We are given three problems. (a) Simplify the expression $(4x+2)(x-2)-3x^2$. (b) Find the value of $x$ given that the angles $40^\circ, 60^\circ, 100^\circ, 3x^\circ$ and $5x^\circ$ are formed at the center of a circle. (c) (i) Find the cost of producing a book with 220 pages, given the cost function $C = 25 + 0.6x$. (ii) Find the number of pages in a book produced at a cost of GH¢145.00, given the same cost function.

1. Problem Description

We are given three problems.

(a) Simplify the expression $(4x+2)(x-2)-3x^2$.

(b) Find the value of $x$ given that the angles $40^\circ, 60^\circ, 100^\circ, 3x^\circ$ and $5x^\circ$ are formed at the center of a circle.

(c)

(i) Find the cost of producing a book with 220 pages, given the cost function $C = 25 + 0.6x$.

(ii) Find the number of pages in a book produced at a cost of GH¢145.00, given the same cost function.

2. Solution Steps

(a) Simplify $(4x+2)(x-2)-3x^2$

First, expand the product $(4x+2)(x-2)$:

$(4x+2)(x-2) = 4x(x-2) + 2(x-2) = 4x^2 - 8x + 2x - 4 = 4x^2 - 6x - 4$

Now subtract $3x^2$:

$(4x^2 - 6x - 4) - 3x^2 = 4x^2 - 6x - 4 - 3x^2 = x^2 - 6x - 4$

(b) Find the value of $x$.

The sum of angles at the center of a circle is $360^\circ$.

$40 + 60 + 100 + 3x + 5x = 360$

$200 + 8x = 360$

$8x = 360 - 200$

$8x = 160$

$x = \frac{160}{8} = 20$

(c)

(i) Find the cost of producing a book with 220 pages.

$C = 25 + 0.6x$, where $x = 220$.

$C = 25 + 0.6(220) = 25 + 132 = 157$

(ii) Find the number of pages in a book produced at a cost of GH¢145.

0

0. $C = 25 + 0.6x$, where $C = 145$.

$145 = 25 + 0.6x$

$145 - 25 = 0.6x$

$120 = 0.6x$

$x = \frac{120}{0.6} = \frac{1200}{6} = 200$

3. Final Answer

(a) $x^2 - 6x - 4$

(b) $x = 20$

(c)

(i) GH¢157.00

(ii) 200 pages