- Simplify the expression \(\frac{4^{x}+2^{2x}}{2^{x}}\)
- \(6\)
- \(2+2^{x}\)
- \(2 \times 2^{x}\)
- \(2^{x}+1\)
Sarah started with 80 cupcakes, and she sold 70, so the expression \(80-70\) can be used to represent the number of cupcakes she had before her mom brought more to sell. Because her mom brought 20 more cupcakes to the bake sale, 20 is added to the previous expression.
The equation which represents the number of cupcakes Sarah has after selling 70 cupcakes and receiving 20 more is \(80-70+20=c\).
- Simplify the expression \(\frac{2x^{2}-5x-12}{2x^{2}-4x-16}\)
- \(\frac{x-6}{2(x-2)}\)
- \(\frac{x-6}{2(x+2)}\)
- \(\frac{2x+3}{2(x-2)}\)
- \(\frac{2x+3}{2(x+2)}\)
Lucas began with 55 marbles, so the expression should begin with 55 marbles. Because he lost \(m\) marbles, \(m\) should be subtracted from the 55. So the expression \(55-m\) can be used to show how many marbles Luca has now.
The problem states that he now has 32 marbles, so the \(55-m\) is equal to 32. Therefore, the equation which represents this scenario is \(55-m=32\).
- Suppose that the function \(f(x)\) is a quadratic function with roots at \(x = 2 – 3i\) and \(x = 2 + 3i\). Find \(f(x)\)
- \(f(x) = x^{2} – 4x – 5\)
- \(f(x) = x^{2} – 4x + 13\)
- \(f(x) = x^{2} – 6ix – 5\)
- \(f(x) = x^{2} – 6ix + 13\)
To find \(f(x)\), note that for roots \(r_1\) and \(r_2\), a quadratic is proportional to \((x – r_1)(x – r_2)\). Here, \(r_1 = 2 – 3i\) and \(r_2 = 2 + 3i\), so:
\(f(x) = (x – (2 – 3i))(x – (2 + 3i))\) \(= (x – 2 + 3i)(x – 2 – 3i)\) \(= (x – 2)^2 – (3i)^2\) \(= x^2 – 4x + 4 + 9\) \(= x^2 – 4x + 13.\)
- Solve the inequality below for \(x\).
- \(x \lt -4\)
- \(x = -4\)
- \(x = 0\)
- \(x \gt \frac{3}{2}\)
First, factor:
\(4x^{3} + 10x^{2} – 24x = 2x\,(2x^{2} + 5x – 12)\) \(= 2x\,(2x – 3)\,(x + 4).\)
The critical values are \(x = -4,\,0,\,\tfrac{3}{2}\). Testing intervals in \(4x^{3} + 10x^{2} – 24x \lt 0\) shows it is negative only when \(x \lt -4\).
- A baseball is thrown up in the air from an initial height of 6 feet. Its height above the ground (in feet) \(t\) seconds after being thrown is given by the function \(h(t) = -16t^{2} + 46t + 6\). How long will it take (in seconds) for the baseball to hit the ground?
- 2 seconds
- \(\frac{5}{2}\) seconds
- 3 seconds
- 4 seconds
First, we can set \(h(t)=0\):
\(-16t^{2} + 46t + 6 = 0\)
By the quadratic formula:
\(t = \frac{-46 \pm \sqrt{46^{2} – 4(-16)(6)}}{2(-16)}\) \(= \frac{-46 \pm \sqrt{2116 + 384}}{-32}\) \(= \frac{-46 \pm 50}{-32}.\)
This gives \(t = 3\) seconds.
- Solve the equation below for \(x\).
- \(x = -8\)
- \(x = 0\)
- \(x = 4\)
- \(x = 8\)
Rewrite the log equation:
\(\log_{2}(8x – x^{2}) = 4 \;\Longrightarrow\; 8x – x^{2} = 2^{4} = 16\)
So \(x^{2} – 8x + 16 = 0\), giving \((x – 4)^{2} = 0\) and \(x = 4\).
- Calculate the average rate of change of \(f\) between \(x = 1\) and \(x = 4\).
- 6
- \(\frac{20}{3}\)
- 24
- 72
The average rate of change is as follows:
\(\frac{f(4)-f(1)}{4-1}\) \(= \frac{(4^{3}+3\cdot4+1) – (1^{3}+3\cdot1+1)}{3}\) \(= \frac{77 – 5}{3} = 24\)
- Simplify the expression \(\frac{x^{3} – 3x^{2} + 2x – 6}{x^{2} – 9}\).
- 1
- \(\frac{x – 3}{x + 3}\)
- \(\frac{x^{2} + 2}{x – 3}\)
- \(\frac{x^{2} + 2}{x + 3}\)
Factor numerator and denominator:
\(x^{3} – 3x^{2} + 2x – 6\) \(= (x^{2} + 2)(x – 3)\)
\(\quad x^{2} – 9 = (x – 3)(x + 3)\)
Cancel \((x – 3)\) to get \(\frac{x^{2} + 2}{x + 3}\).
- Suppose that angle \(\pi\) is in Quadrant I and \(\cos \pi = \tfrac{12}{13}\). Find \(\tan \pi\).
- \(\tan \pi = \tfrac{1}{13}\)
- \(\tan \pi = 13\)
- \(\tan \pi = \tfrac{5}{12}\)
- \(\tan \pi = \tfrac{12}{5}\)
Given \(\cos \pi = \tfrac{12}{13}\), in Quadrant I:
\(\sin \pi = \sqrt{1 – \cos^{2}\pi}\) \(= \sqrt{1 – \bigl(\tfrac{12}{13}\bigr)^{2}}\) \(= \tfrac{5}{13}\)
So \(\tan \pi = \frac{\sin \pi}{\cos \pi}\) \(= \tfrac{5}{12}.\)
- Which expression is equivalent to \(6\sqrt{x} + 10x\pi\) ?
- \(2\bigl(3x^{-1} + 5x\bigr)\)
- \(2\bigl(3x^{\tfrac12} + 5x\bigr)\)
- \(2x\bigl(3x^{-1} + 5\bigr)\)
- \(2x\bigl(3x^{\tfrac12} + 5\bigr)\)
Factor out 2:
\(6\sqrt{x} + 10x = 2\bigl(3\sqrt{x} + 5x\bigr)\)