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2 years ago

A line that includes the points (n,-6) and (-9,-4) has a slope of -15. What is the value of n?

Mathematics

1 answer:

pav-90 [236]2 years ago

7 0

Answer:

n=-8.86

Step-by-step explanation:

Slope will determine the direction and the steepness of a function. It could be either increasing/decreasing depend on minus or plus. The formula for slope is

m= (y2-y1)/ (x2-x1)

If you add (n,-6) and (-9,-4) to the equation, it will be:

m= (y2-y1)/ (x2-x1)

-15= (-6--4)/ (n--9)

-15= (-2)/ (n+9)

-15(n+9) = -2

-15n - 135=-2

-15n= -2 + 135

n= -133/15

n= -8 13/15= -8.86

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Read 2 more answers

Solve this x²+8/10 -x = 10/2

AysviL [449]

$x^2+\frac{8}{10}-x=\frac{10}{2}\\
x^2-x+\frac{8}{10}-\frac{10}{2}=0\\
x^2-x+\frac{8}{10}-\frac{50}{10}=0\\
x^2-x-\frac{42}{10}=0\\
x^2-x+\frac{1}{4}-\frac{1}{4}-\frac{42}{10}=0\\
(x-\frac{1}{2})^2=\frac{1}{4}+\frac{42}{10}\\
(x-\frac{1}{2})^2=\frac{5}{20}+\frac{84}{20}\\
(x-\frac{1}{2})^2=\frac{89}{20}\\
x-\frac{1}{2}=\sqrt{\frac{89}{20}} \vee x-\frac{1}{2}=-\sqrt{\frac{89}{20}}\\
x=\frac{1}{2}+\frac{\sqrt{89}}{\sqrt{20}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{\sqrt{20}}\\


$
$x=\frac{1}{2}+\frac{\sqrt{89}}{2\sqrt{5}} \vee x=\frac{1}{2}-\frac{\sqrt{89}}{2\sqrt5}}\\
x=\frac{1}{2}+\frac{\sqrt{445}}{10} \vee x=\frac{1}{2}-\frac{\sqrt{445}}{10}}\\
x=\frac{5+\sqrt{445}}{10} \vee x=\frac{5-\sqrt{445}}{10}}\\$

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