SAT Advanced Algebra and Functions Quiz - Real Hard College Board SAT Practice Questions

Substitute \( x = 1 \) into the equation: \( 2(1)^2 - 3(1)y + y^2 = 1 \). Simplifying gives \( 2 - 3y + y^2 = 1 \), or \( y^2 - 3y + 1 = 0 \). Solving this quadratic equation gives \( y = \frac{3 \pm \sqrt{5}}{2} \).

Use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Plugging in the points (1, 4) and (3, 10) gives \( m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3 \).

Calculate each term and sum them up: \( (2^1 + 1) + (2^2 + 2) + (2^3 + 3) + (2^4 + 4) + (2^5 + 5) = 3 + 6 + 11 + 20 + 37 = 77 \).

To find the inverse, swap \( x \) and \( y \) in \( y = \frac{x+1}{x-2} \), and solve for \( y \): \( x = \frac{y+1}{y-2} \), solving for \( y \) yields \( y = \frac{2x+1}{x-1} \).

The nth term of a geometric sequence is given by \( a_n = a_1 \cdot r^{n-1} \). For the fifth term, \( a_5 = 2 \cdot 3^{4} = 162 \).

Solve the system by substitution or elimination. Multiplying the first equation by 3 and subtracting the second equation gives \( 10y = 20 \), so \( y = 2 \). Substituting \( y = 2 \) into the first equation yields \( x = 3 \). Therefore, \( x - y = 1 \).

Apply the chain rule to each term: \( h'(x) = 2e^{2x} - 3e^{-x} \).

The expression under the square root must be non-negative, so \( x^2 - 9 \geq 0 \). This means \( x \leq -3 \) or \( x \geq 3 \). The range of \( g(x) \) starts from \( 0 \) and extends to infinity.

First solve the equality \( 2x^2 - 5x - 3 = 0 \) to find critical points, then test intervals around these points. The critical points are \( x = -\frac{1}{2} \) and \( x = 3 \). Testing intervals gives the solution \( x \leq -\frac{1}{2} \) or \( x \geq 3 \).

The domain is determined by ensuring both arguments of the logarithms are positive: \( x + 2 > 0 \) and \( x - 2 > 0 \). Solving these inequalities gives \( x > 2 \).