Here are the solutions to the questions. 15. Calculus (Differentiation) a) Differentiate $y = (x^2 + 3x)(x - 2)$ Step 1: Expand the expression for $y$. $$y = x^2(x - 2) + 3x(x - 2)$$ $$y = x^3 - 2x^2 + 3x^2 - 6x$$ $$y = x^3 + x^2 - 6x$$ Step 2: Differentiate $y$ with respect to $x$ using the power rule $\frac{d}{dx}(ax^n) = nax^{n-1}$. $$\frac{dy}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2) - \frac{d}{dx}(6x)$$ $$\frac{dy}{dx} = 3x^{3-1} + 2x^{2-1} - 6x^{1-1}$$ $$\frac{dy}{dx} = 3x^2 + 2x - 6$$ The derivative is $3x^2 + 2x - 6$. $$\boxed{\frac{dy}{dx} = 3x^2 + 2x - 6}$$ b) Find the gradient at $x = 1$ Step 1: Substitute $x=1$ into the derivative found in part (a). $$\frac{dy}{dx}\Big|_{x=1} = 3(1)^2 + 2(1) - 6$$ $$\frac{dy}{dx}\Big|_{x=1} = 3(1) + 2 - 6$$ $$\frac{dy}{dx}\Big|_{x=1} = 3 + 2 - 6$$ $$\frac{dy}{dx}\Big|_{x=1} = 5 - 6$$ $$\frac{dy}{dx}\Big|_{x=1} = -1$$ The gradient at $x=1$ is $-1$. $$\boxed{\text{Gradient} = -1}$$ 16. Calculus (Integration) a) Integrate: $\int (3x^2 + 4x)\,dx$ Step 1: Apply the power rule for integration, $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$, to each term. $$\int 3x^2\,dx + \int 4x\,dx$$ $$3\left(\frac{x^{2+1}}{2+1}\right) + 4\left(\frac{x^{1+1}}{1+1}\right) + C$$ $$3\left(\frac{x^3}{3}\right) + 4\left(\frac{x^2}{2}\right) + C$$ Step 2: Simplify the expression. $$x^3 + 2x^2 + C$$ The integral is $x^3 + 2x^2 + C$. $$\boxed{x^3 + 2x^2 + C}$$ b) Find area under curve from $x = 0$ to $x = 2$ for $y = 3x^2 + 4x$ Step 1: Set up the definite integral for the area. $$\text{Area} = \int_0^2 (3x^2 + 4x)\,dx$$ Step 2: Integrate the function. From part (a), the indefinite integral is $x^3 + 2x^2$. Step 3: Evaluate the definite integral using the limits of integration. $$\text{Area} = \left[x^3 + 2x^2\right]_0^2$$ $$\text{Area} = (2^3 + 2(2^2)) - (0^3 + 2(0^2))$$ $$\text{Area} = (8 + 2(4)) - (0 + 0)$$ $$\text{Area} = (8 + 8) - 0$$ $$\text{Area} = 16$$ The area under the curve is $16$ square units. $$\boxed{\text{Area} = 16 \text{ square units}}$$ 17. Exponents a) Solve: $2^{x+1} = 16$ Step 1: Express 16 as a power of 2. $$16 = 2^4$$ Step 2: Substitute this back into the equation. $$2^{x+1} = 2^4$$ Step 3: Since the bases are the same, equate the exponents. $$x+1 = 4$$ Step 4: Solve for $x$. $$x = 4 - 1$$ $$x = 3$$ The solution is $x=3$. $$\boxed{x = 3}$$ b) Express your answer in simplest form The answer from part (a) is $x=3$, which is already in its simplest form. $$\boxed{x = 3}$$ 18. Probability a) A bag contains 4 red and 6 blue balls. Find probability of picking red. Step 1: Find the total number of balls. $$\text{Total balls} = 4 (\text{red}) + 6 (\text{blue}) = 10$$ Step 2: Calculate the probability of picking a red ball. $$P(\text{red}) = \frac{\text{Number of red balls}}{\text{Total number of balls}}$$ $$P(\text{red}) = \frac{4}{10}$$ $$P(\text{red}) = \frac{2}{5}$$ The probability of picking a red ball is $\frac{2}{5}$. $$\boxed{P(\text{red}) = \frac{2}{5}}$$ b) If two balls are picked, find probability both are red. This implies picking without replacement. Step 1: Probability of the first ball being red. $$P(\text{1st red}) = \frac{4}{10}$$ Step 2: After picking one red ball, there are 3 red balls left and a total of 9 balls. Probability of the second ball being red. $$P(\text{2nd red | 1st red}) = \frac{3}{9}$$ Step 3: Multiply the probabilities to find the probability of both being red. $$P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red | 1st red})$$ $$P(\text{both red}) = \frac{4}{10} \times \frac{3}{9}$$ $$P(\text{both red}) = \frac{2}{5} \times \frac{1}{3}$$ $$P(\text{both red}) = \frac{2}{15}$$ The probability of picking two red balls is $\frac{2}{15}$. $$\boxed{P(\text{both red}) = \frac{2}{15}}$$ 19. Circles a) Find the equation of a circle with center $(1, -2)$ and radius 4. Step 1: Use the standard equation of a circle: $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Given center $(h, k) = (1, -2)$ and radius $r = 4$. Step 2: Substitute the values into the equation. $$(x - 1)^2 + (y - (-2))^2 = 4^2$$ $$(x - 1)^2 + (y + 2)^2 = 16$$ The equation of the circle is $(x - 1)^2 + (y + 2)^2 = 16$. $$\boxed{(x - 1)^2 + (y + 2)^2 = 16}$$ b) Find the area of the circle. Step 1: Use the formula for the area of a circle: $A = \pi r^2$. Given radius $r = 4$. Step 2: Substitute the value of $r$ into the formula. $$A = \pi (4)^2$$ $$A = 16\pi$$ The area of the circle is $16\pi$ square units. $$\boxed{\text{Area} = 16\pi \text{ square units}}$$ 20. Sequences (Arithmetic Progression) a) A sequence has first term 5 and common difference 3. Find the 15th term. Step 1: Use the formula for the $n$-th term of an arithmetic sequence: $a_n = a_1 + (n-1)d$. Given $a_1 = 5$, $d = 3$, and $n = 15$. Step 2: Substitute the values into the formula. $$a_{15} = 5 + (15 - 1)(3)$$ $$a_{15} = 5 + (14)(3)$$ $$a_{15} = 5 + 42$$ $$a_{15} = 47$$ The 15th term is $47$. $$\boxed{a_{15} = 47}$$ b) Find the sum of first 15 terms. Step 1: Use the formula for the sum of the first $n$ terms of an arithmetic sequence: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$. Given $a_1 = 5$, $d = 3$, and $n = 15$. Step 2: Substitute the values into the formula. $$S_{15} = \frac{15}{2}(2(5) + (15 - 1)(3))$$ $$S_{15} = \frac{15}{2}(10 + (14)(3))$$ $$S_{15} = \frac{15}{2}(10 + 42)$$ $$S_{15} = \frac{15}{2}(52)$$ $$S_{15} = 15 \times 26$$ $$S_{15} = 390$$ The sum of the first 15 terms is $390$. $$\boxed{S_{15} = 390}$$ That's 2 down. 3 left today — send the next one.