Digital SAT Math Practice Test 10: Advanced Module

For infinitely many solutions, the equations must be proportional. Comparing the constants: \(14 = 2 \times 7\). Thus, every term in the second equation must be twice the corresponding term in the first.

\[ \dfrac{k}{2} = 2 \Rightarrow k = 4 \]

\[ \dfrac{5}{p} = 2 \Rightarrow 2p = 5 \Rightarrow p = 2.5 \]

Sum: \(k + p = 4 + 2.5 = 6.5\).

An absolute value inequality \(|x - a| < b\) represents all \(x\) within distance \(b\) from center \(a\).

Step 1: Find the center (a)

\[ a = \dfrac{-2 + 8}{2} = 3 \]

Step 2: Find the distance (b)

The distance from center 3 to boundary 8 is: \[ b = 8 - 3 = 5 \]

Step 3: Calculate the sum

\[ a + b = 3 + 5 = 8 \]

The rate of change indicates the slope (\(m\)):

\[ m = \dfrac{\Delta y}{\Delta x} = \dfrac{8}{2} = 4 \]

Using the form \(f(x) = 4x + b\) and the point \((3, 10)\):

\[ 10 = 4(3) + b \Rightarrow 10 = 12 + b \Rightarrow b = -2 \]

The \(y\)-intercept is -2.

Let the roots be \(r\) and \(3r\). By Vieta's formulas, the sum of roots is \(-b/a\):

\[ r + 3r = -(-8)/2 = 4 \Rightarrow 4r = 4 \Rightarrow r = 1 \]

The roots are 1 and 3. The product of roots is \(c/a\):

\[ (1)(3) = k/2 \Rightarrow 3 = k/2 \Rightarrow k = 6 \]

Divisibility by \(x^2-4\) implies roots at \(x=2\) and \(x=-2\).

1. \(P(2) = 8 - 4a + 2b - 4 = 0 \Rightarrow -4a + 2b = -4\)
2. \(P(-2) = -8 - 4a - 2b - 4 = 0 \Rightarrow -4a - 2b = 12\)

Adding the equations: \(-8a = 8 \Rightarrow a = -1\). Substituting into (1) gives \(b = -4\). Thus, \(a + b = -5\).

Set them equal: \(x^2 - 4x + c = 2x - 1 \Rightarrow x^2 - 6x + (c + 1) = 0\).

Exactly one intersection means the discriminant is zero:

\[ \Delta = (-6)^2 - 4(1)(c+1) = 0 \Rightarrow 36 - 4c - 4 = 0 \Rightarrow c = 8 \]

Equation: \(1.40x \times (1 - d/100) = x\).

\[ 1 - d/100 = 1/1.4 \approx 0.7143 \]

\[ d/100 = 0.2857 \Rightarrow d = 28.6 \]

Weighted average equation: \(82b + 90g = 85(b+g)\).

\[ 5g = 3b \Rightarrow g = \dfrac{3}{5}b \]

Fraction of boys = \(\dfrac{b}{b + \frac{3}{5}b} = \dfrac{1}{8/5} = \dfrac{5}{8}\).

Adding a constant doesn't change the standard deviation. Multiplying by a constant \(C\) scales the SD by \(|C|\).

\[ \text{New SD} = |-2| \times S = 2S \]

Complete the square: \((x-2)^2 + (y+3)^2 = 25\). Radius \(r = 5\), so Diameter \(d = 10\).

For an inscribed square, the diagonal is the diameter. Area = \(d^2/2\):

\[ \text{Area} = 10^2 / 2 = 50 \]

\[ V_{new} = \dfrac{1}{3}\pi (2r)^2 (\dfrac{1}{2}h) = \dfrac{1}{3}\pi (4r^2) (\dfrac{1}{2}h) = 2 \left( \dfrac{1}{3}\pi r^2 h \right) = 2V \]

If \(\sin(P) = a/b\), then \(RQ = a\) and \(PR = b\). By Pythagoras, \(PQ = \sqrt{b^2 - a^2}\).

\[ \tan(R) = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{PQ}{RQ} = \dfrac{\sqrt{b^2 - a^2}}{a} \]