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

Expanding the right side results in \( 10a - 3 \), which is identical to the left side. Therefore, the equation holds for all values of a, implying infinitely many solutions.

To find \( a \), set \( r(x) = 0 \): \( 8x - 16 = 0 Rightarrow x = 2 \). Thus, \( a = 2 \). For \( b \), evaluate \( r(0) = -16 \). Therefore, \( a + b = 2 - 16 = -14 \).

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.

Since the side lengths of square C are 8 times those of square D, the area of square C is \( 8^2 = 64 \) times the area of square D. Therefore, \( k = 64 \).

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.

Solving by elimination or substitution, we find that \( y = 14 \).

Find the slope using the points, then derive the line's equation. Set \(x = n + 2\) to find \(b = 48\).

Dividing the equation by 9, completing the square, and simplifying gives n = 6.

The values and frequencies are symmetric across the two data sets, which means the mean values are the same.

The vertex of the function \( f(x) = \frac{1}{8}(x - 6)^2 + 5 \) occurs at \( (6, 5) \). This represents the minimum height of the kite. The vertex form of a quadratic function, \( a(x - h)^2 + k \), indicates the vertex as \( (h, k) \). Since \( h = 6 \) and \( k = 5 \), the minimum height is 5 feet above the ground.

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 \(q = 40\).

Rewrite \(f(x)\) in vertex form as \(5(x - 2)^2 + 75\). The minimum of \(f(x)\) is at \(x = 2\), so \(g(x) = f(x + 3)\) shifts this to \(x = -1\).

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.

Set up the equation \( 24,000 = 1,500(2)^{12r} \). Divide both sides by 1,500 to obtain \( 16 = (2)^{12r} \). Taking the logarithm base 2 of both sides yields \( 4 = 12r \). Therefore, \( r = \frac{1}{3} \).

To isolate \(7m + 5n\), multiply both sides by \((7m + 5n)\) to get \(k = p(7m + 5n)\), then divide by \(p\) to find \(7m + 5n = \frac{k}{p}\).

Dividing the equation by 25, completing the square, and simplifying gives n = 5.

Factor out the greatest common factor, \(2x\), from the expression: \(2x(x^2 + 9x + 16)\). Then factor \(x^2 + 9x + 16\) as \((x + 4)(x + 9)\). The smallest possible value of \(b\) is 4.

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.

To have no real solution, the discriminant of \(x(p x - 90) + 49 = 0\) must be negative. Solving gives \(\)p^2 < 49[/latex], so the smallest integer for [latex]p[/latex] is 7.

Isolate \(10a + 3b\) by multiplying both sides by \((10a + 3b)\), resulting in \(m = q(10a + 3b)\). Dividing by \(q\) gives \(10a + 3b = \frac{m}{q}\).

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

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 \( m(\frac{1}{4}) \): \( m(\frac{1}{4}) = 16 \cdot 4^{\frac{1}{4}} = 16 \cdot \sqrt[4]{4} = 16 \cdot \sqrt{2} = 16 \sqrt{2} \).

Rewrite \(f(x)\) as \((x - 5)^2\). The minimum of \(f(x)\) is at \(x = 5\), so \(g(x) = f(x - 2)\) shifts this minimum to \(x = 7\).

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.

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.

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

For the system to have no solution, the slopes must be equal. Converting both equations to slope-intercept form reveals that k must equal -14.

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.

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.

Since the triangles are congruent, angle \( Z \) is equal to angle \( C \), so the measure of angle \( C \) is 70°.

Only \( (3, 3) \) is a solution that 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.

To find the surface area, first find the side length using \( V = s^3 \). Given \( V = 27000 \), find \( s = \sqrt[3]{27000} = 30 \). The surface area is given by \( 6s^2 \). Therefore, the surface area is \( 6(30)^2 = 5400 \) square units.

The spill rate is 6 ounces per second. Over 30 seconds (half a minute), the volume spilled is 6 * 30 = 180 ounces.

Solving for \( y \) using substitution or elimination, we find that \( y = 30 \).

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

To find \( a \), set \( g(x) = 0 \) and solve: \( -2x + 10 = 0 Rightarrow x = 5 \), so \( a = 5 \). For the y-intercept \( b \), set \( x = 0 \): \( g(0) = 10 \). Therefore, \( a + b = 5 + 10 = 15 \).

Since \(g(x) = f(x+4) = 7(2)^{x+4} = 7 \cdot 16 \cdot (2)^x = 112(2)^x\), the correct answer is g(x) = 112(2)^x.

In the equation 4𝑛 + 25𝑚 = 1,150, the coefficient 4 corresponds to the bird sanctuary, meaning 𝑛 represents the number of nests per hectare in the bird sanctuary.

To have no real solution, the discriminant of \(x(q x - 64) + 20 = 0\) must be negative. Solving gives \(\)q^2 < 16[/latex], so the smallest integer for [latex]q[/latex] is 4.

To find the number of votes Angel Cruz is expected to win by, first determine the percentage of votes each candidate received. Angel received \( \frac{483}{803} \approx 0.601 \), or 60.1%. Terry received \( \frac{320}{803} \approx 0.399 \), or 39.9%. If 6,424 people vote, Angel would receive \( 0.601 \times 6424 \approx 3864 \) votes and Terry would receive \( 0.399 \times 6424 \approx 2560 \) votes. The difference is \( 3864 - 2560 = 1304 \) votes.

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 percentage can be calculated using the formula: \( \frac{\text{Part}}{\text{Whole}} \times 100 \). Here, \( \frac{50}{200} \times 100 = 25\% \).

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 \( j(0.5) \): \( j(0.5) = 50 \cdot 5^{0.5} = 50 \cdot \sqrt{5} = 50 \sqrt{5} \).