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Sure! Let's break down the problems step-by-step:
### Part B: Mr. Yeboah's monthly salary
Mr. Yeboah spent fractions of his monthly salary on different expenses:
- [tex]\(\frac{1}{4}\)[/tex] on rent
- [tex]\(\frac{2}{5}\)[/tex] on food
- [tex]\(\frac{1}{6}\)[/tex] on books
He still has GHC 55.00 left after these expenses. We need to calculate his monthly salary.
1. First, let's find the total fraction spent:
[tex]\[ \text{Total fraction spent} = \frac{1}{4} + \frac{2}{5} + \frac{1}{6} \][/tex]
2. To add these fractions, find a common denominator. The least common multiple (LCM) of 4, 5, and 6 is 60.
[tex]\[ \frac{1}{4} = \frac{15}{60}, \quad \frac{2}{5} = \frac{24}{60}, \quad \frac{1}{6} = \frac{10}{60} \][/tex]
3. Adding these fractions together
[tex]\[ \frac{15}{60} + \frac{24}{60} + \frac{10}{60} = \frac{49}{60} \][/tex]
4. So, the total fraction spent is [tex]\(\frac{49}{60}\)[/tex]. This means Mr. Yeboah spends [tex]\(\frac{49}{60}\)[/tex] of his salary and has [tex]\(\frac{11}{60}\)[/tex] of his salary left.
[tex]\[ \frac{11}{60} \text{ of the salary is left which equals GHC 55.00} \][/tex]
5. Let's denote his monthly salary as [tex]\( x \)[/tex].
[tex]\[ \frac{11}{60} x = 55 \][/tex]
6. To find [tex]\(x\)[/tex], solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{55 \times 60}{11} = \frac{3300}{11} = 300 \][/tex]
So, Mr. Yeboah’s monthly salary is GHC 300.
### Part A: Factorize [tex]\(x^2 + 10x + 25\)[/tex]
To factorize this quadratic expression, we can recognize it as a perfect square trinomial.
[tex]\[ x^2 + 10x + 25 \][/tex]
Consider the expression in the form [tex]\((x + a)^2\)[/tex], where [tex]\(a\)[/tex] equals the constant term divided by 2 (from the middle term of the trinomial).
[tex]\[ (x + 5)^2 = x^2 + 2 \times 5x + 5^2 = x^2 + 10x + 25 \][/tex]
Thus, the factorized form is:
[tex]\[ (x + 5)^2 \][/tex]
### Part B: Evaluate [tex]\(\sqrt{\frac{0.0048 \times 0.81 \times 10^{-7}}{0.027 \times 0.04 \times 10^6}}\)[/tex]
First, simplify the fraction inside the square root:
1. Combine the numerical values:
[tex]\[ \frac{0.0048 \times 0.81}{0.027 \times 0.04} \][/tex]
2. Simplify each component step-by-step:
[tex]\[ 0.0048 \div 0.027 = 0.18 \quad (\text{since } 0.0048/0.027 = \frac{48}{2700} = \frac{4.8}{270} \approx 0.18) \][/tex]
[tex]\[ 0.81 \div 0.04 = 20.25 \quad (\text{since } 0.81/0.04 = \frac{81}{4} = 20.25) \][/tex]
3. Combine them:
[tex]\[ 0.18 \times 20.25 = 3.645 \][/tex]
4. Handle the exponents of 10:
[tex]\[ \frac{10^{-7}}{10^6} = 10^{-7-6} = 10^{-13} \][/tex]
So, the fraction inside the square root simplifies to:
[tex]\[ \frac{3.645 \times 10^{-13}} \][/tex]
Now, calculate the square root:
[tex]\[ \sqrt{3.645 \times 10^{-13}} = \sqrt{3.645} \times \sqrt{10^{-13}} \][/tex]
5. Simplify each part:
[tex]\[ \sqrt{3.645} \approx 1.91 \quad (\text{using approximate square root calculation}) \][/tex]
[tex]\[ \sqrt{10^{-13}} = 10^{-6.5} \][/tex]
Combine them:
[tex]\[ 1.91 \times 10^{-6.5} = 1.91 \times 10^{-7} \quad (\text{since } 10^{-6.5} = 10^{-7}) \][/tex]
So, the final answer in standard form is:
[tex]\[ 1.91 \times 10^{-7} \][/tex]
### Part B: Mr. Yeboah's monthly salary
Mr. Yeboah spent fractions of his monthly salary on different expenses:
- [tex]\(\frac{1}{4}\)[/tex] on rent
- [tex]\(\frac{2}{5}\)[/tex] on food
- [tex]\(\frac{1}{6}\)[/tex] on books
He still has GHC 55.00 left after these expenses. We need to calculate his monthly salary.
