Answer :
Below is a detailed step‐by‐step explanation for each calculation:
1. For the calculation
[tex]$$41 \cdot 25,$$[/tex]
we write 41 as [tex]$40+1$[/tex]. Then, using the distributive property,
[tex]$$ (40+1) \cdot 25 = 40\cdot 25 + 1\cdot 25. $$[/tex]
Evaluating each term:
[tex]$$ 40\cdot 25 = 1000 \quad \text{and} \quad 1\cdot 25 = 25. $$[/tex]
Adding these results gives:
[tex]$$ 1000 + 25 = 1025. $$[/tex]
2. For the subtraction
[tex]$$79-13,$$[/tex]
we can represent 79 as [tex]$80-1$[/tex]. Thus, we have:
[tex]$$ 79-13 = (80-1)-13. $$[/tex]
Group the [tex]$-1$[/tex] and [tex]$-13$[/tex] together:
[tex]$$ (80-1)-13 = 80 - (1+13). $$[/tex]
Since
[tex]$$ 1+13 = 14, $$[/tex]
it follows that
[tex]$$ 80-14 = 66. $$[/tex]
3. For the multiplication
[tex]$$0.42 \cdot 60,$$[/tex]
first express 0.42 as [tex]$0.40+0.02$[/tex]. Then,
[tex]$$ 0.42 \cdot 60 = (0.40+0.02) \cdot 60 = 0.40\cdot 60 + 0.02\cdot 60. $$[/tex]
Calculate each term:
[tex]$$ 0.40\cdot 60 = 24.0 \quad \text{and} \quad 0.02\cdot 60 = 1.2. $$[/tex]
Adding the results yields:
[tex]$$ 24.0 + 1.2 = 25.2. $$[/tex]
4. For the expression
[tex]$$(80-1) \cdot 13,$$[/tex]
apply the distributive property by multiplying each term by 13:
[tex]$$ (80-1) \cdot 13 = 80 \cdot 13 - 1 \cdot 13. $$[/tex]
Compute each part:
[tex]$$ 80 \cdot 13 = 1040 \quad \text{and} \quad 1 \cdot 13 = 13. $$[/tex]
Thus, the result is:
[tex]$$ 1040 - 13 = 1027. $$[/tex]
5. For the calculation
[tex]$$(0.4+0.02) \cdot 60,$$[/tex]
distribute the 60 to each term:
[tex]$$ (0.4+0.02) \cdot 60 = 0.4 \cdot 60 + 0.02 \cdot 60. $$[/tex]
Evaluate the multiplications:
[tex]$$ 0.4 \cdot 60 = 24.0 \quad \text{and} \quad 0.02 \cdot 60 = 1.2. $$[/tex]
Adding these values gives:
[tex]$$ 24.0 + 1.2 = 25.2. $$[/tex]
In summary, the final results are:
1. [tex]$$41 \cdot 25 = 1025 \quad \text{(obtained from } 1000 \text{ and } 25\text{)}.$$[/tex]
2. [tex]$$79 - 13 = 66 \quad \text{(using } 80, 1, \text{ and } 13 \text{ to get } 80-14\text{)}.$$[/tex]
3. [tex]$$0.42 \cdot 60 = 25.2 \quad \text{(from } 24.0 \text{ and } 1.2\text{)}.$$[/tex]
4. [tex]$$(80-1) \cdot 13 = 1027 \quad \text{(from } 1040 \text{ and } 13\text{)}.$$[/tex]
5. [tex]$$(0.4+0.02) \cdot 60 = 25.2 \quad \text{(from } 24.0 \text{ and } 1.2\text{)}.$$[/tex]
1. For the calculation
[tex]$$41 \cdot 25,$$[/tex]
we write 41 as [tex]$40+1$[/tex]. Then, using the distributive property,
[tex]$$ (40+1) \cdot 25 = 40\cdot 25 + 1\cdot 25. $$[/tex]
Evaluating each term:
[tex]$$ 40\cdot 25 = 1000 \quad \text{and} \quad 1\cdot 25 = 25. $$[/tex]
Adding these results gives:
[tex]$$ 1000 + 25 = 1025. $$[/tex]
2. For the subtraction
[tex]$$79-13,$$[/tex]
we can represent 79 as [tex]$80-1$[/tex]. Thus, we have:
[tex]$$ 79-13 = (80-1)-13. $$[/tex]
Group the [tex]$-1$[/tex] and [tex]$-13$[/tex] together:
[tex]$$ (80-1)-13 = 80 - (1+13). $$[/tex]
Since
[tex]$$ 1+13 = 14, $$[/tex]
it follows that
[tex]$$ 80-14 = 66. $$[/tex]
3. For the multiplication
[tex]$$0.42 \cdot 60,$$[/tex]
first express 0.42 as [tex]$0.40+0.02$[/tex]. Then,
[tex]$$ 0.42 \cdot 60 = (0.40+0.02) \cdot 60 = 0.40\cdot 60 + 0.02\cdot 60. $$[/tex]
Calculate each term:
[tex]$$ 0.40\cdot 60 = 24.0 \quad \text{and} \quad 0.02\cdot 60 = 1.2. $$[/tex]
Adding the results yields:
[tex]$$ 24.0 + 1.2 = 25.2. $$[/tex]
4. For the expression
[tex]$$(80-1) \cdot 13,$$[/tex]
apply the distributive property by multiplying each term by 13:
[tex]$$ (80-1) \cdot 13 = 80 \cdot 13 - 1 \cdot 13. $$[/tex]
Compute each part:
[tex]$$ 80 \cdot 13 = 1040 \quad \text{and} \quad 1 \cdot 13 = 13. $$[/tex]
Thus, the result is:
[tex]$$ 1040 - 13 = 1027. $$[/tex]
5. For the calculation
[tex]$$(0.4+0.02) \cdot 60,$$[/tex]
distribute the 60 to each term:
[tex]$$ (0.4+0.02) \cdot 60 = 0.4 \cdot 60 + 0.02 \cdot 60. $$[/tex]
Evaluate the multiplications:
[tex]$$ 0.4 \cdot 60 = 24.0 \quad \text{and} \quad 0.02 \cdot 60 = 1.2. $$[/tex]
Adding these values gives:
[tex]$$ 24.0 + 1.2 = 25.2. $$[/tex]
In summary, the final results are:
1. [tex]$$41 \cdot 25 = 1025 \quad \text{(obtained from } 1000 \text{ and } 25\text{)}.$$[/tex]
2. [tex]$$79 - 13 = 66 \quad \text{(using } 80, 1, \text{ and } 13 \text{ to get } 80-14\text{)}.$$[/tex]
3. [tex]$$0.42 \cdot 60 = 25.2 \quad \text{(from } 24.0 \text{ and } 1.2\text{)}.$$[/tex]
4. [tex]$$(80-1) \cdot 13 = 1027 \quad \text{(from } 1040 \text{ and } 13\text{)}.$$[/tex]
5. [tex]$$(0.4+0.02) \cdot 60 = 25.2 \quad \text{(from } 24.0 \text{ and } 1.2\text{)}.$$[/tex]
