Ask question

Ask question

All categories

3 years ago

10

Complete the point slope equation of the line through (-5,7) and (-4,0)y-7=

2 answers:

mojhsa [17]3 years ago

7 0

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{7})\qquad
(\stackrel{x_2}{-4}~,~\stackrel{y_2}{0})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-7}{-4-(-5)}\implies \cfrac{0-7}{-4+5}\implies \cfrac{-7}{1}\implies -7
\\\\\\
\begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-7=-7[x-(-5)]\implies y-7=-7(x+5)$

deff fn [24]3 years ago

3 0

Point slope form: y-7= -7(x+5)

You might be interested in

What is the value of 4 to the power of 3 divided by 2 to the power of 3

nydimaria [60]

Answer:

8 is the answer

Step-by-step explanation:

4 to the power of 3 is basically 4x4x4 which equals 64, and 2 to the power of 3 is 2x2x2 which is 8, and 64 divided by 8 is 8

7 0

3 years ago

Read 2 more answers

Write an algebraic expression

Brums [2.3K]

Answer:

Hi

Step-by-step explanation:

hi

hi

hi______________________

4 0

3 years ago

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

b.

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

c.

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

3 years ago

Read 2 more answers

Help is for my daughter

marysya [2.9K]

Answer:

c.

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The volume of a sphere is 36π in³. What is the radius of the sphere?

Inessa05 [86]

36pi = 4/3 * pi * r^3
27 = r^3
r = 3 in

7 0

3 years ago

Read 2 more answers