Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
