Answer:
If a certain cone with a height of 9 inches has volume V = 3πx2 + 42πx + 147π, what is the cone’s radius r in terms of x?
Step-by-step explanation:
V = 3πx2 + 42πx + 147π
V=3π(x2 + 14x +49)
9.42(x2 + 14x +49)
9.42(x2 + 14x +14) -14 + 49= 0
9.42(x + 7)^2 + 35= 0
9.42(9.42(x + 7)^2 = - 35)9.42
(x + 7)^2 = - 35/9.42)
√(x + 7)^2=√- 35/9.42
x + 7 = - 1.927
x= - 1.927 - 7
x= - 8.927
V = 3π(- 8.927)^2 + 42π(- 8.927) + 147π
V=750.69 - 1177.29 + 461.58
<u>V=34.98</u>
h= 9 inches
V = 13πr2h
34.98 = 13(3.14) (r^2) (h)
34.98 = 40.82 (r^2) 9
34.98 = 367.38 r^2
34.98/ 367.38 = 367.38 r^2/ 367.38
0.095= r^2
Is it possible with only letters
0.05 is the least among 0.5, 0.05, and 0.625
Hope that helps :)
The two numbers are 30 and 11
<em><u>Solution:</u></em>
Given that we have to separate the number 41 into two parts
Let the second number be "x"
<em><u>Given that first number is eight more than twice the second number</u></em>
first number = eight more than twice the second number
first number = 8 + twice the "x"
first number = 8 + 2x
So we can say first number added with second number ends up in 41
first number + second number = 41
8 + 2x + x = 41
8 + 3x = 41
3x = 41 - 8
3x = 33
x = 11
first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30
Thus the two numbers are 30 and 11
