Solve the equation 3x2 + x − 2 = 0

We need two numbers that multiply to $(3 \times -2) = -6$ and add up to 1. These numbers are 3 and $-2$.
Rewrite the middle term:

$$3x^2 + 3x - 2x - 2 = 0$$

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

$$3x-2 = 0 \implies 3x = 2 \implies x = \frac{2}{3}$$ $$x+1 = 0 \implies x = -1$$

The roots are $x = -1$ and $x = \frac{2}{3}$.

The correct option is D.

$$\boxed{\text{D. } x = -1, x = \frac{2}{3}}$$

: Find the circumference of latitude $60^\circ$N to the nearest Km. (take the radius of the earth to be 6400km and $\pi = 3.142$)

The radius of a circle of latitude ($r$) is given by $r = R \cos \theta$, where $R$ is the radius of the Earth and $\theta$ is the latitude.

$$r = 6400 \text{ km} \times \cos(60^\circ)$$ $$r = 6400 \text{ km} \times 0.5$$ $$r = 3200 \text{ km}$$

The circumference ($C$) is given by $C = 2\pi r$.

$$C = 2 \times 3.142 \times 3200 \text{ km}$$ $$C = 6.284 \times 3200 \text{ km}$$ $$C = 20108.8 \text{ km}$$

$$C \approx 20109 \text{ km}$$

The correct option is A.

$$\boxed{\text{A. } 20,109\text{km}}$$

: Find the quadratic equation whose roots are $x = \frac{1}{2}, x = \frac{1}{3}$

Let $\alpha = \frac{1}{2}$ and $\beta = \frac{1}{3}$.

$$\alpha + \beta = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$

$$6 \left(x^2 - \frac{5}{6}x + \frac{1}{6}\right) = 6 \times 0$$ $$6x^2 - 5x + 1 = 0$$

The correct option is C.

$$\boxed{\text{C. } 6x^2 - 5x + 1 = 0}$$