Help PLEASEE!!!!!!!!!!

A certain dodgeball court is a circle with a square perfectly inscribed inside it. the square represents the playing field, whil

cluponka [151]

Exactly 8*pi - 16
Approximately 9.132741229 For this problem, we need to subtract the area of the square from the area of the circle. In order to get the area of the circle, we need to calculate its radius, which will be half of its diameter. And the diameter will be the length of the diagonal for the square. And since the area of the square is 16, that means that each side has a length of 4. And the Pythagorean theorem will allow us to easily calculate the diagonal. So: sqrt(4^2 + 4^2) = sqrt(16 + 16) = sqrt(32) = 4*sqrt(2) Therefore the radius of the circle is 2*sqrt(2). And the area of the circle is pi*r^2 = pi*(2*sqrt(2)) = pi*8 So the area of the rest areas is exactly 8*pi - 16, or approximately 9.132741229

6 0

3 years ago

Find the slope of the line that passes through (6, 7) and (2, 10). and simplify if needed thx guys.

tekilochka [14]

Answer:

$-\frac{3}{4}$

Step-by-step explanation:

the equation for finding the slope of a line when given two points is $\frac{y_2-y_1}{x_2-x_1}$ , aka the change in y over the change in x.

pick one of your coordinate pairs to be $y_2\\$ and $x_2$ . it doesn't matter which coordinate pair you choose as long as you keep them as $y_2\\$ and $x_2$ . the remaining coordinate pair will be $y_1$ and $x_1$ .

for this example, i'll use (2, 10) for $y_2\\$ and $x_2$ and (6, 7) for $y_1$ and $x_1$ .

**before i begin, i just want to note that you can do these next four steps in any order that you want. i personally prefer to plug in my y-values first and then my x-values, but you can choose to instead plug in the values of each coordinate pair (like starting by plugging in the coordinate pair (2, 10) with 10 for $y_2\\$ and 2 for $x_2$ ). it's up to you. i'm going to explain the steps by plugging in my y-values first and then my x-values because that's the way i normally do it.

first, start by plugging in the y-value from the coordinate pair of your choosing in for $y_2\\$ . since i chose (2, 10) for $y_2\\$ and $x_2$ , i'll plug in 10 for $y_2\\$ .

$\frac{y_2-y_1}{x_2-x_1}$ ⇒ $\frac{10-y_1}{x_2-x_1}$