Calculate the area of the region bounded by y = 16 − x2, the x-axis, and the y-axis in the first quadrant. Also, find the value of x for which the matrix [[1, 4], [x, x-3]] is singular.

The shaded region is bounded by the curve $y = 16 - x^2$, the x-axis ($y=0$), and the y-axis ($x=0$).
To find the x-intercept, set $y=0$:

$$0 = 16 - x^2$$ $$x^2 = 16$$ $$x = \pm 4$$

Since the region is in the first quadrant, the relevant x-intercept is $x=4$.
The area is given by the definite integral of the function from $x=0$ to $x=4$.

$$\text{Area} = \int_{0}^{4} (16 - x^2)\,dx$$

$$\text{Area} = \left[ 16x - \frac{x^3}{3} \right]_{0}^{4}$$ $$\text{Area} = \left( 16(4) - \frac{4^3}{3} \right) - \left( 16(0) - \frac{0^3}{3} \right)$$ $$\text{Area} = \left( 64 - \frac{64}{3} \right) - (0 - 0)$$ $$\text{Area} = \frac{192}{3} - \frac{64}{3}$$ $$\text{Area} = \frac{128}{3}$$

The exact area of the shaded region is $\boxed{\frac{128}{3} \text{ square units}}$.

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A matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is singular if its determinant $ad - bc = 0$.
For the given matrix $\begin{pmatrix} 1 & 4 \\ x & x-3 \end{pmatrix}$, the determinant is:

$$\text{det} = (1)(x-3) - (4)(x)$$ $$\text{det} = x - 3 - 4x$$ $$\text{det} = -3x - 3$$

For the matrix to be singular:

$$-3x - 3 = 0$$ $$-3x = 3$$ $$x = \frac{3}{-3}$$ $$x = -1$$

The value of $x$ for which the matrix is singular is $\boxed{-1}$.

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$$2 \sin^2 \theta - 5 \cos \theta + 1 = 0$$ $$2(1 - \cos^2 \theta) - 5 \cos \theta + 1 = 0$$ $$2 - 2 \cos^2 \theta - 5 \cos \theta + 1 = 0$$ $$-2 \cos^2 \theta - 5 \cos \theta + 3 = 0$$

Multiply by -1 to make the leading coefficient positive:

$$2 \cos^2 \theta + 5 \cos \theta - 3 = 0$$

Let $y = \cos \theta$. The equation becomes $2y^2 + 5y - 3 = 0$.
Factor the quadratic equation:

$$(2y - 1)(y + 3) = 0$$

So, $2y - 1 = 0$ or $y + 3 = 0$.

$$2y = 1 \implies y = \frac{1}{2}$$ $$y = -3$$

Substitute back $y = \cos \theta$:

$$\cos \theta = \frac{1}{2} \quad \text{or} \quad \cos \theta = -3$$

Since the range of $\cos \theta$ is $[-1, 1]$, $\cos \theta = -3$ has no solution.
Therefore, we only consider $\cos \theta = \frac{1}{2}$.

The basic angle for $\cos \theta = \frac{1}{2}$ is $60^\circ$.
Since cosine is positive, $\theta$ is in the first and fourth quadrants.

• First quadrant: $\theta = 60^\circ$.
• Fourth quadrant: $\theta = 360^\circ - 60^\circ = 300^\circ$. The values of $\theta$ are $\boxed{60^\circ, 300^\circ}$.

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The total amount to be shared is Sh. 12 000.
The ratio for Kamau : Njoroge : Mwangi is 5 : 6 : $x$.
Kamau's share is Sh. 4 000.
Let $k$ be the value of one ratio unit.
Kamau's share = $5k = 4000$.

$$k = \frac{4000}{5}$$ $$k = 800$$

So, one ratio unit is Sh. 800.

The total amount is Sh. 12 000.
Total ratio units = Sum of ratio parts = $5 + 6 + x = 11 + x$.
Total amount = (Total ratio units) $\times$ (Value of one ratio unit)

$$12000 = (11 + x) \times 800$$

$$\frac{12000}{800} = 11 + x$$ $$15 = 11 + x$$ $$x = 15 - 11$$ $$x = 4$$

The value of $x$ is $\boxed{4}$.

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The expression is $2x^2y^2 - 5xy - 12$.
Substitute $A = xy$:

$$2A^2 - 5A - 12$$

We look for two numbers that multiply to $2 \times (-12) = -24$ and add up to $-5$. These numbers are 3 and $-8$.
Rewrite the middle term:

$$2A^2 + 3A - 8A - 12$$

Factor by grouping:

$$A(2A + 3) - 4(2A + 3)$$ $$(A - 4)(2A + 3)$$

$$(xy - 4)(2xy + 3)$$

The factorised expression is $\boxed{(xy - 4)(2xy + 3)}$.