Show the steps taken to complete the problems:

volume of a cylinder: 34m diameter and 27m height
use 3.14 for pi.

find the surface area of the cone. 8in. diameter and 7in. height.

find the lateral area of the square pyramid: 8in. base and 22in. slant

Volume cylinder = 24501.42 m³

The surface area of the cone = 151.5 in²

The lateral area of the square pyramid = 352 in²

Step-by-step explanation:

∵ The volume of the cylinder = area base × height

∵ Its base is a circle

∴ V = πr² × h = 3.14 × (34/2)² × 27 = 24501.42 m³

∵ The surface area of the cone = 1/2 perimeter base ×  Slant height + area base

∵ Its base is a circle

∴ S.A = (1/2) × 2πr × l + πr²

∵ The slant height =

∴ S.A = 3.14 × (8/2) × √65 + 3.14 × (8/2)² = 151.5 in²

∵ The lateral area of the square pyramid = 1/2 perimeter base × slant height

∵ Its base is a square

∴ L.A = 1/2 × (8 × 4) × 22 = 352 in²

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}