7-12 is like a negative number so it will be -5. And 13-18 would be -5 also.
Let x be the original radius. Given that a semicircle will have an arc length of half of a circumference and C=2pr, the arc length will be just pr.
Now the difference of arc is then going to be:
p(6+x)-px
p(6+x-x)
6p inches
So the arc will be 6p inches longer. Or if you wish to approximate...
≈18.85 inches
N³ - 2n² - 15n = 0
"n"
n (n² + 2n - 15) = 0
n = 0
n² + 2n - 15 = 0
Δ = (2)² - 4(1)(-15)
Δ = 4 + 60 = 64
n' = (-2+8) / 2 = 6/2 =3
n'' = (-2-8) / 2 = -10/2 = -5
Solution:
S {-5 , 0 , 3 }
Answer:
option B is the correct answer = TSA = 339.29 in²
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