Answer :
Sure, let's break down and solve each part step by step.
### 1. a) By how much is the sum of [tex]\(3 \frac{2}{3}\)[/tex] and [tex]\(2 \frac{1}{5}\)[/tex] less than 7?
First, convert the mixed numbers to improper fractions.
[tex]\[ 3 \frac{2}{3} = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \][/tex]
[tex]\[ 2 \frac{1}{5} = 2 + \frac{1}{5} = \frac{10}{5} + \frac{1}{5} = \frac{11}{5} \][/tex]
Next, find their sum:
[tex]\[ \frac{11}{3} + \frac{11}{5} \][/tex]
To add these, find a common denominator:
[tex]\[ \text{LCM of 3 and 5 is 15} \][/tex]
Convert each fraction:
[tex]\[ \frac{11}{3} = \frac{11 \times 5}{3 \times 5} = \frac{55}{15} \][/tex]
[tex]\[ \frac{11}{5} = \frac{11 \times 3}{5 \times 3} = \frac{33}{15} \][/tex]
Add the fractions:
[tex]\[ \frac{55}{15} + \frac{33}{15} = \frac{88}{15} \][/tex]
Convert 7 to a fraction with a denominator of 15:
[tex]\[ 7 = \frac{7 \times 15}{1 \times 15} = \frac{105}{15} \][/tex]
Calculate how much is the sum less than 7:
[tex]\[ \frac{105}{15} - \frac{88}{15} = \frac{17}{15} \][/tex]
Thus, the sum of [tex]\(3 \frac{2}{3}\)[/tex] and [tex]\(2 \frac{1}{5}\)[/tex] is [tex]\(\frac{17}{15}\)[/tex] less than 7.
### 1. b) Number of students who play:
Let's use set theory and Venn diagrams to solve this.
Given:
- Total students, [tex]\( n = 31 \)[/tex]
- Students playing football, [tex]\( F = 16 \)[/tex]
- Students playing table-tennis, [tex]\( T = 12 \)[/tex]
- Students playing both games, [tex]\( B = 5 \)[/tex]
i) At least one of the games:
[tex]\[ F \cup T = F + T - B = 16 + 12 - 5 = 23 \][/tex]
Thus, 23 students play at least one game.
ii) None of the games:
[tex]\[ \text{Total students} - F \cup T = 31 - 23 = 8 \][/tex]
Thus, 8 students play none of the games.
### 1. c) Simplify [tex]\(8 \sqrt{20} - 12 \sqrt{5} + 2 \sqrt{45}\)[/tex]
Break down each term:
[tex]\[ 8 \sqrt{20} = 8 \sqrt{4 \times 5} = 8 \times 2 \sqrt{5} = 16 \sqrt{5} \][/tex]
[tex]\[ 2 \sqrt{45} = 2 \sqrt{9 \times 5} = 2 \times 3 \sqrt{5} = 6 \sqrt{5} \][/tex]
Sum the terms:
[tex]\[ 16 \sqrt{5} - 12 \sqrt{5} + 6 \sqrt{5} = (16 - 12 + 6) \sqrt{5} = 10 \sqrt{5} \][/tex]
Thus, the simplified expression is [tex]\( 10 \sqrt{5} \)[/tex].
### 2. a) Find the width of the rectangular garden.
Let the width be [tex]\( w \)[/tex] yards.
Then the length is:
[tex]\[ \text{Length} = 2w + 0.5 \][/tex]
The perimeter of the rectangle is given by:
[tex]\[ 2(\text{Length} + \text{Width}) = 55 \][/tex]
[tex]\[ 2 (2w + 0.5 + w) = 55 \][/tex]
[tex]\[ 2 (3w + 0.5) = 55 \][/tex]
[tex]\[ 6w + 1 = 55 \][/tex]
[tex]\[ 6w = 54 \][/tex]
[tex]\[ w = 9 \][/tex]
Thus, the width of the garden is 9 yards.
