Answer :
Certainly! Let's break down each part of the problem step-by-step.
### 4. a. Find the difference between the sum of 3.85 and 5.19 and the product of 4.7 and 6.2.
First, calculate the sum of 3.85 and 5.19:
[tex]\[ \text{Sum} = 3.85 + 5.19 = 9.04 \][/tex]
Next, calculate the product of 4.7 and 6.2:
[tex]\[ \text{Product} = 4.7 \times 6.2 = 29.14 \][/tex]
Now, determine the difference between the sum and the product:
[tex]\[ \text{Difference} = 9.04 - 29.14 = -20.1 \][/tex]
### 4. b. Evaluate [tex]\(\frac{5.25 \times 0.004}{0.2 \times 0.5}\)[/tex] leaving the answer in standard form.
First, calculate the numerator:
[tex]\[ \text{Numerator} = 5.25 \times 0.004 = 0.021 \][/tex]
Next, calculate the denominator:
[tex]\[ \text{Denominator} = 0.2 \times 0.5 = 0.1 \][/tex]
Now, divide the numerator by the denominator to get the result in standard form:
[tex]\[ \frac{0.021}{0.1} = 0.21 \][/tex]
### 4. c. Factorize [tex]\(5xy - 15y - 3x + 9\)[/tex] completely.
Let's factor by grouping:
[tex]\[ 5xy - 15y - 3x + 9 \][/tex]
Group the terms:
[tex]\[ (5xy - 15y) - (3x - 9) \][/tex]
Factor out common terms in each group:
[tex]\[ 5y(x - 3) - 3(x - 3) \][/tex]
Now, factor out the common binomial factor [tex]\((x - 3)\)[/tex]:
[tex]\[ (x - 3)(5y - 3) \][/tex]
So, the factorized form is:
[tex]\[ (x - 3)(5y - 3) \][/tex]
### 4. d. Kofi, Koo, and Kesse shared an amount of GH₵ 270,000.00 in the ratio 4:2:12 respectively. Find Kesse's share.
First, find the total ratio:
[tex]\[ \text{Total ratio} = 4 + 2 + 12 = 18 \][/tex]
Next, find Kesse's share in the ratio:
[tex]\[ Kesse's \ share = \frac{12}{18} \times 270,000.00 \][/tex]
[tex]\[ Kesse's \ share = \frac{2}{3} \times 270,000.00 = 180,000.00 \][/tex]
Kesse's share is GH₵ 180,000.00.
### 5. a. A ladder leans against a wall of height 12 m, the foot of the ladder is 5 m away from the wall. Find the length of the ladder.
Using the Pythagorean theorem:
[tex]\[ \text{Length}^2 = \text{Height}^2 + \text{Distance}^2 \][/tex]
[tex]\[ \text{Length}^2 = 12^2 + 5^2 \][/tex]
[tex]\[ \text{Length}^2 = 144 + 25 \][/tex]
[tex]\[ \text{Length}^2 = 169 \][/tex]
Thus, the length of the ladder:
[tex]\[ \text{Length} = \sqrt{169} = 13 \ \text{m} \][/tex]
### 5. b. If [tex]\(3(t-1)-1=\frac{t}{3}\)[/tex], find the value of [tex]\(t\)[/tex].
First, expand and simplify the equation:
[tex]\[ 3(t - 1) - 1 = \frac{t}{3} \][/tex]
[tex]\[ 3t - 3 - 1 = \frac{t}{3} \][/tex]
[tex]\[ 3t - 4 = \frac{t}{3} \][/tex]
To clear the fraction, multiply through by 3:
[tex]\[ 9t - 12 = t \][/tex]
Bring [tex]\(t\)[/tex] to one side:
[tex]\[ 9t - t = 12 \][/tex]
[tex]\[ 8t = 12 \][/tex]
[tex]\[ t = \frac{12}{8} \][/tex]
[tex]\[ t = \frac{3}{2} \][/tex]
### 5. c.
In the diagram, given [tex]\(WX = WY\)[/tex], [tex]\(\angle WZY = 40^\circ\)[/tex] and [tex]\(\angle WYZ = 14x - 15\)[/tex].
