Questions 11 - 20
Question 11: Which property is used to write \( a(x + y) = a x + a y \)?
This algebraic property shows multiplication distributing over addition.
Answer: Distributive Property (Distributivity)
Question 12: Simplify \( \frac{8x^3}{2x^{-3}} \).
Divide the coefficients: \( 8 / 2 = 4 \).
Use the quotient rule for exponents \( \frac{x^a}{x^b} = x^{a - b} \):
\( x^{3 - (-3)} = x^{3 + 3} = x^6 \)
Answer: \( 4x^6 \)
Question 13: Simplify \( (-a^2b^3)^2(c^2)^0 \).
First, recall that any non-zero base raised to the power of 0 is 1. So, \( (c^2)^0 = 1 \).
Square the first term by multiplying the exponents:
\( (-1)^2 \cdot (a^2)^2 \cdot (b^3)^2 = 1 \cdot a^4 \cdot b^6 \)
Answer: \( a^4b^6 \)
Question 14: For what value of \( k \) is the point \( (-2, k) \) on the line \( -3 x + 3 y = 4 \)?
Substitute \( x = -2 \) and \( y = k \) into the equation:
\( -3(-2) + 3(k) = 4 \)
\( 6 + 3k = 4 \)
Subtract 6 from both sides:
\( 3k = -2 \)
Divide by 3:
Answer: \( k = -2/3 \)
Question 15: For what value of \( a \) will the system have no solutions? \( \{2x+6y=-2; -3x+ay=4\} \)
A system of linear equations has no solutions if the lines are parallel, meaning their slopes are equal but their y-intercepts are different.
Find the slope of the first line: \( 6y = -2x - 2 \Rightarrow y = -1/3x - 1/3 \) (Slope is \(-1/3\)).
Find the slope of the second line: \( ay = 3x + 4 \Rightarrow y = (3/a)x + 4/a \) (Slope is \(3/a\)).
Set the slopes equal to each other:
\( -1/3 = 3/a \)
Cross multiply to solve for \(a\): \( -a = 9 \Rightarrow a = -9 \).
Answer: \( a = -9 \)
Question 16: Find the equation for the \( (x,y) \) pairs: (0,-4), (4,-20), (-4,12), (8,-36).
First, find the slope (\(m\)) using the points (0, -4) and (4, -20):
\( m = \frac{-20 - (-4)}{4 - 0} = \frac{-16}{4} = -4 \).
The y-intercept (\(b\)) is given when \(x = 0\), which is -4.
Using \( y = mx + b \):
Answer: C) \( y = -4x - 4 \)
Question 17: Which formula best represents the area of the rectangle shown?
The diagram shows a rectangle with a length of \( (x + 1) \) and a width of \( (x - 1) \), you multiply them together to find the area:
\( \text{Area} = (x + 1)(x - 1) \)
Expanding this yields the difference of squares:
Answer: \( \text{area} = x^2 - 1 \)
Question 18: Which line contains the points \( (1, -1) \) and \( (3, 5) \)?
Find the slope (\(m\)):
\( m = \frac{5 - (-1)}{3 - 1} = \frac{6}{2} = 3 \).
Use point-slope form with (1, -1):
\( y - y_1 = m(x - x_1) \)
\( y - (-1) = 3(x - 1) \)
\( y + 1 = 3x - 3 \)
\( y = 3x - 4 \)
Multiply the entire equation by 2 to match the standard multiple-choice formats commonly presented:
Answer: \( 2y = 6x - 8 \)
Question 19: Solve the equation \( 2|3x - 2| - 3 = 7 \).
Isolate the absolute value expression:
\( 2|3x - 2| = 10 \)
\( |3x - 2| = 5 \)
Set up the two possible equations:
Case 1: \( 3x - 2 = 5 \Rightarrow 3x = 7 \Rightarrow x = 7/3 \)
Case 2: \( 3x - 2 = -5 \Rightarrow 3x = -3 \Rightarrow x = -1 \)
Answer: Solution set: \( \{ 7/3, -1 \} \)
Question 20: Solve for \( x \) the equation \( (1/2) x^2 + mx - 2 = 0 \).
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1/2 \), \( b = m \), and \( c = -2 \):
\( x = \frac{-m \pm \sqrt{m^2 - 4(1/2)(-2)}}{2(1/2)} \)
\( x = \frac{-m \pm \sqrt{m^2 + 4}}{1} \)
Answer: \( \{ -m \pm \sqrt{m^2 + 4} \} \)