Show that if 2m + 1 is an odd prime, then m = 2n for some nonnegative integer n. [Hint: First show that the polynomial identity xm + 1 = (xk + 1)(xk(t−1) − xk(t−2) + ⋯ − xk + 1) holds, where m = kt and t is odd.]

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

~

the equation is not included but the answer is (1.50, 2.00, 2.75)

step-by-step explanation:

step-by-step explanation:

i think its about 8