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Rick wants to use multiplication to solve {k} = 7 \div \dfrac13k=7÷ 3 1 ​ k, equals, 7, divided by, start fraction, 1, divided by, 3, end fraction. Which multiplication equation can Rick use? Choose 1 answer: Choose 1 answer: (Choice A) A \dfrac13\times{k} =7 3 1 ​ ×k=7start fraction, 1, divided by, 3, end fraction, times, k, equals, 7 (Choice B) B \dfrac13={k}\times7 3 1 ​ =k×7start fraction, 1, divided by, 3, end fraction, equals, k, times, 7 (Choice C) C \dfrac13\times7 = {k} 3 1 ​ ×7=kstart fraction, 1, divided by, 3, end fraction, times, 7, equals, k

User Shawn Wang
by
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2 Answers

1 vote

Answer:

C

Explanation:

ITS C BECAUSE I WASTED A MISTAKE BC OF THE OTHER GUY ON TOP

DONT USE HIS

JUST USE MINE

DONT NEED AN EXPLANATION NOOOOB

User Vdwees
by
7.8k points
2 votes

Answer: (a)

Explanation:

Given

Equation is


k=(7)/((1)/(3))

Write the cross-multiplication of the given equation i.e.


\Rightarrow (1)/(3)* k=7

This corresponds to the option (a).

User Schleis
by
8.7k points
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