Area of the large semicircle
Area of the small semicircle
Area of the figure
14r+20<14r+16 or 8−10r>15−9r
14r+20<14r+16
14r+20−14r<14r+16−14r(Subtract 14r from both sides)
20<16
20−20<16−20(Subtract 20 from both sides)
0<−4
8−10r>15−9r
−10r+8>−9r+15(Simplify both sides of the inequality)
−10r+8+9r>−9r+15+9r(Add 9r to both sides)
−r+8>15
−r+8−8>15−8(Subtract 8 from both sides)
−r>7
−r/−1 > 7/−1(Divide both sides by -1)
r<−7
Ur answer is:
r < - 7 OR 0 < - 4
As we know that
<span>h = 60 - 16t^2 </span>
<span>t = sqrt((60 - h)/16) </span>
<span>= sqrt((60 - 11)/16) </span>
<span>= 1.75
hope it helps</span>
Answer:
1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
Step-by-step explanation:
