SAT Randomized Questions - 1 Full Math Practice Test - Answers and Detailed Explanations at the END

Both sides simplify to \( 8z + 4 \), making the equation true for all values of z. Therefore, there are infinitely many solutions.

We use the Pythagorean theorem: \( c^2 = 16^2 + 30^2 = 256 + 900 = 1156 \). Therefore, \( c = \sqrt{1156} = 34 \). The hypotenuse can be written as \( 2\sqrt{289} \), so \( d = 289 \).

To isolate \(5x + 6y\), multiply both sides by \((5x + 6y)\), resulting in \(n = t(5x + 6y)\). Dividing by \(t\) gives \(5x + 6y = \frac{n}{t}\).

Let the number of voters who voted in favor of the bill be w. According to the problem, 3 times as many voters voted against, so the number against is 3w. The report states that 18,000 more people voted against, so the equation is 3w = w + 18,000. Solving for w gives 9,000, and the number against is 3(9,000) = 27,000.

Since \( j(5) = 15625 \), substitute into \( j(x) = 50 \cdot k^x \): \( 50 \cdot k^5 = 15625 \Rightarrow k^5 = 312.5 \Rightarrow k = 5 \). Now find \( j(0.5) \): \( j(0.5) = 50 \cdot 5^{0.5} = 50 \cdot \sqrt{5} = 50 \sqrt{5} \).

To find the doubling time, set \( h(t) = 2 \cdot h(0) \) and solve for \( t \): \( 45,000 \cdot (1.04)^{t/200} = 90,000 \). Use logarithms to solve for \( t \) and convert to hours.

The two lines have the same slope but different intercepts, which means they are parallel and do not intersect at any point. Thus, the answer is zero points of intersection.

Substitute \( (1, 5) \) into both inequalities. Both inequalities are satisfied.

To find \( p \): \( 75 = \frac{p}{100} \times 150 \implies p = \frac{75 \times 100}{150} = 50 \). Calculate \( 50 \% \) of \( 75 \): \( \frac{50}{100} \times 75 = 37.5 \).

To convert from square yards to square miles, divide the area by \(1760^2\): \(5,904,900 / 3,097,600 = 1.91\) square miles.

Substitute \( s = 10 \) into the equation: \( 4p + 2(10) = 100 \Rightarrow 4p + 20 = 100 \Rightarrow 4p = 80 \Rightarrow p = 20 \).

We use the Pythagorean theorem: \( c^2 = 8^2 + 15^2 = 64 + 225 = 289 \). Therefore, \( c = \sqrt{289} = 17 \). The hypotenuse can be written as \( \sqrt{289} \), so \( d = 289 \).

The percentage can be calculated using the formula: \( \frac{120}{400} \times 100 = 30\% \).

Since the triangles are similar, corresponding angles have the same trigonometric ratios. Thus, \( \sin(D) = \sin(A) = \frac{120}{125} \).

Rewrite \(f(x)\) in vertex form as \(-3(x - 2)^2 + 7\). The maximum of \(f(x)\) is at \(x = 2\), so \(g(x) = f(x - 6)\) shifts this maximum to \(x = 8\).

Start by setting up the equation: \( 32,000 = 4,000(2)^{3r} \). Divide both sides by 4,000 to obtain \( 8 = (2)^{3r} \). Taking the logarithm base 2 of both sides yields \( 3 = 3r \). Therefore, \( r = 1 \).

The two lines are parallel because their slopes are equal. Dividing the first equation by 2 yields \( 3.5x + y = 7.5 \). Since this differs from the second equation, the lines are parallel and do not intersect.

Start by setting up the equation with the given values: \( 64,000 = 8,000(2)^{6r} \). Divide both sides by 8,000 to get \( 8 = (2)^{6r} \). Taking the logarithm base 2 of both sides gives \( 3 = 6r \). Therefore, \( r = \frac{1}{2} \).

Expanding the equation, we obtain \( 2x^2 + 3x - 9 = 4x^2 - 24x \). Rearranging terms, we get \( -2x^2 + 27x - 9 = 0 \). Using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -2 \) and \( b = 27 \), which yields \( 27/2 \).

Multiplying the first equation by -2 results in \( -2x + 8y = -6 \), which matches the second equation. Thus, the lines are the same, intersecting at infinitely many points.

With side lengths of square X being 50 times those of square Y, the area of square X is \( 50^2 = 2500 \) times the area of square Y. Thus, \( k = 2500 \).

In the function j(x) = -0.5x + 10, the y-intercept (where x = 0) represents the initial amount of water the runner had before starting the marathon. When x = 0, j(0) = 10, indicating that the runner started with 10 liters of water.

Rewrite both equations in slope-intercept form. To have no solution, the slopes must be equal but the y-intercepts must differ. Solving, we find m = -6.

To find the surface area, find the side length first. The volume of the cube is \( V = s^3 \). Given \( V = 729 \), \( s = \sqrt[3]{729} = 9 \). The surface area formula is \( 6s^2 \). Therefore, the surface area is \( 6(9)^2 = 486 \) square units.

To have no real solution, the discriminant of \(x(m x - 40) + 25 = 0\) must be negative. Solving gives \(\)m^2 < 25[/latex], so the smallest integer for [latex]m[/latex] is 5.

Since \(\frac{c}{d} = 2\), we have \(c = 2d\). Substitute into \(\frac{40c}{qd} = 2\) to get \(\frac{40(2d)}{qd} = 2\). Simplify to \(\frac{80d}{qd} = 2\), cancel \(d\) and solve to find \(q = 40\).

The second equation is just a multiple of the first. Dividing the second equation by -2 yields \( 5x + 4y = 16 \). Thus, the two lines are the same and intersect at infinitely many points.

For the system to have no solution, the lines must be parallel. This means their slopes must be equal. Rewrite each equation in slope-intercept form to compare slopes, leading to k = 5.

For a quadratic equation, the minimum value occurs at \( x = \frac{-b}{2a} \). Here, \( x = \frac{16}{2} = 8 \).

Using the points, calculate the slope and equation of the line, then substitute \(x = q + 5\) to find \(b = -55\).

Since \(g(x) = f(x+3) = 5(3)^{x+3} = 5 \cdot 27 \cdot (3)^x = 135(3)^x\), the correct answer is g(x) = 135(3)^x.

Substitute values from each answer choice into both equations. Only \( (5, 10) \) satisfies both equations.

The half-life is 3 years, so the remaining quantity after \( x \) minutes is \( Q = 5 \cdot \left(\frac{1}{2}\right)^{\frac{x}{3 \cdot 365 \cdot 24 \cdot 60}} \).

To cover the walls twice, double the total area \(w\) to get \(2w\). Since each gallon covers 250 square feet, divide \(2w\) by 250 to get \(P = \frac{w}{125}\).

Since the side lengths of square P are 20 times the side lengths of square Q, the area of square P is \( 20^2 = 400 \) times the area of square Q. Therefore, \( k = 400 \).

We use the Pythagorean theorem: \( c^2 = 5^2 + 12^2 = 25 + 144 = 169 \). Therefore, \( c = \sqrt{169} = 13 \). Since the length is already in the form \( \sqrt{169} [/latex}, the value of [latex] d \) is 169.

To ensure no solution, we set the slopes equal, indicating b = 48 based on intercept calculations.

Let the number of people who voted in favor of the measure be z. According to the problem, 4 times as many people voted against, so the number against is 4z. The survey states that 12,000 more people voted against, so the equation is 4z = z + 12,000. Solving for z gives 4,000.

(1,1) works since for the first equation, subbing in 1 gives us []

Since the function increases by 50%, it becomes 1.50 times its previous value at each step. Thus, the multiplier for growth is 1.50. Therefore, the correct equation is \( k(x) = 80(1.50)^x \).

Since \(g(x) = f(x+2) = 8(5)^{x+2} = 8 \cdot 25 \cdot (5)^x = 200(5)^x\), the correct answer is g(x) = 200(5)^x.

To determine the doubling time, solve \( p(t) = 2 \cdot p(0) \), which gives \( 75,000 \cdot (1.02)^{t/250} = 150,000 \). Take the logarithm of both sides, isolate \( t \), and convert to hours.

Rewrite \(f(x)\) in vertex form as \(2(x + 2)^2 - 2\). The minimum of \(f(x)\) is at \(x = -2\), so \(g(x) = f(x + 1)\) shifts this to \(x = -3\).

Data set A is skewed toward lower values, while data set B is skewed toward higher values. Therefore, the mean of Data set A is less than the mean of Data set B.