Advanced SAT Math Challenge Practice Quiz - Real Challenging College Board SAT Math Questions

First, find g(3): \( g(3) = 3 - 2 = 1 \). Then, substitute this into f(x): \( f(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0 \).

Rewrite 32 as a power of 2: \( 32 = 2^5 \). Thus, \( 2^{x+3} = 2^5 \). Since the bases are the same, we equate the exponents: \( x + 3 = 5 \), giving \( x = 2 \).

Since A and B are complementary angles in a right triangle, \( \cos(B) = \sin(A) \). Using the identity \( \sin^2(A) + \cos^2(A) = 1 \) and knowing \( \tan(A) = \frac{3}{4} \), we can deduce \( \sin(A) = \frac{3}{5} \) and thus \( \cos(B) = \frac{3}{5} \).

Substitute \( b = 2c \) into the expression for a: \( a = (2c)^2 + c^2 = 4c^2 + c^2 = 5c^2 \).

Solve the system using substitution or elimination. Multiply the first equation by 3 to eliminate y: \( 6x - 3y = 15 \) and add it to the second equation: \( 6x - 3y + x + 3y = 15 + 4 \), simplifying to \( 7x = 19 \) or \( x = \frac{19}{7} \).

To solve \( 2x + 3 < 7x - 5 \), rearrange the terms to isolate x: \( 3 + 5 < 7x - 2x \) leading to \( 8 < 5x \) or \( x > \frac{8}{5} \).

The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where (h, k) is the center and r is the radius. Substituting the given values, we get \( (x - 2)^2 + (y - 3)^2 = 25 \).

From the definition of logarithms, if \( \log_{10}(a) = 3 \), then \( 10^3 = a \), which means \( a = 1000 \).

Substitute x = -2 into the function: \( f(-2) = 3(-2)^2 - 2(-2) + 1 = 3(4) + 4 + 1 = 12 + 4 + 1 = 17 \).

The formula for the slope between two points (x1, y1) and (x2, y2) is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substituting the values, we get \( m = \frac{5 - (-3)}{-4 - 2} = \frac{8}{-6} = -\frac{4}{3} \).