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1. (06.07 HC) An expression is shown below: 4(m + 3 + 5m) Part A: Write two expressions that are equivalent to the given expression. (3 points) Part B: Show that one of your expressions in Part A is equivalent to the given expression using algebraic properties. Explain which properties you used Part C: Show that your other expression from Part A is equivalent to the given expression by substituting a number for m. (3 points) B i U Font Family AA AOS - 4(m + 5 + 3m) 3(m + 4 + 5m) part a parib partc

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The given expression is

[tex]4(m+3+5m)[/tex]

Part A.

We get the first equivalent expression by combining the like terms

[tex]4(6m+3)[/tex]

We get the second equivalent expression by using the distributive property

[tex]24m+12[/tex]

Part B.

To show the equivalence between the given expression and one of the part A expressions, we just have to extract the greatest common factor of 24m + 12, which is 4.

[tex]4(6m+3)[/tex]

Then, we separate the term 6m into m+5m

[tex]4(m+5m+3)[/tex]

At last, we use the commutative property

[tex]4(m+3+5m)[/tex]

There you have it, the equivalence has been demonstrated.

Part C.

Let's evaluate the expressions when m = 1.

[tex]4(m+3+5m)=4(1+3+5\cdot1)=4(1+3+5)=4(9)=36[/tex][tex]24m+12=24\cdot1+12=24+12=36[/tex]

As you can see, using m = 1, we proved that the expressions were equivalent.

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