Digital SAT Math Practice Test 13: The Final Advanced Module

Step 1: Find the slope from the equation

The slope $m$ can be calculated by rewriting the equation in the form $y = mx + k$.

$$2y + 3x = 4 \implies 2y = -3x + 4 \implies y = -\frac{3}{2}x + 2$$

The slope $m$ is $-\frac{3}{2}$.

Step 2: Use the slope formula

Use the coordinates of the two points to express the slope $m$:

$$m = \frac{d - b}{c - a}$$

Equate this to the slope found above:

$$-\frac{3}{2} = \frac{d - b}{c - a}$$

Step 3: Solve for the target expression

We are given $a - c = 3$, which means $c - a = -3$. Substitute this into the equation:

$$d - b = -\frac{3}{2}(c - a) = -\frac{3}{2}(-3) = \frac{9}{2}$$

Answer: $\frac{9}{2}$

Find the midpoint (center) of the range and the distance from the center to the boundaries.

Midpoint: $\frac{1}{2}\left(\frac{3}{2} + \frac{5}{2}\right) = \frac{8}{4} = 2$.

Distance from center to boundary: $\frac{5}{2} - 2 = 0.5$.

An absolute value inequality $|x - \text{center}| \le \text{distance}$ describes this range.

Answer: $|x - 2| \le 0.5$

Notice the proportional relationship between the target expression and the given equation. Multiply the given equation by $0.05$:

$$0.05(5x + 3y) = 0.05(2)$$

$$0.25x + 0.15y = 0.1$$

Answer: $0.1$

Let $x$ be the number to find. Write the algebraic equation:

$$x = \frac{1}{4} \sqrt[3]{x}$$

Multiply all terms by 4 and cube both sides:

$$4x = \sqrt[3]{x} \implies (4x)^3 = (\sqrt[3]{x})^3$$

$$64x^3 = x$$

Rewrite with zero on the right, factor, and solve:

$$64x^3 - x = 0 \implies x(64x^2 - 1) = 0$$

Solutions: $x = 0$ and $64x^2 - 1 = 0 \implies x^2 = \frac{1}{64} \implies x = \pm \frac{1}{8}$.

Since we are looking for a positive number, we take the positive root.

Answer: $x = \frac{1}{8}$

Multiply all terms of the first equation by 3 to obtain:

$$x^2 - y^2 = 21$$

Multiply all terms of the second equation by 100 to obtain:

$$x + y = 5$$

Factor the left side of the squared equation as a difference of squares:

$$(x + y)(x - y) = 21$$

Substitute $(x + y) = 5$ into the equation and solve for $x - y$:

$$5(x - y) = 21 \implies x - y = \frac{21}{5}$$

Answer: $\frac{21}{5}$

According to the Polynomial Remainder Theorem, the remainder of the division of $f(x)$ by $(x - c)$ is $f(c)$.

$$f(1) = 1^2 + a(1) + b = -2 \implies a + b = -3$$

$$f(-2) = (-2)^2 + a(-2) + b = -5 \implies 4 - 2a + b = -5 \implies -2a + b = -9$$

Subtract the second equation from the first to eliminate $b$:

$$(a - (-2a)) = -3 - (-9) \implies 3a = 6 \implies a = 2$$

Substitute $a = 2$ back into $a + b = -3$ to find $b$:

$$2 + b = -3 \implies b = -5$$

Answer: $a = 2$ and $b = -5$

Expand the left side and group like terms:

$$-4x^2 - 20x - 12x - 6 = ax^2 + bx + c$$

$$-4x^2 - 32x - 6 = ax^2 + bx + c$$

Equate the corresponding coefficients.

Answer: $a = -4, b = -32, c = -6$

Set $y = -\frac{1}{4}$ and solve for $x$:

$$-\frac{1}{4} = \frac{2x - 1}{x + 3}$$

Cross multiply:

$$-(x + 3) = 4(2x - 1) \implies -x - 3 = 8x - 4 \implies -9x = -1$$

Answer: $x = \frac{1}{9}$

Square both sides of the equation to eliminate the radical and absolute value:

$$(|x - 3|)^2 = (\sqrt{x + 17})^2 \implies (x - 3)^2 = x + 17$$

$$x^2 - 6x + 9 = x + 17 \implies x^2 - 7x - 8 = 0$$

Factor the quadratic equation:

$$(x + 1)(x - 8) = 0$$

Answer: $\{-1, 8\}$

Rewrite the equation as $x^2 - 7|x| + 10 = 0$. Since $x^2 = |x|^2$, we can let $u = |x|$:

$$u^2 - 7u + 10 = 0 \implies (u - 5)(u - 2) = 0$$

This gives $u = 5$ and $u = 2$. Re-substitute $|x|$ for $u$:

$$|x| = 5 \implies x = \pm 5$$

$$|x| = 2 \implies x = \pm 2$$

Answer: $\{-5, -2, 2, 5\}$

Rewrite the last two terms as $-(x - 2)$ to factor out the common binomial:

$$(x - 2)(x^2 - 7x + 13) - (x - 2) = 0$$

$$(x - 2)(x^2 - 7x + 13 - 1) = 0 \implies (x - 2)(x^2 - 7x + 12) = 0$$

Factor the resulting quadratic expression:

$$(x - 2)(x - 3)(x - 4) = 0$$

Answer: $\{2, 3, 4\}$

Isolate the absolute value expression:

$$-2|x - 4| = k + 3 \implies |x - 4| = \frac{k + 3}{-2}$$

For an absolute value equation to have two solutions, the right-hand side must be strictly positive ($> 0$).

$$\frac{k + 3}{-2} > 0 \implies k + 3 < 0 \implies k < -3$$

Answer: $k < -3$

From the linear equation $y - 2 = x$, we can isolate $y = x + 2$.

