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Complete each calculation using the distributive property.

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

Answer :

We can use the distributive property to break the multiplication into easier parts. Here is a step‐by‐step explanation.

1. Calculating [tex]$\boldsymbol{79\cdot 13}$[/tex]

Rewrite [tex]$79$[/tex] as [tex]$(80-1)$[/tex]. Then,
[tex]$$
79\cdot 13 = (80-1)\cdot 13.
$$[/tex]
Distribute the [tex]$13$[/tex]:
[tex]$$
(80-1)\cdot 13 = 80\cdot 13 - 1\cdot 13.
$$[/tex]
We have:
[tex]$$
80\cdot 13 = 1040 \quad \text{and} \quad 1\cdot 13 = 13.
$$[/tex]
Therefore,
[tex]$$
1040 - 13 = 1027.
$$[/tex]

2. Calculating [tex]$\boldsymbol{0.42\cdot 60}$[/tex]

Write [tex]$0.42$[/tex] as [tex]$(0.4+0.02)$[/tex]. Thus,
[tex]$$
0.42\cdot 60 = (0.4+0.02)\cdot 60.
$$[/tex]
Use the distributive property:
[tex]$$
(0.4+0.02)\cdot 60 = 0.4\cdot 60 + 0.02\cdot 60.
$$[/tex]
We calculate:
[tex]$$
0.4\cdot 60 = 24 \quad \text{and} \quad 0.02\cdot 60 = 1.2.
$$[/tex]
Adding these results,
[tex]$$
24 + 1.2 = 25.2.
$$[/tex]

3. Calculating [tex]$\boldsymbol{41\cdot 25}$[/tex]

Express [tex]$41$[/tex] as [tex]$(40+1)$[/tex]. Then,
[tex]$$
41\cdot 25 = (40+1)\cdot 25.
$$[/tex]
Distributing the [tex]$25$[/tex]:
[tex]$$
(40+1)\cdot 25 = 40\cdot 25 + 1\cdot 25.
$$[/tex]
We have:
[tex]$$
40\cdot 25 = 1000 \quad \text{and} \quad 1\cdot 25 = 25.
$$[/tex]
Thus,
[tex]$$
1000 + 25 = 1025.
$$[/tex]

4. Other Calculations:

Notice that:

- The expression [tex]$(80-1)\cdot 13$[/tex] is the same as in step 1, so it equals [tex]$1027$[/tex].
- The expression [tex]$(0.4+0.02)\cdot 60$[/tex] is the same as in step 2, so it equals [tex]$25.2$[/tex].
- The expression [tex]$(40+1)\cdot 25$[/tex] is the same as in step 3, so it equals [tex]$1025$[/tex].

5. Final Results:

- For [tex]$79\cdot 13$[/tex], the intermediate steps were
[tex]$$
80\cdot 13 = 1040 \quad \text{and} \quad 1\cdot 13 = 13,
$$[/tex]
giving the result [tex]$1027$[/tex].

- For [tex]$0.42\cdot 60$[/tex], the intermediate results were
[tex]$$
0.4\cdot 60 = 24 \quad \text{and} \quad 0.02\cdot 60 = 1.2,
$$[/tex]
giving the result [tex]$25.2$[/tex].

- For [tex]$41\cdot 25$[/tex], the intermediate steps were
[tex]$$
40\cdot 25 = 1000 \quad \text{and} \quad 1\cdot 25 = 25,
$$[/tex]
giving the result [tex]$1025$[/tex].

- The expressions [tex]$(80-1)\cdot 13$[/tex], [tex]$(0.4+0.02)\cdot 60$[/tex], and [tex]$(40+1)\cdot 25$[/tex] repeat the above calculations and produce the same intermediate results and final totals.

So the final numerical results can be summarized as:
[tex]$$
\begin{aligned}
79\cdot 13 &: \quad (1040,\;13,\;1027)\\[1mm]
0.42\cdot 60 &: \quad (24,\;1.2,\;25.2)\\[1mm]
41\cdot 25 &: \quad (1000,\;25,\;1025)\\[1mm]
(80-1)\cdot 13 &: \quad (1040,\;13,\;1027)\\[1mm]
(0.4+0.02)\cdot 60 &: \quad (24,\;1.2,\;25.2)\\[1mm]
(40+1)\cdot 25 &: \quad (1000,\;25,\;1025)
\end{aligned}
$$[/tex]

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