Question 1

Solve. Express your answer using interval notation. $x + 6 < 1$&#10;A) $( - \infty , 5 )$&#10;B) $( - \infty , 5 ]$&#10;C) $( - \infty , - 5 ]$&#10;D) $( - 5 , \infty )$&#10;E) $( - \infty , - 5 )$

Accepted Answer

Subtracting 6 from both sides gives us x+6-6 < 1-6 which simplifies to x < -5. Expressing this in interval notation gives us (-∞, -5). Therefore, the correct answer is option E.

Question 2

Solve the equation. $2 y ^ { 2 } - 3 y - 3 = 0$&#10;A) $y = \frac { 3 } { 4 } \pm \sqrt { 33 }$&#10;B) $y = \frac { - 3 \pm \sqrt { 33 } } { 6 }$&#10;C) $y = \frac { \sqrt { 35 } \pm 3 } { 2 }$&#10;D) $y = \frac { 3 \pm \sqrt { 35 } } { 3 }$&#10;E) $y = \frac { 3 \pm \sqrt { 33 } } { 4 }$

Accepted Answer

$y = \frac { 3 \pm \sqrt { 33 } } { 4 }$&#10;

Question 3

Find all real solutions. $t ^ { \frac { 3 } { 2 } } = 125$&#10;A) $t = 25$&#10;B) $t = \pm 625$&#10;C) $t = \pm 5$&#10;D) $t = \pm 25$&#10;E) $t = 5$

Accepted Answer

$t = 25$&#10;

Question 4

Data from the Apollo 11 moon mission in July Error in evaluating expression: helpvar2 showed temperature readings on the lunar surface varying over the interval $- 185 ^ { \circ } \leq C \leq 111 ^ { \circ }$ on the Celsius scale. What is the corresponding interval on the Fahrenheit scale? (Round the numbers you obtain to the nearest integers.) Express your answer using interval notation.

A)

\left[ - 301 ^ { \circ } \mathrm { F } , 301 ^ { \circ } \mathrm { F } \right]

B)

\left[ - 275 ^ { \circ } \mathrm { F } , 257 ^ { \circ } \mathrm { F } \right]

C)

\left[ - 333 ^ { \circ } \mathrm { F } , 333 ^ { \circ } \mathrm { F } \right]

D)

\left[ - 135 ^ { \circ } \mathrm { F } , 94 ^ { \circ } \mathrm { F } \right]

E)

\left[ - 301 ^ { \circ } \mathrm { F } , 232 ^ { \circ } \mathrm { F } \right]

Accepted Answer

The answer of Data from the Apollo 11 moon mission...

Question 5

Solve. Express your answer using interval notation. $\frac { 9 } { 10 } \leq \frac { 7 x - 1 } { - 4 } \leq \frac { 91 } { 100 }$&#10;A) $\left[ \frac { 13 } { 35 } , \frac { 66 } { 175 } \right)$&#10;B) $\left[ \frac { 13 } { 35 } , \frac { 66 } { 175 } \right]$&#10;C) $\left( - \infty , - \frac { 66 } { 175 } \right] \cup \left[ - \frac { 13 } { 35 } , \infty \right)$&#10;D) $\left[ - \frac { 66 } { 175 } , - \frac { 13 } { 35 } \right]$&#10;E) $\left( - \frac { 66 } { 175 } , - \frac { 13 } { 35 } \right)$

Accepted Answer

To solve the inequality, multiply through by -4 (which reverses the inequality signs) and solve for x. The solution is $[- \frac{66}{175}, - \frac{13}{35}]$.

Question 6

Solve the equation. $5 y ^ { 2 } - 5 y - 2 = 0$&#10;A) $y = \frac { 5 \pm \sqrt { 65 } } { 10 }$&#10;B) $y = \frac { 5 \pm \sqrt { 67 } } { 20 }$&#10;C) $y = \frac { - 5 \pm \sqrt { 65 } } { 5 }$&#10;D) $y = 5 \pm \sqrt { 65 }$&#10;E) $y = \frac { \sqrt { 67 } \pm 5 } { 10 }$

Accepted Answer

To solve the quadratic equation $5y^2 - 5y - 2 = 0$, we use the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 5$, $b = -5$, and $c = -2$. Plugging these values into the formula gives $y = \frac{5 \pm \sqrt{(-5)^2 - 4(5)(-2)}}{2(5)} = \frac{5 \pm \sqrt{25 + 40}}{10} = \frac{5 \pm \sqrt{65}}{10}$, which matches option A.

Question 7

Find a quadratic equation with the roots of $r _ { 1 } = 1 - \sqrt { 2 } \text { and } r _ { 2 } = 1 + \sqrt { 2 }$ . Write the answer in the form $a x ^ { 2 } + b x + c = 0$ where a, b, and c are integers and $a = 1$ .

A) x² + 4x + 1 = 0
B) x² - x-1 = 0
C) x² - 2x-1 = 0
D) x² + x-2 = 0
E) x² + 2x -1 = 0

Accepted Answer

The sum of the roots $ (1 - \sqrt{2}) + (1 + \sqrt{2}) = 2 $ and the product of the roots $ (1 - \sqrt{2})(1 + \sqrt{2}) = 1 - 2 = -1 $. The quadratic equation is $ x^2 - (sum)x + (product) = 0 $, which simplifies to $ x^2 - 2x - 1 = 0 $.

