What is the equation of a line that passes through the point (6,-8) and is perpendicular to the line whose equation is Y=2/3 x+3

Mathematics

1 answer:

sertanlavr [38]3 years ago

6 0

Question-

What is the equation of a line that passes through the point (6,-8) and is perpendicular to the line whose equation is Y=2/3 x+3

Answer-

y = x/2 + 3......x/2 is the same as 1/2x....so u have a slope of 1/2.

A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal, flip the slope and change the sign. So we have 1/2....and we flip it making it 2/1....and we change the sign making it -2/1 or just -2. So ur perpendicular line will have a slope of -2.

y = mx + b

slope(m) = -2

(1,8)...x = 1 and y = 8

now we sub and find b, the y int

8 = -2(1) + b

8 = -2 + b

8 + 2 = b

10 = b

so ur perpendicular equation is : y = -2x + 10 <==

Hope i helped have a wonderful day...

You might be interested in

What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?

german

For this case we have the following function:
$f(x) = tan(\frac{x^2}{9})$
The average rate of change is given by:
$AVR = \frac{f(x2) - f(x1)}{x2- x1}$
Evaluating the function for the given interval we have:
For x = 1.25:
$f(1.25) = tan(\frac{1.25^2}{9})$
$f(1.25) = 0.18$
For x = 2:
$f(2) = tan(\frac{2^2}{9})$
$f(2) = 0.48$
Then, replacing values we have:
$AVR = \frac{0.48 - 0.18}{2 - 1.25}$
$AVR = 0.4$
Answer:
the average value of on the closed interval [1.25, 2] is:
$AVR = 0.4$

3 0

3 years ago

2. A bicycle trail is 65 miles long. Rashid

Tresset [83]

Answer:

35 more miles

5 0

3 years ago

Read 2 more answers

Given that f(x)= x2+3 and g(x)= x+4/3 solve for (f(g(x)) when x =2

Savatey [412]

$\bf \begin{cases}
f(x)=x^2+3\\\\
g(x)=x+\cfrac{4}{3}
\end{cases}
\\\\\\
\boxed{x=2}\qquad g(2)=(2)+\cfrac{4}{3}\implies \boxed{g(2)=\cfrac{10}{3}}
\\\\\\
f(~~g(x)~~)=[g(x)]^2+3\implies f(~~g(2)~~)=[g(2)]^2+3
\\\\\\
f(~~g(2)~~)=\left[ \boxed{\frac{10}{3}} \right]^2+3\implies f(~~g(2)~~)=\cfrac{10^2}{3^2}+3
\\\\\\
f(~~g(2)~~)=\cfrac{100}{9}+3\implies f(~~g(2)~~)=\cfrac{100+27}{9}
\\\\\\
f(~~g(2)~~)=\cfrac{127}{9}$