What is the area of a sector of a circle with a radius of 8 inches and formed by a central angle that measures 60°?

1 answer:

7 0

$\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad 
\begin{cases}
r=radius\\
\theta=\textit{angle in degrees}
\end{cases}$

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What is the equation of the line that passes through the point (-2,5) and has a<br> slope of -4?

adoni [48]

Answer:

$\boxed {\boxed {\sf y= -4x -3}}$

Step-by-step explanation:

We are asked to find the equation of a line given a point and the slope. We will use the point-slope formula.

$y-y_1= m(x-x_1)$

In this formula, m is the slope and (x₁, y₁) is the point the line passes through. The slope of this line is -4 and the line passes through (-2,5).

m= -4
x₁= -2
y₁ = 5

Substitute the values into the formula.

$y-5 = -4 (x--2)$

$y-5 = -4 (x+2)$

We are finding the equation of the line, so we should find the slope-intercept form or y=mx+b ( where m is the slope and b is the y-intercept).We must isolate the variable y on one side of the equation.

First, distribute the -4 on the right side of the equation. Multiply each term inside the parentheses by -4.

$y-5=( -4 *x ) + (-4* 2 )$

$y-5 =(-4x)+ (-8)\\y-5= -4x -8$

5 is being subtracted from y. The inverse operation of subtraction is addition, so we will add 5 to both sides of the equation.

$y-5 =-4x -8 + 5$

$y= -4x -3$

The equation of the line is y= -4x -3. The slope of the line is -4 and the y-intercept is -3.

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dem82 [27]

Answer:

4.75 or round up to 5.

Step-by-step explanation:

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