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

The decay factor is 0.68 every half-year, which means it decays by 32% each six months. This translates to approximately a 55% annual decay rate, so \(p = 55\).

To find \( p \), use the equation: \( 39 = \frac{p}{100} \times 65 \Rightarrow p = \frac{39 \times 100}{65} = 60 \). Then, calculate \( 60 \% \) of \( 39 \): \( \frac{60}{100} \times 39 = 23.4 \).

In the equation 3𝑠 + 50𝑡 = 8,000, the coefficient 3 corresponds to the trail, meaning 𝑠 represents the total number of streetlights per kilometer along the trail.

The minimum value of \( y \) occurs at \( x = \frac{-b}{2a} \), giving \( x = \frac{6}{2} = 3 \).

The lines have the same slope but different intercepts, indicating that they are parallel and do not intersect.

To find the doubling time, set \( b(t) = 2 \cdot b(0) \) and solve for \( t \): \( 15,000 \cdot (1.06)^{t/100} = 30,000 \). Use logarithms to isolate \( t \) and convert to hours.

To have no real solution, the discriminant of \(x(r x - 120) + 64 = 0\) must be negative. Solving gives \(\)r^2 < 64[/latex], so the smallest integer for [latex]r[/latex] is 8.

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.

With the slope \( \frac{7}{8} \) and point \( (4, -1) \), apply the point-slope formula to find the equation \( y = \frac{7}{8}x - \frac{23}{8} \).

Substitute \( (-14, 0) \) into both inequalities. Both inequalities are satisfied.

Since \( f(3) = 1024 \), substitute into \( f(x) = 8 \cdot d^x \): \( 8 \cdot d^3 = 1024 \Rightarrow d^3 = 128 \Rightarrow d = 4 \). Now find \( f(-2) \): \( f(-2) = 8 \cdot 4^{-2} = 8 \cdot \frac{1}{16} = 0.5 \).

Using the quadratic formula, \(x = \frac{12 \pm \sqrt{144 - 44}}{2} = 6 \pm \sqrt{25}\). So, the solution can be written as \(6 + \sqrt{25}\) and the correct value of \(k\) is 25.

Dividing the equation by 16, completing the square, and simplifying gives n = 8.

Dividing the equation by 4, completing the square, and simplifying gives n = 4.

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

Calculate percentages: Candidate X received \( \frac{350}{800} = 0.4375 \), or 43.75%. Candidate Y received \( \frac{450}{800} = 0.5625 \), or 56.25%. If 5,600 people vote, Candidate X would receive \( 0.4375 \times 5600 = 2450 \) votes, and Candidate Y \( 0.5625 \times 5600 = 3150 \) votes. The difference is \( 3150 - 2450 = 700 \) votes.

The vertex of the function \( f(x) = \frac{1}{9}(x - 7)^2 + 3 \) occurs at \( (7, 3) \). This represents the minimum height of the metal ball. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 7 \) and \( k = 3 \), the minimum height is 3 inches above the ground.

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 \(x = 8\).

The change in degrees Fahrenheit is determined by multiplying the temperature increase in kelvins by \( \frac{9}{5} \), since the \( x \) term in the function is being multiplied by this factor. Therefore, the increase in Fahrenheit temperature is \( \frac{9}{5} \times 2.10 = 3.78 \).

To find the doubling time, solve \( j(t) = 2 \cdot j(0) \), leading to \( 100,000 \cdot (1.03)^{t/350} = 200,000 \). Taking logarithms of both sides allows solving for \( t \) and then converting to hours.

Let the number of small candles be represented by \( x \), and the number of large candles by \( y \). The total number of candles is at least 200, so we have the inequality:

\( x + y geq 200 \).

The budget constraint is that the total cost must not exceed $2,200. Therefore, we have the equation:

\( 4.90x + 11.60y leq 2,200 \).

