Answer:
X + 8 = 8 + X is the correct answer
Hello from MrBillDoesMath!
Answer:
+2
Discussion:
-8 (17 - 12) / ( -2 (8 - (-2)) ) -
= -8 ( 5) / ( -2 ( 8+2)) as 17 -12 = 5 and -(-2) = +2
= -40/ ( -2 (10))
= -40/(-20) =
= +2
which is the third choice.
Thank you,
MrB
Answer:
Step-by-step explanation:
We are given a circle with a partially shaded region. First, we need to determine the area of the whole circle. To do this, we need the measurement of the radius of the circle:
Use the Pythagorean theorem to solve for the other leg of the right triangle inside the circle:
5^2 = 3^2 + x^2
x = 4
The radius is 4 + 1 cm = 5 cm
So the area of the circle is A = pi*r^2
A = 3.14 * (5)^2
A = 25pi cm^2
To solve for the area of the shaded region:
Ashaded = Acircle - Atriangles
we need to solve for the area of the triangles:
A = 1/2 *b*h
A = 1/2 *6 * 5
A = 15 cm^2
Atriangles = 2 * 15
Atriangles = 30 cm^2
Ashaded = 25pi - 30
<h3><u>Solution</u></h3>
<u>Given </u><u>:</u><u>-</u>
- Perimeter of rectangle = 72 cm
- The length is 3 more than twice the width.
<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>:</u><u>-</u>
<h3 /><h3>
<u>Explantion</u></h3>
<u>Using </u><u>Formula</u>

<u>Let,</u>
- Length of Rectangle = x cm
- Breadth of Rectangle = y cm
<u>According</u><u> to</u><u> question</u><u>,</u>
==> perimeter of Rectangle = 72
==> 2(x+y) = 72
==> x + y = 72/2
==> x + y = 36_________________(1)
<u>Again,</u>
==> x = 2y + 3
==> x - 2y = 3__________________(2)
<u>Subtract</u><u> </u><u>equ(</u><u>1</u><u>)</u><u> </u><u>&</u><u> </u><u>equ(</u><u>2</u><u>)</u>
==> y + 2y = 36 - 3
==> 3y = 33
==> y = 33/3
==> y = 11
<u>keep </u><u>in </u><u>equ(</u><u>1</u><u>)</u>
==> x - 2×11 = 3
==> x = 3 + 22
==> x = 25
<h3><u>Hence</u></h3>
- <u>Length</u><u> of</u><u> </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u> </u><u>cm</u>
- <u>Width </u><u>of </u><u>Rectangle</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u> </u><u>cm</u>
<h3>
<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u></h3>
<h3 />