Targeted Review Solutions

Determine the perimeter of the right triangle.

$\begin{matrix} a = 5, c = 13 \\ a^{2} + b^{2} = c^{2} & P = a + b + c \\ b = \sqrt{c^{2} - a^{2}} & P = 5 + 12 + 13 \\ b = \sqrt{13^{2} - 5^{2}} \\ b = 12 \end{matrix}$

30 units

Note

You are encouraged to show your work, even if you notice this Pythagorean triple.

Determine the area of the triangle. Round to the nearest hundredth.

$\begin{matrix} b = 8, c = 13 \\ a^{2} + b^{2} = c^{2} & b = 8, h = 10.25 \\ a = \sqrt{c^{2} - b^{2}} & A = \frac{1}{2} b h \\ a = \sqrt{13^{2} - 8^{2}} & A = \frac{1}{2} (8) (10.25) \\ a = 10.25 \end{matrix}$

41 square units

Determine the value of Q and R.
${(7 x + 1)}^{2} = 49 x^{2} + Q x + R$

$(7 x + 1) (7 x + 1) = 49 x^{2} + Q x + R 49 x^{2} + 14 x + 1 = 49 x^{2} + Q x + R \begin{matrix} 14 x = Q x & 1 = R \end{matrix}$

$Q = 14, R = 1$

Determine the value of n.
${(2 x - n)}^{2} = 4 x^{2} - 36 x + 81$

$\begin{array}{rcl} (2 x - n) (2 x - n) & = & 4 x^{2} - 36 x + 81 \\ 4 x^{2} - 4 x n - n^{2} & = & 4 x^{2} - 36 x + 81 \\ n^{2} & = & 81 \\ n & = & \pm 9 \end{array}$

$n = 9$

Note

The value must be $n = 9,$ because –9 would make the middle term positive.

Note

Problem 5–6

The rules for simplifying square roots are only true when the radicand is positive. Therefore, $\sqrt{- 1} = i$ must be completed in the first step.

Simplify.
  $\sqrt{- 6} (i + \sqrt{3})$

$i \sqrt{6} (i + \sqrt{3}) i^{2} \sqrt{6} + i \sqrt{18} (- 1) \sqrt{6} + i \sqrt{18}$

$- \sqrt{6} + 3 i \sqrt{2}$

Simplify.
  $\sqrt{- 10} (\sqrt{- 10} + 5 \sqrt{2}) - 3 i \sqrt{5}$

$i \sqrt{10} (i \sqrt{10} + 5 \sqrt{2}) - 3 i \sqrt{5} i^{2} \sqrt{10^{2}} + 5 i \sqrt{20} - 3 i \sqrt{5} (- 1) (10) + 5 i \sqrt{20} - 3 i \sqrt{5} - 10 + 5 i (2) \sqrt{5} - 3 i \sqrt{5} - 10 + 10 i \sqrt{5} - 3 i \sqrt{5}$

$- 10 + 7 i \sqrt{5}$

A parabola is reflected over the x-axis and translated 4 spaces up and left 2 spaces as compared to the parent graph. Write the equation of the new parabola in vertex form.

$a = - 1, h = - 2, k = 4 y = a {(x - h)}^{2} + k y = (- 1) {(x - (- 2))}^{2} + 4$

$y = - {(x + 2)}^{2} + 4$

Graph the equation you wrote in the previous problem.

Select the equation that best matches the graph.

$y = \{\begin{matrix} - 2 x + 3 & x > 1 \\ x - 2 & - 2 \leq x \leq 1 \\ - 3 & x < - 2 \end{matrix}$
$y = \{\begin{matrix} - 2 x + 3 & x > - 2 \\ x - 2 & - 2 \leq x \leq 1 \\ - 3 & x < 1 \end{matrix}$
$y = \{\begin{matrix} - 2 x + 1 & x > 1 \\ x - 2 & - 2 \leq x \leq 1 \\ - 3 & x < - 2 \end{matrix}$
$y = \{\begin{matrix} 2 x + 3 & x \geq 1 \\ x - 2 & - 2 \leq x < 1 \\ - 3 & x < - 2 \end{matrix}$

The left-most piece of the graph is a horizontal line at $y = - 3$ for all values less than –2. The middle line segment has a slope of 1 and y-intercept at –2 when x is between and including –2 to 1. The right-most equation has a slope of –2 and a y-intercept of 3 for all i-values greater than 1.

Note

The inequalities for the first and last parts of the piecewise function are reversed.
The y-intercept for the top expression should be 3.
The slope for the top expression should be –2.

Determine the missing value, M.
$(8 i - 5) + {(2 i + 7)}^{2} = 4 (M + 10)$

9
2i
9i
36i

$\begin{array}{rcl} 8 i - 5 + 4 i^{2} + 28 i + 49 & = & 4 M + 40 \\ 36 i + 44 + 4 (- 1) & = & 4 M + 40 \\ 36 i + 40 & = & 4 M + 40 \\ 36 i & = & 4 M \\ M & = & 9 i \end{array}$

Note

This value is missing the imaginary unit i.
This option is the answer when the binomial is not written as a trinomial.
This option is the value of 4M.

Determine the graph that best represents the piecewise function.
$y = \{\begin{matrix} - {(x + 2)}^{2} - 1 & x < - 1 \\ \frac{1}{3} x & - 1 < x \leq 4 \\ \sqrt{x - 4} & x > 4 \end{matrix}$

Note

The direction and vertex of the parabola do not match the given equation.
The graph of the square root should move right 4, not up 4. This is not a function.
The slope of this linear function is 3.

Determine the expression that represents the total area, including the x inch wide frame, when the picture is a 5 inch by 7 inch photo.

$x^{2} + 12 x + 35 square units$
$4 x^{2} + 28 x + 35 square units$
$x^{2} - 12 x + 35 square units$
35 square units

$\begin{matrix} l = x + x + 5, & w = x + x + 7 \\ l = 2 x + 5 & w = 2 x + 7 \end{matrix} A = l w A = (2 x + 5) (2 x + 7) A = 4 x^{2} + 28 x + 35$

Note

This option is the area using $(x + 5)$ and $(x + 7)$ for the sides.
This option is the answer if the entire side is treated as x, and $(x - 5) (x - 7)$ is used.
This option is the area of the picture only.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	–	–	L04	L04	L12	L12	L18	L18	L21	L16	L21	A1

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics