Answer:
800 in²
Step-by-step explanation:
Let’s find the area of all the shapes in the net.
10 · 10 = 100
100 · 2 = <u>200</u>
15 · 10 = 150
150 · 4 = <u>600</u>
200 + 600 = <u><em>800</em></u>
we know that
The measurement of the exterior angle is the semi-difference of the arcs which comprises
In this problem
∠FGH is the exterior angle
∠FGH=
∠FGH=
-----> equation A

--------> equation B
Substitute equation B in equation A
![100\°=(arc\ FEH-[360\°-arc\ FEH])](https://tex.z-dn.net/?f=100%5C%C2%B0%3D%28arc%5C%20FEH-%5B360%5C%C2%B0-arc%5C%20FEH%5D%29)



therefore
<u>The answer is</u>
The measure of arc FEH is equal to 
The park formed like a rectangle so :
Area = length × width
Area = ( 9/10 ) × ( 4/5 )
Area = 9 × 4 / 10 × 5
Area = 9 × 2 × 2 / 5 × 5 × 2
Area = 18 / 25
Answer:
x=7
Step-by-step explanation:
4(x-8)= -2(x-5)
4x+32= -2x+10
+2x +2x
6x-32=10
+32 +32
6x= 42
divide by 6 on both sides
x=7