Question

1. use a graph. what is the solution of the inequality $-x^2 + 16 > 0$?

a.

$-4 < x < 4$
b.

$-4 < x < 4$
c.

$x < -4$ or $x > 4$
d.

$x < -4$ or $x > 4$

1. use graphing technology to approximate the solutions of the equation $x^2 - 3x + 11 = |x - 5| + 13$ to the nearest tenth.

a. $x \approx 1.0$ and $x \approx 3.0$
b. $x \approx 14.2$ and $x \approx 19.8$
c. $x \approx 9.0$ and $x \approx 11.0$
d. $x \approx -1.8$ and $x \approx 3.8$
what are the solutions of the following systems?

1. $\

$$\begin{cases} x - 3y = 9 \\\\ -x + 3y = -9 \\end{cases}$$

$
a. $(5, -6)$
b. no solutions
c. $(-5, 6)$
d. infinitely many solutions

Explanation:

Question 10

Step1: Analyze the inequality

The inequality is \(-x^{2}+16>0\), which can be rewritten as \(x^{2}<16\) (by multiplying both sides by - 1 and reversing the inequality sign). Solving \(x^{2}<16\) gives \(- 4

Step2: Match with the graph

For the inequality \(y=-x^{2}+16>0\), we are looking for the region where the graph of the parabola is above the x - axis. Since the parabola opens downwards, the region above the x - axis is between the two x - intercepts (\(-4\) and \(4\)). Among the given graphs, graph (a) and (c) are parabolas opening upwards (so they represent \(y = x^{2}-16\) or similar functions), and graph (b) and (d) are parabolas opening downwards. Graph (b) has the solution \(-4 < x < 4\) (the region between the x - intercepts where the parabola is above the x - axis) and graph (d) has the solution \(x < - 4\) or \(x>4\) (the region where the parabola is below the x - axis, which is incorrect for our inequality). So the correct graph is (b).

Question 11

Step1: Define the functions

Let \(y_1=x^{2}-3x + 11\) and \(y_2=\vert x - 5\vert+13\). We need to find the x - values where \(y_1=y_2\) (the solutions of the equation \(x^{2}-3x + 11=\vert x - 5\vert+13\)).

Step2: Analyze the functions

• The function \(y_1=x^{2}-3x + 11\) is a parabola opening upwards with vertex at \(x=-\frac{b}{2a}=\frac{3}{2}=1.5\), \(y_1=(1.5)^{2}-3\times1.5 + 11=2.25-4.5 + 11 = 8.75\)
• The function \(y_2=\vert x - 5\vert+13\) is a V - shaped graph with vertex at \((5,13)\).

Step3: Approximate the solutions

We can test the given options:

• Option a: For \(x = 1\), \(y_1=1^{2}-3\times1 + 11=1 - 3+11 = 9\), \(y_2=\vert1 - 5\vert+13=4 + 13=17\), \(y_1

eq y_2\). For \(x = 3\), \(y_1=3^{2}-3\times3+11=9 - 9 + 11=11\), \(y_2=\vert3 - 5\vert+13=2 + 13=15\), \(y_1
eq y_2\)

• Option b: For \(x = 14.2\), \(y_1=(14.2)^{2}-3\times14.2+11=201.64-42.6 + 11=170.04\), \(y_2=\vert14.2 - 5\vert+13=9.2+13 = 22.2\), \(y_1

eq y_2\)

• Option c: For \(x = 9\), \(y_1=9^{2}-3\times9+11=81-27 + 11=65\), \(y_2=\vert9 - 5\vert+13=4 + 13=17\), \(y_1

eq y_2\)

• Option d: For \(x=-1.8\), \(y_1=(-1.8)^{2}-3\times(-1.8)+11=3.24 + 5.4+11=19.64\), \(y_2=\vert-1.8 - 5\vert+13=\vert-6.8\vert+13=6.8 + 13=19.8\) (close, considering approximation). For \(x = 3.8\), \(y_1=(3.8)^{2}-3\times3.8+11=14.44-11.4 + 11=14.04\), \(y_2=\vert3.8 - 5\vert+13=\vert-1.2\vert+13=1.2 + 13=14.2\) (close, considering approximation). So the approximate solutions are \(x\approx - 1.8\) and \(x\approx3.8\)

Question 12

Step1: Analyze the system of equations

The system of equations is \(

$$\begin{cases}x-3y = 9\\-x + 3y=-9\end{cases}$$

\)
If we add the two equations together: \((x-3y)+(-x + 3y)=9+(-9)\)

Step2: Simplify the sum

Simplifying the left - hand side: \(x-3y - x + 3y=0\)
Simplifying the right - hand side: \(9-9 = 0\)
We get \(0 = 0\), which is a true statement. This means that the two equations are equivalent. The first equation \(x-3y = 9\) can be rewritten as \(-x + 3y=-9\) (by multiplying both sides by - 1). So the two equations represent the same line. When two equations in a system represent the same line, there are infinitely many solutions (all the points on the line).

Answer:

b. The graph (b) with solution \(-4 < x < 4\)