Answer :
Sure, let's break down and solve each part of the question step-by-step.
### (i) Use distributive property to solve [tex]\( 395 \times 12 \)[/tex]
To solve [tex]\( 395 \times 12 \)[/tex] using the distributive property, we can break down the multiplication as follows:
[tex]\[ 395 \times 12 = 395 \times (10 + 2) = (395 \times 10) + (395 \times 2) \][/tex]
Calculating each term:
[tex]\[ 395 \times 10 = 3950 \][/tex]
[tex]\[ 395 \times 2 = 790 \][/tex]
Adding the results:
[tex]\[ 3950 + 790 = 4740 \][/tex]
Thus, [tex]\( 395 \times 12 = 4740 \)[/tex].
### (ii) Round 4,398 and 256 to the nearest tens. After rounding off, add them. Round off the sum to the nearest hundreds.
First, round each number to the nearest tens:
- 4398 rounds to 4400 (since the units digit is 8, which is 5 or more)
- 256 rounds to 260 (since the units digit is 6, which is 5 or more)
Next, add the rounded numbers:
[tex]\[ 4400 + 260 = 4660 \][/tex]
Finally, round 4660 to the nearest hundreds:
- 4660 rounds to 4700 (since the tens digit is 6, which is 5 or more)
So, the results are:
- Rounded values: 4400 and 260
- Sum: 4660
- Rounded sum: 4700
### Fred is [tex]\((n-1)\)[/tex] years old now. How old:
(i) Was he 4 years ago?
Fred was [tex]\( (n - 1) - 4 = n - 5 \)[/tex] years old 4 years ago.
(ii) Will he be 8 years from now?
Fred will be [tex]\( (n - 1) + 8 = n + 7 \)[/tex] years old 8 years from now.
(iii) Is he now, if his age in 8 years' time will be three times his age 4 years ago?
We are given:
[tex]\[ n + 7 = 3(n - 5) \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ n + 7 = 3n - 15 \][/tex]
[tex]\[ 7 + 15 = 3n - n \][/tex]
[tex]\[ 22 = 2n \][/tex]
[tex]\[ n = \frac{22}{2} \][/tex]
[tex]\[ n = 11 \][/tex]
So, Fred is currently 11 years old.
### Find the sum of [tex]\( \left( 6m^2n + 3mn - 2n^2 \right), \left( n^2 - mn - m^2n \right) \)[/tex] and [tex]\( \left( mn - 3n^2 - 4m^2n \right) \)[/tex]
Let’s add these three expressions step-by-step:
[tex]\[ \left( 6m^2n + 3mn - 2n^2 \right) + \left( n^2 - mn - m^2n \right) + \left( mn - 3n^2 - 4m^2n \right) \][/tex]
Combine like terms:
- For [tex]\( m^2n \)[/tex]: [tex]\( 6m^2n - m^2n - 4m^2n = 1m^2n \)[/tex]
- For [tex]\( mn \)[/tex]: [tex]\( 3mn - mn + mn = 3mn \)[/tex]
- For [tex]\( n^2 \)[/tex]: [tex]\( -2n^2 + n^2 - 3n^2 = -4n^2 \)[/tex]
Thus, the sum of the expressions is:
[tex]\[ m^2n + 3mn - 4n^2 = n(m^2 + 3m - 4n) \][/tex]
### (d) Find the sum of [tex]\( 2483.65, 701.532 \)[/tex] and [tex]\( 102.7 \)[/tex], leaving your answer in one decimal place.
First, add the numbers:
[tex]\[ 2483.65 + 701.532 + 102.7 = 3287.882 \][/tex]
Round to one decimal place:
[tex]\[ 3287.882 \approx 3287.9 \][/tex]
So, the sum to one decimal place is [tex]\( 3287.9 \)[/tex].
Thus, we have solved all parts of the question step-by-step.
### (i) Use distributive property to solve [tex]\( 395 \times 12 \)[/tex]
To solve [tex]\( 395 \times 12 \)[/tex] using the distributive property, we can break down the multiplication as follows:
[tex]\[ 395 \times 12 = 395 \times (10 + 2) = (395 \times 10) + (395 \times 2) \][/tex]
Calculating each term:
[tex]\[ 395 \times 10 = 3950 \][/tex]
[tex]\[ 395 \times 2 = 790 \][/tex]
Adding the results:
[tex]\[ 3950 + 790 = 4740 \][/tex]
Thus, [tex]\( 395 \times 12 = 4740 \)[/tex].
### (ii) Round 4,398 and 256 to the nearest tens. After rounding off, add them. Round off the sum to the nearest hundreds.
First, round each number to the nearest tens:
- 4398 rounds to 4400 (since the units digit is 8, which is 5 or more)
- 256 rounds to 260 (since the units digit is 6, which is 5 or more)
Next, add the rounded numbers:
[tex]\[ 4400 + 260 = 4660 \][/tex]
Finally, round 4660 to the nearest hundreds:
- 4660 rounds to 4700 (since the tens digit is 6, which is 5 or more)
So, the results are:
- Rounded values: 4400 and 260
- Sum: 4660
- Rounded sum: 4700
### Fred is [tex]\((n-1)\)[/tex] years old now. How old:
(i) Was he 4 years ago?
Fred was [tex]\( (n - 1) - 4 = n - 5 \)[/tex] years old 4 years ago.
(ii) Will he be 8 years from now?
Fred will be [tex]\( (n - 1) + 8 = n + 7 \)[/tex] years old 8 years from now.
(iii) Is he now, if his age in 8 years' time will be three times his age 4 years ago?
We are given:
[tex]\[ n + 7 = 3(n - 5) \][/tex]
Solving for [tex]\( n \)[/tex]:
[tex]\[ n + 7 = 3n - 15 \][/tex]
[tex]\[ 7 + 15 = 3n - n \][/tex]
[tex]\[ 22 = 2n \][/tex]
[tex]\[ n = \frac{22}{2} \][/tex]
[tex]\[ n = 11 \][/tex]
So, Fred is currently 11 years old.
### Find the sum of [tex]\( \left( 6m^2n + 3mn - 2n^2 \right), \left( n^2 - mn - m^2n \right) \)[/tex] and [tex]\( \left( mn - 3n^2 - 4m^2n \right) \)[/tex]
Let’s add these three expressions step-by-step:
[tex]\[ \left( 6m^2n + 3mn - 2n^2 \right) + \left( n^2 - mn - m^2n \right) + \left( mn - 3n^2 - 4m^2n \right) \][/tex]
Combine like terms:
- For [tex]\( m^2n \)[/tex]: [tex]\( 6m^2n - m^2n - 4m^2n = 1m^2n \)[/tex]
- For [tex]\( mn \)[/tex]: [tex]\( 3mn - mn + mn = 3mn \)[/tex]
- For [tex]\( n^2 \)[/tex]: [tex]\( -2n^2 + n^2 - 3n^2 = -4n^2 \)[/tex]
Thus, the sum of the expressions is:
[tex]\[ m^2n + 3mn - 4n^2 = n(m^2 + 3m - 4n) \][/tex]
### (d) Find the sum of [tex]\( 2483.65, 701.532 \)[/tex] and [tex]\( 102.7 \)[/tex], leaving your answer in one decimal place.
First, add the numbers:
[tex]\[ 2483.65 + 701.532 + 102.7 = 3287.882 \][/tex]
Round to one decimal place:
[tex]\[ 3287.882 \approx 3287.9 \][/tex]
So, the sum to one decimal place is [tex]\( 3287.9 \)[/tex].
Thus, we have solved all parts of the question step-by-step.
