If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed? CuCl2 + NaNO3 → Cu(NO3)2 + NaCl 12.4 g NaCl 15.7 g NaCl 16.2 g NaCl 28.4 g NaCl

15.7 g

Step-by-step explanation:

18.8 cubic inches.

step-by-step explanation:

volume of a cone = πr²h

r = 1.5 inches

h = 8 inches

pi = 3.14

volume of our cone = × 3.14 × 1.5² × 8 = 18.84 square inches

or equals to 18.8 square inches rounded up to nearest tenth.

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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