HELP HOW DO I DO THIS

Step-by-step explanation:

The sum of an infinite geometric series is expressed according to the formula;

a 1 r where;

a is the first term of the series

r is the common ratio

If the sum of an infinite geometric series is three times the first term, this is expressed as

S∞ = 3a

Substitute S = 3a into the formula above to get the common ratio r;

a

3a 1-r

cross multiply

3a(1-r) = a

3(1-r) = 1

open the parenthesis

3- 3r 1

subtract 3 from both sides

33r31-3

-3r = -2

2|3 r =

Hence the common ratio of this series is 2 upon 3

= 2\3

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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