20 POINTS AND BRAINLIEST FOR CORRECT ANSWER! A cone with height h and radius r has volume V = 13πr2h. 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?

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

V=34.98

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  

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