Use logarithms table to evaluate (0.7214 x 20.17)/(608) and express the length and width of the seedbed in terms of x.

Here are the solutions to the questions from the image:

11. Use logarithms table to evaluate $\frac{0.7214 \times 20.17}{608}$.

Step 1: Let $N = \frac{0.7214 \times 20.17}{608}$.

Step 2: Take the logarithm of both sides.

$$\log N = \log(0.7214) + \log(20.17) - \log(608)$$

Step 3: Find the logarithms of each number using a logarithm table.

• $\log(0.7214) = \bar{1}.8582$
• $\log(20.17) = 1.3047$
• $\log(608) = 2.7839$

Step 4: Perform the addition and subtraction.

$$\log N = \bar{1}.8582 + 1.3047 - 2.7839$$ $$\log N = (-1 + 0.8582) + (1 + 0.3047) - (2 + 0.7839)$$ $$\log N = 0.8582 + 0.3047 - 2.7839$$ $$\log N = 1.1629 - 2.7839$$ $$\log N = -1.6210$$

Step 5: Convert the negative logarithm to the standard form with a negative characteristic and positive mantissa.

$$\log N = -2 + 0.3790 = \bar{2}.3790$$

Step 6: Find the antilogarithm of $\bar{2}.3790$.

$$\text{Antilog}(0.3790) \approx 2.393$$ $$N = 2.393 \times 10^{-2}$$ $$N = \boxed{0.02393}$$

12. During an agricultural project, every learner was to make a seedbed from a square piece of land measuring 3m by 3m leaving a uniform path of x cm all round.

• Original side length of the square land = 3 m = 300 cm.
• Path width = x cm all around.

a) Express the length and the width of the seedbed in terms of x.

The path reduces the length and width from both sides. So, the total reduction is $2x$.

• Length of seedbed = Original length - $2 \times$ path width $$\text{Length} = \boxed{(300 - 2x) \text{ cm}}$$
• Width of seedbed = Original width - $2 \times$ path width $$\text{Width} = \boxed{(300 - 2x) \text{ cm}}$$

b) Express the area of the seedbed in terms of x.

The seedbed is also square, with side length $(300 - 2x)$ cm.
Area of seedbed = Length $\times$ Width

$$\text{Area} = (300 - 2x) \times (300 - 2x)$$ $$\text{Area} = \boxed{(300 - 2x)^2 \text{ cm}^2}$$

13. Cherono's kitchen is in the shape of a square. Determine the area of the kitchen garden if each side measures $(2y + 11)$ using quadratic identities.

Step 1: The side length of the square kitchen garden is $(2y + 11)$.

Step 2: The area of a square is $(\text{side})^2$.

$$\text{Area} = (2y + 11)^2$$

Step 3: Use the quadratic identity $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a = 2y$ and $b = 11$.

$$\text{Area} = (2y)^2 + 2(2y)(11) + (11)^2$$ $$\text{Area} = 4y^2 + 44y + 121$$

The area of the kitchen garden is $\boxed{4y^2 + 44y + 121 \text{ square units}}$.

14. The sides of a rectangular field is 104 m and 297 m wide. Using the quadratic identity $(a+b)(a-b) = a^2 - b^2$ to determine the area of the rectangular field.

Step 1: The sides of the rectangular field are 297 m and 104 m. The area is $297 \times 104$.

Step 2: To use the identity $(a+b)(a-b) = a^2 - b^2$, we need to find $a$ and $b$ such that $a+b = 297$ and $a-b = 104$.

• Adding the two equations: $$(a+b) + (a-b) = 297 + 104$$ $$2a = 401$$ $$a = \frac{401}{2} = 200.5$$
• Subtracting the second equation from the first: $$(a+b) - (a-b) = 297 - 104$$ $$2b = 193$$ $$b = \frac{193}{2} = 96.5$$

Step 3: Substitute $a$ and $b$ into the identity.

$$\text{Area} = (200.5 + 96.5)(200.5 - 96.5)$$ $$\text{Area} = (200.5)^2 - (96.5)^2$$

Step 4: Calculate the squares.

$$(200.5)^2 = 40200.25$$ $$(96.5)^2 = 9312.25$$

Step 5: Subtract the values to find the area.

$$\text{Area} = 40200.25 - 9312.25$$ $$\text{Area} = \boxed{30888 \text{ m}^2}$$

15. Solve: $5^{x+2} = 125^{x-1}$

Step 1: Express both sides of the equation with the same base. We know that $125 = 5^3$.

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