Question

1. \\(\

$$\begin{cases}-x + 2y = 10\\\\-3x + 6y = 11\\end{cases}$$

\\)\
a. infinitely many solutions\
b. \\((-5, 2)\\)\
c. \\((5, -2)\\)\
d. no solutions\
solve the system by graphing.\

1. \\(\

$$\begin{cases}-0.7x - 0.2y = 0.6\\\\0.1x - 0.5y = 1.5\\end{cases}$$

\\)\
a. \
\\(\

$$\begin{tikzpicture}scale=0.5\ \\draw-> (-5,0) -- (5,0) noderight {$x$};\ \\draw-> (0,-5) -- (0,5) nodeabove {$y$};\ \\foreach \\x in {-4,-2,2,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x};\ \\foreach \\y in {-4,-2,2,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {\\y};\ \\drawthick, blue (-2,0) -- (0,-3);\ \\drawthick, red (0,-3) -- (4,-2);\ \\end{tikzpicture}$$

\\)\
\\((-3, 0)\\)\
b. \
\\(\

$$\begin{tikzpicture}scale=0.5\ \\draw-> (-5,0) -- (5,0) noderight {$x$};\ \\draw-> (0,-5) -- (0,5) nodeabove {$y$};\ \\foreach \\x in {-4,-2,2,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x};\ \\foreach \\y in {-4,-2,2,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {\\y};\ \\drawthick, blue (-4,2) -- (0,1);\ \\drawthick, red (0,-2) -- (2,4);\ \\end{tikzpicture}$$

\\)\
\\((3, 0)\\)\
c. \
\\(\

$$\begin{tikzpicture}scale=0.5\ \\draw-> (-5,0) -- (5,0) noderight {$x$};\ \\draw-> (0,-5) -- (0,5) nodeabove {$y$};\ \\foreach \\x in {-4,-2,2,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x};\ \\foreach \\y in {-4,-2,2,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {\\y};\ \\drawthick, blue (-2,0) -- (0,-3);\ \\drawthick, red (0,-3) -- (4,-2);\ \\end{tikzpicture}$$

\\)\
\\((0, -3)\\)\
d. \
\\(\

$$\begin{tikzpicture}scale=0.5\ \\draw-> (-5,0) -- (5,0) noderight {$x$};\ \\draw-> (0,-5) -- (0,5) nodeabove {$y$};\ \\foreach \\x in {-4,-2,2,4} \\draw (\\x,0.1) -- (\\x,-0.1) nodebelow {\\x};\ \\foreach \\y in {-4,-2,2,4} \\draw (0.1,\\y) -- (-0.1,\\y) nodeleft {\\y};\ \\drawthick, blue (-4,4) -- (0,3);\ \\drawthick, red (0,-2) -- (2,4);\ \\end{tikzpicture}$$

\\)\
\\((0, 3)\\)\

1. what steps transform the graph of \\(y = x^2\\) to \\(y = -(x + 3)^2 + 5\\)?\

a. translate 3 units to the right, translate down 5 units\
b. translate 3 units to the left, translate up 5 units\
c. reflect across the x-axis, translate 3 units to the left, translate up 5 units\
d. reflect across the x-axis, translate 3 units to the right, translate down 5 units

Explanation:

Question 13

Step1: Analyze the first equation

The first equation is \(-x + 2y=10\). We can rewrite it in slope - intercept form (\(y = mx + b\)):
Add \(x\) to both sides: \(2y=x + 10\), then divide by 2: \(y=\frac{1}{2}x + 5\). The slope (\(m_1\)) of this line is \(\frac{1}{2}\) and the y - intercept (\(b_1\)) is 5.

Step2: Analyze the second equation

The second equation is \(-3x + 6y = 11\). Rewrite it in slope - intercept form:
Add \(3x\) to both sides: \(6y=3x + 11\), then divide by 6: \(y=\frac{3x + 11}{6}=\frac{1}{2}x+\frac{11}{6}\). The slope (\(m_2\)) of this line is \(\frac{1}{2}\) and the y - intercept (\(b_2\)) is \(\frac{11}{6}\).

Step3: Compare the two lines

Since the slopes of the two lines (\(m_1 = m_2=\frac{1}{2}\)) are equal and the y - intercepts (\(b_1 = 5=\frac{30}{6}\) and \(b_2=\frac{11}{6}\)) are different, the two lines are parallel. Parallel lines do not intersect, so the system of equations has no solutions.

Question 14

Step1: Rewrite the first equation in slope - intercept form

The first equation is \(-0.7x-0.2y = 0.6\).
Add \(0.7x\) to both sides: \(-0.2y=0.7x + 0.6\).
Divide both sides by \(- 0.2\): \(y=\frac{0.7x + 0.6}{-0.2}=\frac{0.7x}{-0.2}+\frac{0.6}{-0.2}=-3.5x - 3\).

Step2: Rewrite the second equation in slope - intercept form

The second equation is \(0.1x-0.5y = 1.5\).
Subtract \(0.1x\) from both sides: \(-0.5y=-0.1x + 1.5\).
Divide both sides by \(-0.5\): \(y=\frac{-0.1x + 1.5}{-0.5}=\frac{-0.1x}{-0.5}+\frac{1.5}{-0.5}=0.2x-3\).

Step3: Find the intersection point

To find the solution of the system, we can set the two equations equal to each other: \(-3.5x - 3=0.2x-3\).
Add \(3.5x\) to both sides: \(-3 = 3.7x-3\).
Add 3 to both sides: \(0 = 3.7x\), so \(x = 0\).
Substitute \(x = 0\) into \(y=-3.5x - 3\) (we could also use the other equation), we get \(y=-3\). So the solution of the system is \((0,-3)\), which corresponds to option d.

Question 15

Step1: Recall the transformation rules for quadratic functions

• For a quadratic function of the form \(y=a(x - h)^2+k\) (vertex form), where \((h,k)\) is the vertex of the parabola:
• The transformation of \(y = x^{2}\) (where \(a = 1\), \(h = 0\), \(k = 0\)) to \(y=-x^{2}\) is a reflection across the x - axis (because the coefficient of \(x^{2}\) changes from positive to negative).
• The transformation of \(y=-x^{2}\) to \(y=-(x + 3)^{2}\) is a horizontal translation. The formula for horizontal translation is: if we have \(y = f(x)\), then \(y=f(x + c)\) is a translation of \(c\) units to the left (when \(c>0\)). Here, \(c = 3\), so it is a translation of 3 units to the left.
• The transformation of \(y=-(x + 3)^{2}\) to \(y=-(x + 3)^{2}+5\) is a vertical translation. The formula for vertical translation is: if we have \(y = f(x)\), then \(y=f(x)+d\) is a translation of \(d\) units up (when \(d>0\)). Here, \(d = 5\), so it is a translation of 5 units up.

Step2: Match with the options

The steps to transform \(y = x^{2}\) to \(y=-(x + 3)^{2}+5\) are: reflect across the x - axis, translate 3 units to the left, translate up 5 units.

Answer:

d. no solutions