We want to maximize the number of large candles, \( y \). First, solve for \( x \) in terms of \( y \) using the inequality:

\( x = 200 - y \).

Substitute this into the budget constraint:

\( 4.90(200 - y) + 11.60y leq 2,200 \)

Simplify:

\( 980 - 4.90y + 11.60y leq 2,200 \)

Combine like terms:

\( 980 + 6.70y leq 2,200 \)

Subtract 980 from both sides:

\( 6.70y leq 1,220 \)

Solve for \( y \):

\( y leq rac{1,220}{6.70} approx 182.09 \)

Since the number of candles must be an integer, the maximum number of large candles is 182.

Find the sum of the roots for each factor separately, then combine them and compare it to the given total of 8 to find the value of \( p \).

With the side lengths of square R being 12 times those of square S, the area of square R is \( 12^2 = 144 \) times the area of square S. Thus, \( k = 144 \).

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.

Since \( m(3) = 4096 \), substitute into \( m(x) = 16 \cdot q^x \): \( 16 \cdot q^3 = 4096 \Rightarrow q^3 = 256 \Rightarrow q = 4 \). Now find \( m(\frac{1}{4}) \): \( m(\frac{1}{4}) = 16 \cdot 4^{\frac{1}{4}} = 16 \cdot \sqrt[4]{4} = 16 \cdot \sqrt{2} = 16 \sqrt{2} \).

To find \( f(0) \), substitute \( 0 \) for \( x \) in the function: \( f(0) = 300(0.2)^0 \). Since any non-zero number raised to the power of \( 0 \) is \( 1 \), the expression becomes \( 300 imes 1 = 300 \).

The triangles are similar, so corresponding angles have the same trigonometric ratios. Therefore, \( \sin(P) = \sin(X) = \frac{5}{13} \).

In the equation 10𝑎 + 30𝑏 = 1,800, the coefficient 10 corresponds to the orchard, meaning 𝑎 represents the number of apple trees per acre in the orchard.

The percentage can be calculated using the formula: \( \frac{250}{500} \times 100 = 50\% \).

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, \( k = 2500 \).

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

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

Using elimination or substitution, we solve for \( y = 5 \).

The drainage rate is 10 liters per minute. Over 2 hours (120 minutes), the volume drained is 10 * 120 = 1200 liters. Converting to milliliters (1 liter = 1000 mL), the answer is 1,200,000 mL.

The two equations are actually multiples of each other. Dividing the second equation by 2 yields \( 3x + 5y = 25 \). Thus, both lines coincide, which means they intersect at infinitely many points.

Only \( (2, 6) \) satisfies both equations.

In the function f(x) = -5x + 30, the y-intercept (where x = 0) represents the initial amount of juice Caleb had before he made any popsicles. When x = 0, f(0) = 30, indicating that Caleb started with 30 fluid ounces of juice.

With \(p(x) = 0.75x\), the function decreases linearly as \(x\) increases, representing a consistent reduction in value. Hence, the answer is 'Decreasing Linear.'

Factor out the greatest common factor, \(3x\), from the expression: \(3x(x^2 + 9x + 18)\). Then factor \(x^2 + 9x + 18\) as \((x + 3)(x + 6)\). The smallest possible value of \(b\) is 3.

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

To isolate \(2p + 9q\), multiply both sides by \((2p + 9q)\), resulting in \(g = w(2p + 9q)\). Dividing by \(w\) gives \(2p + 9q = \frac{g}{w}\).

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

Since the rate of drainage is 3 liters per second, over one minute (60 seconds), the volume drained would be 3 * 60 = 180 liters. Converting to milliliters (1 liter = 1000 mL), the answer is 180,000 mL.

The vertex of the function \( f(x) = \frac{1}{4}(x - 8)^2 + 6 \) occurs at \( (8, 6) \). This represents the minimum height of the drone. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 8 \) and \( k = 6 \), the minimum height is 6 meters above the ground.