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Luba_88 [7]
3 years ago
14

Consider the expression below. Assume the variable m represents an integer. 6m(3m + 21) Enter an expression in the box that uses

the variable m and makes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.) 6m(3m + 21) = 9 Given that m represents an integer, is 6m(3m + 21) divisible by 9?
Mathematics
1 answer:
Monica [59]3 years ago
4 0

Answer:

(a)\ 6m(3m + 21) = 9(2m^2 + 14m)

(b) Yes, it is divisible by 9

Step-by-step explanation:

Given

6m(3m + 21)

Solving (a): Complete the blanks

6m(3m + 21) = 9 [\ ]

Expand the bracket

6m(3[m + 7]) = 9 [\ ]

6m*3(m + 7) = 9 [\ ]

Express 6m as 2m * 3

2m*3*3(m + 7) = 9 [\ ]

2m*9(m + 7) = 9 [\ ]

Rewrite as:

9 * 2m(m + 7) = 9 [\ ]

Multiply the bracket by 2m

9 * (2m^2 + 14m) = 9 [\ ]

Divide both sides by 9

2m^2 + 14m = [\ ]

<em>Hence, the bracket will be filled with: </em>2m^2 + 14m<em></em>

<em>So:</em>

6m(3m + 21) = 9(2m^2 + 14m)

Solving (b): Is 6m(3m + 21) divisible by 9?

In (a), we have:

6m(3m + 21) = 9(2m^2 + 14m)

The leading factor "9" implies that the expression is divisible by 9

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