SAT Math Randomized Questions - 2 Full Math Practice Tests - Answers and Detailed Explanations at the END

To find \( f(0) \), substitute \( 0 \) for \( x \): \( f(0) = 180(0.3)^0 \). Since \( (0.3)^0 = 1 \), the expression becomes \( 180 imes 1 = 180 \).

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 \( \frac{9}{5} \times 1.50 = 2.70 \).

In the function g(x) = -3x + 50, the y-intercept (where x = 0) represents the initial amount of dough the baker had before making any loaves. When x = 0, g(0) = 50, indicating that the baker started with 50 pounds of dough.

To find the doubling time, set \( g(t) = 2 \cdot g(0) \). Thus, \( 20,000 \cdot (1.05)^{t/300} = 40,000 \). Taking logarithms, isolate \( t \) and convert the answer to hours.

Rewriting the equation: \( x^2 + x - 56 = 4x^2 - 28x \). Rearranging terms, we get \( -3x^2 + 29x - 56 = 0 \). Solving using the quadratic formula, the sum of the solutions is given by \( -b/a \), where \( a = -3 \) and \( b = 29 \), which yields \( 29/3 \).

The investment grows by 8% per month, so the monthly growth factor is 1.08. After \( x \) years, the equation is \( V = 10,000 \cdot 1.08^{12x} \).

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

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 \( \frac{9}{5} \times 3.00 = 5.40 \).

Dividing the equation by 36, completing the square, and simplifying gives n = 12.

To find \( p \): \( 50 = \frac{p}{100} \times 80 \implies p = \frac{50 \times 100}{80} = 62.5 \). Calculate \( 62.5 \% \) of \( 50 \): \( \frac{62.5}{100} \times 50 = 31.25 \).

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 \( d = 289 \).

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.

Rearrange and substitute to find that \( y = 10 \).

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.

Substitute \( c = 12 \) into the equation: \( 1.5a + 4.5(12) = 90 \Rightarrow 1.5a + 54 = 90 \Rightarrow 1.5a = 36 \Rightarrow a = 24 \).

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.

Using the point-slope form \( y - y_1 = m(x - x_1) \) with the slope \( m = \frac{3}{5} \) and point \( (2, -4) \), the resulting equation is \( y = \frac{3}{5}x - \frac{22}{5} \).

The population triples every hour, so the growth factor per hour is 3. Since there are 24 hours in a day, the population after \( x \) days is \( p = 1,000 \cdot 3^{24x} \).

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

Since \(\frac{r}{s} = 6\), we get \(r = 6s\). Substitute into \(\frac{48r}{ks} = 6\) to obtain \(\frac{48(6s)}{ks} = 6\). Simplify to \(\frac{288s}{ks} = 6\), cancel \(s\) and solve for \(k = 48\).

First, find the measure of angle T in radians: \( \frac{2\pi}{3} + \frac{\pi}{4} = \frac{11\pi}{12} \). Then, convert to degrees by multiplying by \( \frac{180}{\pi} \): \( \frac{11\pi}{12} \cdot \frac{180}{\pi} = 165 \) degrees.

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

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

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, \( r = \frac{1}{2} \).

Since the function decreases by 45%, it retains 55% of its value at each step. Thus, the multiplier for the decay is 0.55. Therefore, the correct equation is \( q(x) = 14(0.55)^x \).

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 \( p \).

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.

Since triangles \( \triangle ABC \) and \( \triangle DEF \) are congruent, angle \( C \) corresponds to angle \( F \). Therefore, angle \( F \) must be 55°.

Both equations are identical, which means the lines coincide perfectly and intersect at infinitely many points.

Calculate percentages: Candidate L received \( \frac{600}{1000} = 0.6 \), or 60%. Candidate M received \( \frac{400}{1000} = 0.4 \), or 40%. If 7,500 people vote, Candidate L would receive \( 0.6 \times 7500 = 4500 \) votes, and Candidate M \( 0.4 \times 7500 = 3000 \) votes. The difference is \( 4500 - 3000 = 1500 \) votes.

Using the quadratic formula, \(x = \frac{10 \pm \sqrt{100 - 8}}{2} = 5 \pm \sqrt{23}\). So, the solution can be written as \(5 + \sqrt{23}\) and the correct value of \(k\) is 23.

The mean of each data set is centered around 15, as both data sets have symmetric distributions around this value.

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

The population grows by 2% annually, so the growth factor is 1.02. Since \( x \) is in seconds, we convert it to years by dividing by \( 365 \cdot 24 \cdot 60 \cdot 60 \). The equation is \( p = 50,000 \cdot 1.02^{\frac{x}{365 \cdot 24 \cdot 60 \cdot 60}} \).

Let the number of people who voted in favor of the policy be x. According to the problem, 5 times as many people voted against it, so the number against is 5x. The article states that 40,000 more people voted against it, so the equation is 5x = x + 40,000. Solving for x gives 10,000.

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

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

Since \(f(x) = 1.2x\) shows a constant proportional increase, the function is linear and increases with \(x\). Thus, the correct answer is 'Increasing Linear.'

The sum of the solutions is calculated separately for each factor. Using the given total sum of \( \frac{25}{2} \), solve for the value of \( p \).

Calculate the percentages: Candidate A received \( \frac{600}{1000} = 0.6 \), or 60%. Candidate B received \( \frac{400}{1000} = 0.4 \), or 40%. If 10,000 people vote, Candidate A is expected to receive \( 0.6 \times 10000 = 6000 \) votes, and Candidate B \( 0.4 \times 10000 = 4000 \) votes. The difference is \( 6000 - 4000 = 2000 \) votes.

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 \( 27/2 \).

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 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.