#### i. Find the value of [tex]\(x\)[/tex].
Since [tex]\(WX = WY\)[/tex], triangle [tex]\(WXY\)[/tex] is isosceles, and thus:
[tex]\[ 2 \times \angle WZY + \angle WYZ = 180^\circ \][/tex]
Substituting the given angles:
[tex]\[ 2 \times 40^\circ + 14x - 15 = 180^\circ \][/tex]
[tex]\[ 80 + 14x - 15 = 180 \][/tex]
[tex]\[ 14x + 65 = 180 \][/tex]
[tex]\[ 14x = 115 \][/tex]
[tex]\[ x = \frac{115}{14} \][/tex]
#### ii. Find [tex]\(\angle ZWY\)[/tex].
Using the value of [tex]\(x\)[/tex]:
[tex]\[ \angle ZWY = 14x - 15 \][/tex]
[tex]\[ \angle ZWY = 14 \times \frac{115}{14} - 15 \][/tex]
[tex]\[ \angle ZWY = 115 - 15 \][/tex]
[tex]\[ \angle ZWY = 100 \][/tex]
Putting the answers together, we obtain:
- The difference in part 4a is [tex]\(-20.1\)[/tex].
- The standard form in part 4b is [tex]\(0.21\)[/tex].
- The factorized form in part 4c is [tex]\((x - 3)(5y - 3)\)[/tex].
- Kesse's share in part 4d is GH₵ 180,000.00.
- The length of the ladder in part 5a is 13 m.
- The value of [tex]\(t\)[/tex] in part 5b is [tex]\(\frac{3}{2}\)[/tex].
- The value of [tex]\(x\)[/tex] in part 5ci is [tex]\(\frac{115}{14}\)[/tex].
- The angle [tex]\(ZWY\)[/tex] in part 5cii is [tex]\(100^\circ\)[/tex].
I hope this detailed step-by-step solution helps you understand how to approach and solve each part of the problem!
### 4. a. Find the difference between the sum of 3.85 and 5.19 and the product of 4.7 and 6.2.
First, calculate the sum of 3.85 and 5.19:
[tex]\[ \text{Sum} = 3.85 + 5.19 = 9.04 \][/tex]
Next, calculate the product of 4.7 and 6.2:
[tex]\[ \text{Product} = 4.7 \times 6.2 = 29.14 \][/tex]
Now, determine the difference between the sum and the product:
[tex]\[ \text{Difference} = 9.04 - 29.14 = -20.1 \][/tex]
### 4. b. Evaluate [tex]\(\frac{5.25 \times 0.004}{0.2 \times 0.5}\)[/tex] leaving the answer in standard form.
First, calculate the numerator:
[tex]\[ \text{Numerator} = 5.25 \times 0.004 = 0.021 \][/tex]
Next, calculate the denominator:
[tex]\[ \text{Denominator} = 0.2 \times 0.5 = 0.1 \][/tex]
Now, divide the numerator by the denominator to get the result in standard form:
[tex]\[ \frac{0.021}{0.1} = 0.21 \][/tex]
### 4. c. Factorize [tex]\(5xy - 15y - 3x + 9\)[/tex] completely.
Let's factor by grouping:
[tex]\[ 5xy - 15y - 3x + 9 \][/tex]
Group the terms:
[tex]\[ (5xy - 15y) - (3x - 9) \][/tex]
Factor out common terms in each group:
[tex]\[ 5y(x - 3) - 3(x - 3) \][/tex]
Now, factor out the common binomial factor [tex]\((x - 3)\)[/tex]:
[tex]\[ (x - 3)(5y - 3) \][/tex]
So, the factorized form is:
[tex]\[ (x - 3)(5y - 3) \][/tex]
### 4. d. Kofi, Koo, and Kesse shared an amount of GH₵ 270,000.00 in the ratio 4:2:12 respectively. Find Kesse's share.
First, find the total ratio:
[tex]\[ \text{Total ratio} = 4 + 2 + 12 = 18 \][/tex]
Next, find Kesse's share in the ratio:
[tex]\[ Kesse's \ share = \frac{12}{18} \times 270,000.00 \][/tex]
[tex]\[ Kesse's \ share = \frac{2}{3} \times 270,000.00 = 180,000.00 \][/tex]
Kesse's share is GH₵ 180,000.00.