1. First, let's find the total fraction spent:
[tex]\[ \text{Total fraction spent} = \frac{1}{4} + \frac{2}{5} + \frac{1}{6} \][/tex]
2. To add these fractions, find a common denominator. The least common multiple (LCM) of 4, 5, and 6 is 60.
[tex]\[ \frac{1}{4} = \frac{15}{60}, \quad \frac{2}{5} = \frac{24}{60}, \quad \frac{1}{6} = \frac{10}{60} \][/tex]
3. Adding these fractions together
[tex]\[ \frac{15}{60} + \frac{24}{60} + \frac{10}{60} = \frac{49}{60} \][/tex]
4. So, the total fraction spent is [tex]\(\frac{49}{60}\)[/tex]. This means Mr. Yeboah spends [tex]\(\frac{49}{60}\)[/tex] of his salary and has [tex]\(\frac{11}{60}\)[/tex] of his salary left.
[tex]\[ \frac{11}{60} \text{ of the salary is left which equals GHC 55.00} \][/tex]
5. Let's denote his monthly salary as [tex]\( x \)[/tex].
[tex]\[ \frac{11}{60} x = 55 \][/tex]
6. To find [tex]\(x\)[/tex], solve for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{55 \times 60}{11} = \frac{3300}{11} = 300 \][/tex]
So, Mr. Yeboah’s monthly salary is GHC 300.
### Part A: Factorize [tex]\(x^2 + 10x + 25\)[/tex]
To factorize this quadratic expression, we can recognize it as a perfect square trinomial.
[tex]\[ x^2 + 10x + 25 \][/tex]
Consider the expression in the form [tex]\((x + a)^2\)[/tex], where [tex]\(a\)[/tex] equals the constant term divided by 2 (from the middle term of the trinomial).
[tex]\[ (x + 5)^2 = x^2 + 2 \times 5x + 5^2 = x^2 + 10x + 25 \][/tex]
Thus, the factorized form is:
[tex]\[ (x + 5)^2 \][/tex]
### Part B: Evaluate [tex]\(\sqrt{\frac{0.0048 \times 0.81 \times 10^{-7}}{0.027 \times 0.04 \times 10^6}}\)[/tex]
First, simplify the fraction inside the square root:
1. Combine the numerical values:
[tex]\[ \frac{0.0048 \times 0.81}{0.027 \times 0.04} \][/tex]
2. Simplify each component step-by-step:
[tex]\[ 0.0048 \div 0.027 = 0.18 \quad (\text{since } 0.0048/0.027 = \frac{48}{2700} = \frac{4.8}{270} \approx 0.18) \][/tex]
[tex]\[ 0.81 \div 0.04 = 20.25 \quad (\text{since } 0.81/0.04 = \frac{81}{4} = 20.25) \][/tex]
3. Combine them:
[tex]\[ 0.18 \times 20.25 = 3.645 \][/tex]
4. Handle the exponents of 10:
[tex]\[ \frac{10^{-7}}{10^6} = 10^{-7-6} = 10^{-13} \][/tex]
So, the fraction inside the square root simplifies to:
[tex]\[ \frac{3.645 \times 10^{-13}} \][/tex]
Now, calculate the square root:
[tex]\[ \sqrt{3.645 \times 10^{-13}} = \sqrt{3.645} \times \sqrt{10^{-13}} \][/tex]
5. Simplify each part:
[tex]\[ \sqrt{3.645} \approx 1.91 \quad (\text{using approximate square root calculation}) \][/tex]
[tex]\[ \sqrt{10^{-13}} = 10^{-6.5} \][/tex]
Combine them:
[tex]\[ 1.91 \times 10^{-6.5} = 1.91 \times 10^{-7} \quad (\text{since } 10^{-6.5} = 10^{-7}) \][/tex]
So, the final answer in standard form is:
[tex]\[ 1.91 \times 10^{-7} \][/tex]
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