Question 8

Solve. Express your answer using interval notation. $3 - x \leq 5$&#10;A) $[ 2 , \infty )$&#10;B) $( 2 , \infty )$&#10;C) $[ - 2 , \infty )$&#10;D) $( - \infty , - 2 )$&#10;E) $( - 2 , \infty )$

Accepted Answer

Solving the inequality $3 - x \leq 5$ gives $x \geq -2$, which is represented by the interval $[ - 2 , \infty )$.

Question 9

Find all real solutions. x⁴ + x² - 4 = 0 A) $x = \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }$ B) $x = \sqrt { \frac { 1 - \sqrt { 13 } } { 2 } }$ C) $x = \pm \sqrt { \frac { 1 - \sqrt { 13 } } { 2 } }$ D) $x = \pm \sqrt { \frac { - 1 \pm \sqrt { 17 } } { 2 } }$ E) $x = \pm \sqrt { \frac { - 1 + \sqrt { 17 } } { 2 } }$

Accepted Answer

The equation can be solved by substituting $ y = x^2 $, leading to $ y^2 + y - 4 = 0 $. Solving this quadratic equation gives $ y = \frac{-1 \pm \sqrt{17}}{2} $. Since $ y = x^2 $, the real solutions for $ x $ are $ x = \pm \sqrt{\frac{-1 + \sqrt{17}}{2}} $.

Question 10

Use the graph to solve the inequality. $x ^ { 3 } - 3 x ^ { 2 } - 4 x + 12 \geq 0$  &#10;A) $( - 2,2 ) \cup ( 3 , \infty )$&#10;B) $[ - \infty , 3 ) \cup ( 2,3 ]$&#10;C) $( - 2,2 ]$&#10;D) $[ - 2,2 ] \cup [ 3 , \infty )$&#10;E) no solution

Accepted Answer

To solve the inequality, we first need to find the $x$-values where the function $f(x) = x^3-3x^2-4x+12$ is equal to zero (i.e. the roots of the function). These $x$-values divide the number line into intervals, and we must test a point in each interval to determine whether $f(x)$ is positive or negative in that interval. Using synthetic division or polynomial long division, we can find that the function can be factored as: $$f(x) = (x-2)(x+1)^2$$ Therefore, the roots are $x=2$ and $x=-1$, and the number line is divided into three intervals: $(-\infty,-1)$, $(-1,2)$, and $(2,\infty)$. We can test a point in each interval to determine whether $f(x)$ is positive or negative in that interval. For example, if we test $x=0$ in $f(x)$, we get: $$f(0) = 12 > 0$$ Therefore, $f(x)$ is positive for any $x$ in the interval $(-1,2)$. Using similar reasoning, we can construct the following sign chart: \begin{array}{c|ccc|c} & x<-1 & -12 & x \ \hline x^3-3x^2-4x+12 & - & + & - & \ \end{array} From the sign chart, we see that $f(x)$ is non-negative (i.e. greater than or equal to zero) for $x\in[-1,2]$ and $x\geq 3$. Therefore, the solution to the inequality $f(x)\geq 0$ is: $$x\in[-1,2]\cup[3,\infty)$$ The correct answer is D, $[ - 2,2 ] \cup [ 3 , \infty )$.

Question 11

Solve the equation. $x ^ { 2 } + 6 x - 4 = 0$&#10;A) $x = - 3 \pm \sqrt { 13 }$&#10;B) $x = 1 \pm \sqrt { 17 }$&#10;C) $x = - 2 \pm \sqrt { 13 }$&#10;D) $x = - 3 \pm \sqrt { 17 }$&#10;E) The equation has no real roots.

Accepted Answer

The answer of Solve the equation. \(x ^ { 2...

Question 12

Solve the equation. $81 s ^ { 2 } + 18 s + 1 = 0$&#10;A) $s = - \frac { 1 } { 9 }$&#10;B) $s = - \frac { 1 } { 8 }$&#10;C) $s = - \frac { 2 } { 7 }$&#10;D) $s = - \frac { 1 } { 10 }$&#10;E) The equation has no real roots.

Accepted Answer

The answer of Solve the equation. \(81 s ^ {...

Question 13

Find all real solutions. 3y² = 4 - y⁴ A) y = - 3, 4, 1 B) y = - 4, - 1 C) y = 1, - 1 D) y = 3, 1 E) y = 3, - 4

Accepted Answer

The answer of Find all real solutions. 3y²...

Question 14

Solve. Express your answer using interval notation. $5 x + 7 < 4 ( x - 1 ) - x$&#10;A) $\left( - \infty , - \frac { 2 } { 11 } \right]$&#10;B) $\left( - \infty , - \frac { 2 } { 11 } \right)$&#10;C) $\left( - \frac { 11 } { 2 } , \infty \right)$&#10;D) $\left( - \frac { 2 } { 11 } , 0 \right)$&#10;E) $\left( - \infty , - \frac { 11 } { 2 } \right)$

Accepted Answer