Hello!
The formula for the area of a sector can be written as follows:
Area =


(R)
In the above formula, “r” represents the
radius while “R” represents
the radian measure of a sector. The radius is given to us in the image above as 10 inches. However, we still need the radian measure of the two sectors. To find this measure, we can use the following conversion:
1 degree =

radians
Because the two sectors have a given measure of 72 degrees, we need to multiply both sides of the above conversion by 72:
72 degrees =

Reduce the fraction on the right side of the equation:
72 degrees =

We now have the radian measure of both sectors. Now simply insert this and any other known values into the “area of a sector” formula above:
Area =


(

)
Simplify the right side of the equation to get the following answer:
Area = 20 pi
We have now proven that
the area of one sector is equal to 20 pi.If, however, you need the combined area of the two identical sectors, simply multiply the proven area by 2 to get a total area of
40 pi.I hope this helps!
Answer:
y = -1/2x^2 -5x -45/2
Step-by-step explanation:
Multiply it out and collect terms.
y = -1/2(x^2 +10x +25) -10 . . . . . expand the square
y = -1/2x^2 -5x -25/2 -20/2 . . . eliminate parentheses
y = -1/2x^2 -5x -45/2 . . . . . . . . collect terms
Answer:
-12
Step-by-step explanation:
To find the "quotient" of something is to divide the digits. -84 ÷ 7 = -12.
Answer:
252 Inches cube
Step-by-step explanation:
<em>So for this question you need to find the area of cross section which is the H,</em>
2 Times (3 Inches by 7 Inches) {The 2 same shape}
+
3 Inches by 7 Inches {The middle shape}
<u>For </u><u>the </u><u>H </u><u>Shape </u><u>=</u><u> </u><u>6</u><u>3</u><u> </u><u>Inches </u><u>Square</u>
<em>Then you'll time the lenght which is given 4 Inches</em>
63 Inches square ×4 Inches
252 Inches Cube
<u>The </u><u>volume </u><u>of </u><u>this </u><u>shape </u><u>will </u><u>be </u><u>2</u><u>5</u><u>2</u><u> </u><u>Inches </u><u>cube</u>
<u>Hope </u><u>this </u><u>helps </u><u>d</u><u>ont </u><u>hesitate</u><u> to</u><u> </u><u>ask </u><u>any </u><u>question!</u><u>!</u>
<u>Mark </u><u>me </u><u>as </u><u>brainliest</u><u> </u><u>aslo </u><u>appreciated!</u><u>!</u><u> </u><u>Tq</u>