Solve for theta given that theta is acute and (3theta - 50)^ - (2theta + 10)^ = 0. Miss Ludicy used to spend shs.720 per month on eggs, when the price of eggs went down by sh. 3.00 per egg she found t

Here are the solutions to the problems.

9. Solve for $\theta$ given that $\theta$ is acute and $\sin(3\theta - 50)^\circ - \cos(2\theta + 10)^\circ = 0$.

Step 1: Rewrite the equation.

$$\sin(3\theta - 50)^\circ = \cos(2\theta + 10)^\circ$$

Step 2: Use the trigonometric identity $\sin x = \cos(90^\circ - x)$.
Since $\sin(3\theta - 50)^\circ = \cos(90^\circ - (3\theta - 50)^\circ)$, we can equate the arguments of the cosine functions.

$$90^\circ - (3\theta - 50)^\circ = 2\theta + 10^\circ$$

Step 3: Simplify and solve for $\theta$.

$$90 - 3\theta + 50 = 2\theta + 10$$ $$140 - 3\theta = 2\theta + 10$$

Add $3\theta$ to both sides:

$$140 = 5\theta + 10$$

Subtract 10 from both sides:

$$130 = 5\theta$$

Divide by 5:

$$\theta = \frac{130}{5}$$ $$\theta = 26^\circ$$

Since $\theta = 26^\circ$ is an acute angle, this solution is valid.

The final answer is $\boxed{\text{26}^\circ}$.

10. Miss Ludicy used to spend shs.720 per month on eggs, when the price of eggs went down by sh. 3.00 per egg she found that with the same amount of money she could buy 20 more eggs per month. How much per egg was she buying at the lower price.

Step 1: Define variables.
Let $P_1$ be the original price per egg (in shs).
Let $N_1$ be the original number of eggs bought.
Let $P_2$ be the new price per egg (in shs).
Let $N_2$ be the new number of eggs bought.

Step 2: Formulate equations based on the given information.
Total spending is shs. 720:

$$P_1 N_1 = 720 \quad (1)$$ $$P_2 N_2 = 720 \quad (2)$$

Price went down by sh. 3.00 per egg:

$$P_2 = P_1 - 3 \quad (3)$$

She could buy 20 more eggs:

$$N_2 = N_1 + 20 \quad (4)$$

Step 3: Substitute equations (3) and (4) into equation (2).

$$(P_1 - 3)(N_1 + 20) = 720$$

Expand the equation:

$$P_1 N_1 + 20P_1 - 3N_1 - 60 = 720$$

Step 4: Substitute $P_1 N_1 = 720$ (from equation 1) into the expanded equation.

$$720 + 20P_1 - 3N_1 - 60 = 720$$

Subtract 720 from both sides:

$$20P_1 - 3N_1 - 60 = 0$$ $$20P_1 - 3N_1 = 60 \quad (5)$$

Step 5: Express $N_1$ in terms of $P_1$ from equation (1) and substitute into equation (5).
From (1), $N_1 = \frac{720}{P_1}$.
Substitute into (5):

$$20P_1 - 3\left(\frac{720}{P_1}\right) = 60$$ $$20P_1 - \frac{2160}{P_1} = 60$$

Step 6: Solve the quadratic equation for $P_1$.
Multiply the entire equation by $P_1$ (assuming $P_1 \neq 0$):

$$20P_1^2 - 2160 = 60P_1$$

Rearrange into standard quadratic form $aP_1^2 + bP_1 + c = 0$:

$$20P_1^2 - 60P_1 - 2160 = 0$$

Divide by 20 to simplify:

$$P_1^2 - 3P_1 - 108 = 0$$

Factor the quadratic equation:
We need two numbers that multiply to -108 and add to -3. These are -12 and 9.

$$(P_1 - 12)(P_1 + 9) = 0$$

This gives two possible values for $P_1$:
$P_1 = 12$ or $P_1 = -9$.
Since price cannot be negative, we take $P_1 = 12$ shs/egg.

Step 7: Calculate the lower price ($P_2$).
The question asks for the price per egg at the lower price, which is $P_2$.
Using equation (3):

$$P_2 = P_1 - 3$$ $$P_2 = 12 - 3$$ $$P_2 = 9 \text{ shs/egg}$$

The final answer is $\boxed{\text{9 shs/egg}}$.

11. A line L is perpendicular to the line $\frac{2}{3}x + \frac{5}{7}y = 1$. Given that L passes through $(4, 11)$, find