### 2. b) Evaluate [tex]\( \frac{0.0048 \times 0.81}{0.0027 \times 0.004} \)[/tex], leaving your answer in standard form.
First, calculate the numerator and the denominator:
[tex]\[ 0.0048 \times 0.81 = 0.003888 \][/tex]
[tex]\[ 0.0027 \times 0.004 = 0.0000108 \][/tex]
Now, divide:
[tex]\[ \frac{0.003888}{0.0000108} = 360 \][/tex]
Convert to standard form:
[tex]\[ 360 = 3.6 \times 10^2 \][/tex]
Thus, the answer in standard form is [tex]\( 3.6 \times 10^2 \)[/tex].
### 2. c) Find the vector [tex]\( b \)[/tex] given [tex]\( 3a - 2b = c \)[/tex].
Given vectors:
[tex]\[ a = \begin{pmatrix} -1 \\ 4 \end{pmatrix}, \quad c = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Express the equation:
[tex]\[ 3a - 2b = c \][/tex]
Plug in the values:
[tex]\[ 3 \begin{pmatrix} -1 \\ 4 \end{pmatrix} - 2 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Calculate [tex]\( 3a \)[/tex]:
[tex]\[ 3 \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \end{pmatrix} \][/tex]
Subtract [tex]\( 2b \)[/tex] from [tex]\( 3a \)[/tex]:
[tex]\[ \begin{pmatrix} -3 \\ 12 \end{pmatrix} - \begin{pmatrix} 2x \\ 2y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Equate components:
[tex]\[ \begin{pmatrix} -3 - 2x = 3 \\ 12 - 2y = 2 \end{pmatrix} \][/tex]
Solve each equation:
[tex]\[ -3 - 2x = 3 \implies -2x = 6 \implies x = -3 \][/tex]
[tex]\[ 12 - 2y = 2 \implies -2y = -10 \implies y = 5 \][/tex]
Thus, the vector [tex]\( b = \begin{pmatrix} -3 \\ 5 \end{pmatrix} \)[/tex].
### 1. a) By how much is the sum of [tex]\(3 \frac{2}{3}\)[/tex] and [tex]\(2 \frac{1}{5}\)[/tex] less than 7?
First, convert the mixed numbers to improper fractions.
[tex]\[ 3 \frac{2}{3} = 3 + \frac{2}{3} = \frac{9}{3} + \frac{2}{3} = \frac{11}{3} \][/tex]
[tex]\[ 2 \frac{1}{5} = 2 + \frac{1}{5} = \frac{10}{5} + \frac{1}{5} = \frac{11}{5} \][/tex]
Next, find their sum:
[tex]\[ \frac{11}{3} + \frac{11}{5} \][/tex]
To add these, find a common denominator:
[tex]\[ \text{LCM of 3 and 5 is 15} \][/tex]
Convert each fraction:
[tex]\[ \frac{11}{3} = \frac{11 \times 5}{3 \times 5} = \frac{55}{15} \][/tex]
[tex]\[ \frac{11}{5} = \frac{11 \times 3}{5 \times 3} = \frac{33}{15} \][/tex]
Add the fractions:
[tex]\[ \frac{55}{15} + \frac{33}{15} = \frac{88}{15} \][/tex]
Convert 7 to a fraction with a denominator of 15:
[tex]\[ 7 = \frac{7 \times 15}{1 \times 15} = \frac{105}{15} \][/tex]
Calculate how much is the sum less than 7:
[tex]\[ \frac{105}{15} - \frac{88}{15} = \frac{17}{15} \][/tex]
Thus, the sum of [tex]\(3 \frac{2}{3}\)[/tex] and [tex]\(2 \frac{1}{5}\)[/tex] is [tex]\(\frac{17}{15}\)[/tex] less than 7.
### 1. b) Number of students who play:
Let's use set theory and Venn diagrams to solve this.
