- Simplify the expression \(\frac{4^{x}+2^{2x}}{2^{x}}\)
- \(6\)
- \(2+2^{x}\)
- \(2 \times 2^{x}\)
- \(2^{x}+1\)
First, rewrite \(4^{x}\) using base \(2\): \(4^{x} = (2^{2})^{x} = 2^{2x}\). So the numerator becomes:
\(4^{x} + 2^{2x} = 2^{2x} + 2^{2x} = 2 \cdot 2^{2x} = 2^{2x+1}\)
Now divide by \(2^{x}\):
\(\dfrac{2^{2x+1}}{2^{x}} = 2^{2x+1 – x} = 2^{x+1} = 2 \cdot 2^{x}\)
So the simplified expression is \(2 \cdot 2^{x}\).
- Simplify the expression \(\frac{2x^{2}-5x-12}{2x^{2}-4x-16}\)
- \(\frac{x-6}{2(x-2)}\)
- \(\frac{x-6}{2(x+2)}\)
- \(\frac{2x+3}{2(x-2)}\)
- \(\frac{2x+3}{2(x+2)}\)
Begin by factoring the numerator and denominator. The numerator factors as:
\(2x^{2} – 5x – 12 = (2x + 3)(x – 4)\)
The denominator factors as:
\(2x^{2} – 4x – 16 = 2(x^{2} – 2x – 8) = 2(x – 4)(x + 2)\)
So the fraction becomes:
\(\dfrac{2x^{2} – 5x – 12}{2x^{2} – 4x – 16}\) \(= \dfrac{(2x + 3)(x – 4)}{2(x – 4)(x + 2)}\)
Cancel \((x – 4)\) (for \(x \neq 4\)) to get:
\(\dfrac{2x + 3}{2(x + 2)}\)
Thus the simplified expression is \(\frac{2x + 3}{2(x + 2)}\)
- 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:
\(6\sqrt{x} + 10x = 2\bigl(3\sqrt{x} + 5x\bigr)\)