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4 years ago

Through: (-5, 5), perp. to y = 5/9x - 4

Mathematics

1 answer:

zysi [14]4 years ago

6 0

Answer:

$y=-\frac{9}{5}x-4$

Step-by-step explanation:

<u><em>The complete question is:</em></u>

Find the equation of the line that passes through: (-5, 5), and is perpendicular to the line y = 5/9x - 4

step 1

Find the slope of the line

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

The equation of the given line is

$y=\frac{5}{9}x-4$

The slope of the given line is

$m_1=\frac{5}{9}$

The opposite reciprocal is equal to

$m_2=-\frac{9}{5}$ ----> slope of the perpendicular line

step 2

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=-\frac{9}{5}$

$point\ (-5,5)$

substitute

$y-5=-\frac{9}{5}(x+5)$

step 3

Convert to slope intercept form

Isolate the variable y

$y-5=-\frac{9}{5}(x+5)\\\\y-5=-\frac{9}{5}x-9\\\\y=-\frac{9}{5}x-9+5\\\\y=-\frac{9}{5}x-4$

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Convert 72° into grades.

kumpel [21]

Answer:

80 grades is the answer.

Step-by-step explanation:

72° into grades

1° = 10/9 grades

72° = 10/9 × 72grades

= 720/9 grades

= 80 grades

7 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!!!!!

Montano1993 [528]

<h3>Answer: 7.1</h3>

=========================================================

Explanation:

The two points are (x1,y1) = (1,3) and (x2,y2) = (8,4)

Use the distance formula to get,

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(1-8)^2 + (3-4)^2}\\\\d = \sqrt{(-7)^2 + (-1)^2}\\\\d = \sqrt{49 + 1}\\\\d = \sqrt{50}\\\\d = \sqrt{25*2}\\\\d = \sqrt{25}*\sqrt{2}\\\\d = 5\sqrt{2}\\\\d \approx 7.0710678\\\\d \approx 7.1$

Alternatively, you can plot the two points A(1,3) and B(8,4) on the same xy grid. Then plot point C(8,3). Note that C has the same x coordinate as B, and the same y coordinate as A. Triangle ABC is a right triangle, where the 90 degree angle is at point C. The distance from A to B is the same as finding the length of AB, which is the hypotenuse.

This means we can apply the pythagorean theorem to find the hypotenuse AB. The distance formula is a modified version of the pythagorean theorem. Refer to the diagram below.