Question: Factorize: 1) \( K + 6Ph + 3F + 3Kh \) Expand: 2) \( (2n + 1)^2 \) 3) Make \( V \) the…

Factorize:

1. \( K + 6Ph + 3F + 3Kh \) Expand:
2. \( (2n + 1)^2 \)
3. Make \( V \) the subject of formula in \( K = \frac{1}{2}(v + 1) \)

The formula \( C = 180n + 450 \) gives the weekly cost \( C \) in rand of running a household of \( n \) people.

4. Find the weekly cost of free people.

5. How many people are there if the weekly cost is \( R1170 \)?

Solution

Question 1: Expand \( K + 6n + 2F + 2Kh \) To expand the expression, combine like terms. \[ K + 2Kh = K(1 + 2h) & \] So, the expanded form is: \[ K(1 + 2h) + 6n + 2F & \] --- Question 2: Expand \( (2n + 1)^2 \) Use the formula \( (a + b)^2 = a^2 + 2ab + b^2 \). \[ (2n + 1)^2 = (2n)^2 + 2 \cdot 2n \cdot 1 + 1^2 & \] \[ = 4n^2 + 4n + 1 & \] --- Question 3: Make \( V \) the subject of the formula \( K = \frac{1}{2}(V + r) \) \[ K = \frac{1}{2}(V + r) & \] Multiply both sides by 2: \[ 2K = V + r & \] Subtract \( r \) from both sides: \[ V = 2K - r & \] --- Question 4: Find the weekly cost of five people using \( C = 180n + 450 \) Let \( n = 5 \): \[ C = 180(5) + 450 & \] \[ C = 900 + 450 & \] \[ C = 1350 & \] The weekly cost for five people is \$1350. --- Question 5: How many people are there if the weekly cost is \$1170 using \( C = 180n + 450 \) Given \( C = 1170 \): \[ 1170 = 180n + 450 & \] Subtract 450 from both sides: \[ 720 = 180n & \] Divide both sides by 180: \[ n = \frac{720}{180} & \] \[ n = 4 & \] There are 4 people.