Lines k and n are perpendicular.if the slope of line k is -6,what is the slope of line n?

Mathematics

2 answers:

anastassius [24]2 years ago

7 0

Answer:

1/6.

Step-by-step explanation:

Lines that are perpendicular have slopes that are negative reciprocals. So, if one line has a slope of -6, the negative of the slope would be 6, and the reciprocal would be 1/6.

Hope this helps!

aleksley [76]2 years ago

3 0

Answer:

1/6

Step-by-step explanation:

Well perpendicular lines are reciprocals of each other,

meaning if like k has a slope of -6 then line n will have a slope of positive 1/6.

Thus,

line n has a slope of 1/6.

Hope this helps:)

You might be interested in

Cheng inc. is considering a capital budgeting project that has an expected return of 40% and a standard deviation of 30%. what i

Radda [10]

Coefficient of variation is the ratio of the standard deviation to the mean (average)

Coefficient of variation = (standard deviation/mean) × 100
$CV= \frac{30}{40}$ × $100$ = $75$

5 0

2 years ago

HELP ME QUICK PLEASEEEE solve for x: 2/x-2+7/x^2-4=5/x

Anna35 [415]

$\displaystyle\\\\\frac{2}{x-2}+\frac{7}{x^2-4}=\frac{5}{x}\\\\\frac{^{x+2)}2~~~~~}{x-2}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2(x+2)}{(x-2)(x+2)}+\frac{7}{(x-2)(x+2)}=\frac{5}{x}\\\\\frac{2x+4+7}{x^2-4}=\frac{5}{x}\\\\\frac{2x+11}{x^2-4}=\frac{5}{x}\\\\5(x^2-4)=x(2x+11)\\\\5x^2-20=2x^2+11x\\\\5x^2-2x^2 -11x-20=0$

$\displaystyle\\3x^2-11x-20=0\\\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{11\pm\sqrt{121+4\cdot3\cdot20}}{2\cdot3}=\\\\=\frac{11\pm\sqrt{121+240}}{6}=\frac{11\pm\sqrt{361 }}{6}=\frac{11\pm19}{6}\\\\x_1=\frac{11-19}{6}=\frac{-8}{6}\\\\\boxed{\bf x_1=-\frac{4}{3}}\\\\x_2=\frac{11+19}{6}=\frac{30}{6}\\\\\boxed{\bf x_2=5}$