College

Complete each calculation using the distributive property.

[tex]
\[
\begin{array}{ccc}
41 \cdot 25 & 79 \cdot 13 & 0.42 \cdot 60 \\
(40+1) \cdot 25 & (80-1) \cdot 13 & (0.4+0.02) \cdot 60
\end{array}
\]
[/tex]

Answer :

Sure! Let's solve each of these calculations using the distributive property:

1. Calculate [tex]\(41 \cdot 25\)[/tex] using the distributive property:

We can rewrite 41 as [tex]\(40 + 1\)[/tex]. So, we calculate:
[tex]\[
(40 + 1) \cdot 25 = (40 \cdot 25) + (1 \cdot 25)
\][/tex]
- First, multiply [tex]\(40 \cdot 25 = 1000\)[/tex].
- Then, multiply [tex]\(1 \cdot 25 = 25\)[/tex].
- Add the results: [tex]\(1000 + 25 = 1025\)[/tex].

So, [tex]\(41 \cdot 25 = 1025\)[/tex].

2. Calculate [tex]\(79 \cdot 13\)[/tex] using the distributive property:

We can rewrite 79 as [tex]\(80 - 1\)[/tex]. So, we calculate:
[tex]\[
(80 - 1) \cdot 13 = (80 \cdot 13) - (1 \cdot 13)
\][/tex]
- First, multiply [tex]\(80 \cdot 13 = 1040\)[/tex].
- Then, multiply [tex]\(1 \cdot 13 = 13\)[/tex].
- Subtract the results: [tex]\(1040 - 13 = 1027\)[/tex].

So, [tex]\(79 \cdot 13 = 1027\)[/tex].

3. Calculate [tex]\(0.42 \cdot 60\)[/tex] using the distributive property:

We can rewrite 0.42 as [tex]\(0.4 + 0.02\)[/tex]. So, we calculate:
[tex]\[
(0.4 + 0.02) \cdot 60 = (0.4 \cdot 60) + (0.02 \cdot 60)
\][/tex]
- First, multiply [tex]\(0.4 \cdot 60 = 24.0\)[/tex].
- Then, multiply [tex]\(0.02 \cdot 60 = 1.2\)[/tex].
- Add the results: [tex]\(24.0 + 1.2 = 25.2\)[/tex].

So, [tex]\(0.42 \cdot 60 = 25.2\)[/tex].

These are the solutions using the distributive property for each multiplication given.