### 5. a. A ladder leans against a wall of height 12 m, the foot of the ladder is 5 m away from the wall. Find the length of the ladder.
Using the Pythagorean theorem:
[tex]\[ \text{Length}^2 = \text{Height}^2 + \text{Distance}^2 \][/tex]
[tex]\[ \text{Length}^2 = 12^2 + 5^2 \][/tex]
[tex]\[ \text{Length}^2 = 144 + 25 \][/tex]
[tex]\[ \text{Length}^2 = 169 \][/tex]
Thus, the length of the ladder:
[tex]\[ \text{Length} = \sqrt{169} = 13 \ \text{m} \][/tex]
### 5. b. If [tex]\(3(t-1)-1=\frac{t}{3}\)[/tex], find the value of [tex]\(t\)[/tex].
First, expand and simplify the equation:
[tex]\[ 3(t - 1) - 1 = \frac{t}{3} \][/tex]
[tex]\[ 3t - 3 - 1 = \frac{t}{3} \][/tex]
[tex]\[ 3t - 4 = \frac{t}{3} \][/tex]
To clear the fraction, multiply through by 3:
[tex]\[ 9t - 12 = t \][/tex]
Bring [tex]\(t\)[/tex] to one side:
[tex]\[ 9t - t = 12 \][/tex]
[tex]\[ 8t = 12 \][/tex]
[tex]\[ t = \frac{12}{8} \][/tex]
[tex]\[ t = \frac{3}{2} \][/tex]
### 5. c.
In the diagram, given [tex]\(WX = WY\)[/tex], [tex]\(\angle WZY = 40^\circ\)[/tex] and [tex]\(\angle WYZ = 14x - 15\)[/tex].
#### i. Find the value of [tex]\(x\)[/tex].
Since [tex]\(WX = WY\)[/tex], triangle [tex]\(WXY\)[/tex] is isosceles, and thus:
[tex]\[ 2 \times \angle WZY + \angle WYZ = 180^\circ \][/tex]
Substituting the given angles:
[tex]\[ 2 \times 40^\circ + 14x - 15 = 180^\circ \][/tex]
[tex]\[ 80 + 14x - 15 = 180 \][/tex]
[tex]\[ 14x + 65 = 180 \][/tex]
[tex]\[ 14x = 115 \][/tex]
[tex]\[ x = \frac{115}{14} \][/tex]
#### ii. Find [tex]\(\angle ZWY\)[/tex].
Using the value of [tex]\(x\)[/tex]:
[tex]\[ \angle ZWY = 14x - 15 \][/tex]
[tex]\[ \angle ZWY = 14 \times \frac{115}{14} - 15 \][/tex]
[tex]\[ \angle ZWY = 115 - 15 \][/tex]
[tex]\[ \angle ZWY = 100 \][/tex]
Putting the answers together, we obtain:
- The difference in part 4a is [tex]\(-20.1\)[/tex].
- The standard form in part 4b is [tex]\(0.21\)[/tex].
- The factorized form in part 4c is [tex]\((x - 3)(5y - 3)\)[/tex].
- Kesse's share in part 4d is GH₵ 180,000.00.
- The length of the ladder in part 5a is 13 m.
- The value of [tex]\(t\)[/tex] in part 5b is [tex]\(\frac{3}{2}\)[/tex].
- The value of [tex]\(x\)[/tex] in part 5ci is [tex]\(\frac{115}{14}\)[/tex].
- The angle [tex]\(ZWY\)[/tex] in part 5cii is [tex]\(100^\circ\)[/tex].
I hope this detailed step-by-step solution helps you understand how to approach and solve each part of the problem!
