Digital SAT Math Practice Test 9: Advanced Module

A linear equation has no solution if the coefficients of \(x\) on both sides are equal, but the constant terms are different.

Step 1: Simplify the left side

\[ 3kx - 5k + 2x = px + 10 \]

\[ (3k + 2)x - 5k = px + 10 \]

Step 2: Match coefficients

For there to be no solution, the coefficient of \(x\) must be the same: \[ 3k + 2 = p \]

Step 3: Solve for the target expression

Rearranging the equation: \[ 3k - p = -2 \]

First, find the slope of the original line: \[ m_1 = \dfrac{b - 0}{0 - a} = -\dfrac{b}{a} \]

Perpendicular slopes are negative reciprocals: \[ m = -\left(-\dfrac{a}{b}\right) = \dfrac{a}{b} \]

The center point of an absolute value equation is the midpoint of its solutions:

\[ \text{Midpoint} = \dfrac{2 + 8}{2} = 5 \]

At the center point, the expression inside the absolute value equals zero:

\[ 2(5) - c = 0 \Rightarrow 10 - c = 0 \Rightarrow c = 10 \]

A horizontal line intersects a parabola at one point only if it passes through the vertex. Thus, \(c\) is the y-coordinate of the vertex.

\[ x_{\text{vertex}} = -\dfrac{6}{2(-1)} = 3 \]

\[ y_{\text{vertex}} = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4 \]

Therefore, \(c = 4\).

First, find the minimum value of \(f(x)\). Its vertex is at \(x = 2\):

\[ f(2) = (2)^2 - 4(2) + 3 = -1 \]

The transformation \(f(x+2)\) shifts the graph horizontally, which doesn't change the minimum. The \(-5\) shifts the graph down by 5 units.

\[ \text{Min of } g(x) = -1 - 5 = -6 \]

Using the Difference of Cubes formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\):

\[ x^3 - 2^3 = (x - 2)(x^2 + 2x + 4) \]

Comparing to the given form, \(b = 2\) and \(c = 4\). Thus, \(b + c = 6\).

\[ \left(1 + \dfrac{p}{100}\right)\left(1 - \dfrac{p}{100}\right) = 0.91 \]

\[ 1 - \left(\dfrac{p}{100}\right)^2 = 0.91 \Rightarrow \left(\dfrac{p}{100}\right)^2 = 0.09 \]

\[ \dfrac{p}{100} = 0.3 \Rightarrow p = 30 \]

The standard model is \(A(t) = P(1-r)^{t/d}\) where \(d\) is the time for one cycle. Since the 20% drop occurs every 5 days, the cycle time is 5. Therefore, \(k = 5\).

\[ \text{Total Sum} = (n \times 70) + (2n \times 85) = 70n + 170n = 240n \]

\[ \text{Total Students} = n + 2n = 3n \]

\[ \text{Combined Mean} = 240n / 3n = 80 \]

\[ \text{Area} = \dfrac{\theta}{360} \pi r^2 \Rightarrow \dfrac{5\pi}{3} = \dfrac{150}{360} \pi r^2 \]

\[ \dfrac{5}{3} = \dfrac{5}{12}r^2 \Rightarrow r^2 = 4 \Rightarrow r = 2 \]

Complete the square: \[ (x+3)^2 + (y-4)^2 = c + 9 + 16 \]

The right side is \(r^2 = 10^2 = 100\). \[ c + 25 = 100 \Rightarrow c = 75 \]

Identity: \(\cos(C) = \sin(A)\). For \(\tan(A) = x/1\), opposite is \(x\), adjacent is \(1\), and hypotenuse is \(\sqrt{x^2 + 1}\).

\[ \sin(A) = \dfrac{x}{\sqrt{x^2 + 1}} \]