

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Answer from: BECKX6770

11) D. y=5/2x+5/2 , 12) B. y=8/5x+69/5, 14) A. y=-9/5x-67/5

Step-by-step explanation:

11) The function of the perpendicular line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\perp}\cdot (x-x_{o})$

Where:

$x_{o}$ , $y_{o}$ - Components of the given point, dimensionless.

$m_{\perp}$ - Slope, dimensionless.

Besides, a slope that is perpendicular to original line can be calculated by this expression:

$m_{\perp} = -\frac{1}{m}$

Where is the slope of the original line, dimensionless.

The original slope is determined from the explicitive form of the given line:

$-2\cdot x - 5\cdot y = -19$

$2\cdot x +5\cdot y = 19$

$5\cdot y = 19 - 2\cdot x$

$y = \frac{19}{5} -\frac{2}{5}\cdot x$

The original slope is $-\frac{2}{5}$ , and the slope of the perpendicular line is:

$m_{\perp} = -\frac{1}{\left(-\frac{2}{5}\right) }$

$m_{\perp} = \frac{5}{2}$

If $x_{o} = -3$ , $y_{o} = -5$ and $m_{\perp} = \frac{5}{2}$ , then:

$y-(-5) = \frac{5}{2}\cdot [x-(-3)]$

$y + 5 = \frac{5}{2}\cdot x +\frac{15}{2}$

$y = \frac{5}{2}\cdot x +\frac{5}{2}$

The right answer is D.

12) The function of the parallel line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\parallel}\cdot (x-x_{o})$

Where:

$x_{o}$ , $y_{o}$ - Components of the given point, dimensionless.

$m_{\parallel}$ - Slope, dimensionless.

Its slope is the slope of the given, which must be transformed into its explicitive form:

$-8\cdot x + 5\cdot y = 89$

$5\cdot y = 89 +8\cdot x$

$y = \frac{89}{5}+\frac{8}{5} \cdot x$

The slope of the parallel line is $\frac{8}{5}$ .

If $x_{o} = -8$ , $y_{o} = 1$ and $m_{\parallel} = \frac{8}{5}$ , then:

$y-1 = \frac{8}{5}\cdot [x-(-8)]$

$y-1 = \frac{8}{5}\cdot x +\frac{64}{5}$

$y = \frac{8}{5}\cdot x +\frac{69}{5}$

The correct answer is B.

14) The function of the perpendicular line can be found in terms of its slope and a given point by this formula:

$y-y_{o} = m_{\perp}\cdot (x-x_{o})$

Where:

$x_{o}$ , $y_{o}$ - Components of the given point, dimensionless.

$m_{\perp}$ - Slope, dimensionless.

Besides, a slope that is perpendicular to original line can be calculated by this expression:

$m_{\perp} = -\frac{1}{m}$

Where is the slope of the original line, dimensionless.

The original slope is determined from the explicitive form of the given line:

$-5\cdot x +9\cdot y = 49$

$9\cdot y = 49+5\cdot x$

$y = \frac{49}{9} +\frac{5}{9}\cdot x$

The original slope is $\frac{5}{9}$ , and the slope of the perpendicular line is:

$m_{\perp} = -\frac{1}{m}$

$m_{\perp} = -\frac{1}{\frac{5}{9} }$

$m_{\perp} = -\frac{9}{5}$

If $x_{o} = -8$ , $y_{o} = 1$ and $m_{\perp} = -\frac{9}{5}$ , then:

$y-1 = -\frac{9}{5}\cdot [x-(-8)]$

$y-1 = -\frac{9}{5}\cdot x-\frac{72}{5}$

$y = -\frac{9}{5}\cdot x -\frac{67}{5}$

The correct answer is A.

Answer from: Quest

step-by-step explanation: you take each of the numbers and divide by the amount then you take the range

Which of the following choices is the variance of the sample shown here? 35, 36, 41, 56, 60, 71

Answer from: Quest

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right

Draw lines to match each expression on the left with the correct product on the right.

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11. Through (-3,-5), perpendicular to -2x - 5y = -19 A. y=2/5x+2/5 B. y=3/5x-19/5 C. y=-5/2x+5/2...

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