The surface area of a right circular cone is 15π in.. The cone is enlarged by multiplying both the radius of the base and the slant height by 4. What is the surface area of the new cone? A. 80π in.
B. 120π in.
C. 60π in.
D. 240π in.

Can you pless add the process

The surface area of the new cone is 240π inches² ⇒ answer D

Step-by-step explanation:

If the two cones are similar, then:

1. t1/t2=l1/l2=constant ratio

2. S.A1/S.A2=(R1/R2)^2

3. V1/V2=(R1/R2)3

The surface area of a right circular cone is 15π inches²

 The cone is enlarged by multiplying both the radius of the base  and the slant height by 4

 The two cons are similar

R1/R2=1/4

 The surface area of a right circular cone is 15π inches²

S.A1/S.A2=(R1/R2)^2

SO

15π/S.A2=(1/4)^2

15π/S.A2=1/16

By using cross multiplication:

15πX16=S.A2X1

240π=S.A2

The surface area of the enlarged cone = 240π inches²

The surface area of the new cone is 240π inches²

HOPE THAT HELPS!!

x^6

step-by-step explanation:

this is correct because:

x-6=0

0*x=0

so

x^6

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