Given:
- Total students, [tex]\( n = 31 \)[/tex]
- Students playing football, [tex]\( F = 16 \)[/tex]
- Students playing table-tennis, [tex]\( T = 12 \)[/tex]
- Students playing both games, [tex]\( B = 5 \)[/tex]
i) At least one of the games:
[tex]\[ F \cup T = F + T - B = 16 + 12 - 5 = 23 \][/tex]
Thus, 23 students play at least one game.
ii) None of the games:
[tex]\[ \text{Total students} - F \cup T = 31 - 23 = 8 \][/tex]
Thus, 8 students play none of the games.
### 1. c) Simplify [tex]\(8 \sqrt{20} - 12 \sqrt{5} + 2 \sqrt{45}\)[/tex]
Break down each term:
[tex]\[ 8 \sqrt{20} = 8 \sqrt{4 \times 5} = 8 \times 2 \sqrt{5} = 16 \sqrt{5} \][/tex]
[tex]\[ 2 \sqrt{45} = 2 \sqrt{9 \times 5} = 2 \times 3 \sqrt{5} = 6 \sqrt{5} \][/tex]
Sum the terms:
[tex]\[ 16 \sqrt{5} - 12 \sqrt{5} + 6 \sqrt{5} = (16 - 12 + 6) \sqrt{5} = 10 \sqrt{5} \][/tex]
Thus, the simplified expression is [tex]\( 10 \sqrt{5} \)[/tex].
### 2. a) Find the width of the rectangular garden.
Let the width be [tex]\( w \)[/tex] yards.
Then the length is:
[tex]\[ \text{Length} = 2w + 0.5 \][/tex]
The perimeter of the rectangle is given by:
[tex]\[ 2(\text{Length} + \text{Width}) = 55 \][/tex]
[tex]\[ 2 (2w + 0.5 + w) = 55 \][/tex]
[tex]\[ 2 (3w + 0.5) = 55 \][/tex]
[tex]\[ 6w + 1 = 55 \][/tex]
[tex]\[ 6w = 54 \][/tex]
[tex]\[ w = 9 \][/tex]
Thus, the width of the garden is 9 yards.
### 2. b) Evaluate [tex]\( \frac{0.0048 \times 0.81}{0.0027 \times 0.004} \)[/tex], leaving your answer in standard form.
First, calculate the numerator and the denominator:
[tex]\[ 0.0048 \times 0.81 = 0.003888 \][/tex]
[tex]\[ 0.0027 \times 0.004 = 0.0000108 \][/tex]
Now, divide:
[tex]\[ \frac{0.003888}{0.0000108} = 360 \][/tex]
Convert to standard form:
[tex]\[ 360 = 3.6 \times 10^2 \][/tex]
Thus, the answer in standard form is [tex]\( 3.6 \times 10^2 \)[/tex].
### 2. c) Find the vector [tex]\( b \)[/tex] given [tex]\( 3a - 2b = c \)[/tex].
Given vectors:
[tex]\[ a = \begin{pmatrix} -1 \\ 4 \end{pmatrix}, \quad c = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Express the equation:
[tex]\[ 3a - 2b = c \][/tex]
Plug in the values:
[tex]\[ 3 \begin{pmatrix} -1 \\ 4 \end{pmatrix} - 2 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Calculate [tex]\( 3a \)[/tex]:
[tex]\[ 3 \begin{pmatrix} -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \end{pmatrix} \][/tex]
Subtract [tex]\( 2b \)[/tex] from [tex]\( 3a \)[/tex]:
[tex]\[ \begin{pmatrix} -3 \\ 12 \end{pmatrix} - \begin{pmatrix} 2x \\ 2y \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \][/tex]
Equate components:
[tex]\[ \begin{pmatrix} -3 - 2x = 3 \\ 12 - 2y = 2 \end{pmatrix} \][/tex]
Solve each equation:
[tex]\[ -3 - 2x = 3 \implies -2x = 6 \implies x = -3 \][/tex]
[tex]\[ 12 - 2y = 2 \implies -2y = -10 \implies y = 5 \][/tex]
Thus, the vector [tex]\( b = \begin{pmatrix} -3 \\ 5 \end{pmatrix} \)[/tex].