Let $u = x + 2$. This also means $y = u$. Substitute into the first equation:

$$u^2 + 2u + u = -2 \implies u^2 + 3u + 2 = 0$$

Factor the quadratic:

$$(u + 1)(u + 2) = 0 \implies u = -1 \text{ or } u = -2$$

Substitute back $x = u - 2$:

$$x = -1 - 2 = -3 \text{ or } x = -2 - 2 = -4$$

Answer: $-3$ and $-4$

Multiply the entire equation by the Least Common Multiple (LCM), which is $3(w+2)(w+3)$:

$$6(w + 3) = 12(w + 2) - (w + 2)(w + 3)$$

$$6w + 18 = 12w + 24 - (w^2 + 5w + 6)$$

Move all terms to one side to form a quadratic equation:

$$w^2 - w = 0 \implies w(w - 1) = 0$$

Answer: $\{0, 1\}$

Set the two equations equal to each other to find their intersection:

$$x^2 + Ax + B = -x^2 + Mx + N \implies 2x^2 + (A - M)x + (B - N) = 0$$

For the parabolas to be tangent, the resulting quadratic must have exactly one solution. Therefore, the discriminant $D$ must equal 0:

$$(A - M)^2 - 4(2)(B - N) = 0$$

Substitute $A - M = 2$ into the equation:

$$(2)^2 - 8(B - N) = 0 \implies 4 = 8(B - N)$$

Answer: $B - N = \frac{1}{2}$

Multiply both sides by the denominator $(Mx - N)$:

$$-Kx + L = P(Mx - N) \implies -Kx + L = PMx - PN$$

Group all $x$ terms on one side and constants on the other:

$$L + PN = PMx + Kx \implies L + PN = x(PM + K)$$

Isolate $x$:

$$x = \frac{L + PN}{PM + K}$$

Answer: $x = \frac{L + PN}{PM + K}$

Translate to an algebraic equation: $\sqrt{x} + 2x = 10$.

Let $u = \sqrt{x}$, which means $u^2 = x$. Substitute into the equation:

$$u + 2u^2 = 10 \implies 2u^2 + u - 10 = 0$$

Factor the quadratic equation:

$$(2u + 5)(u - 2) = 0 \implies u = 2 \text{ or } u = -2.5$$

Since a principal square root $u = \sqrt{x}$ must be non-negative, $u = 2$. Therefore, $x = 2^2 = 4$.

Answer: $4$

First, isolate the squared binomials:

$$(x + y)^2 = \frac{4}{0.2} = 20$$

$$(x - y)^2 = \frac{3}{0.5} = 6$$

Subtract the second equation from the first:

$$(x + y)^2 - (x - y)^2 = 20 - 6$$

$$(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = 14$$

$$4xy = 14 \implies xy = \frac{14}{4} = \frac{7}{2}$$

Answer: $\frac{7}{2}$

Calculate the water leaked in the first hour ($60$ minutes):

$$60 \times 0.25 = 15 \text{ L}$$

Calculate the water leaked after the first hour. The remaining time in minutes is $60(t - 1)$.

$$\text{Second leak amount} = 60(t - 1) \times 0.4 = 24(t - 1) \text{ L}$$

Subtract the total leaked amount from the initial 200 L:

$$q = 200 - 15 - 24(t - 1) \implies q = 185 - 24t + 24 \implies q = 209 - 24t$$

Answer: $q = 209 - 24t$

First, convert her writing time to minutes to establish a single unit rate: $2 \times 60 + 35 = 155$ minutes.

Her rate is $\frac{11}{155}$ pages per minute.

Calculate total time needed for 30 pages:

$$\text{Time} = 30 \div \frac{11}{155} = 30 \times \frac{155}{11} \approx 422.73 \text{ minutes}$$

Convert back to hours:

$$\frac{422.73}{60} \approx 7.05 \text{ hours}$$

Answer: $\approx 7$ hours

Calculate the price of the car exactly 2 years after purchase (which is our $t = 0$ point):

$$50000 - 2 \times 4000 = 42000$$

From this point onward, the car depreciates at $\$6000$ per year. Set up the linear equation:

$$P(t) = 42000 - 6000t$$

Answer: $P(t) = 42000 - 6000t$

Rewrite the equation in slope-intercept form ($y = mx + b$):

$$3y = 3x + 3 \implies y = x + 1$$

The line has a slope of $1$ and a y-intercept of $1$. Looking at the provided graph options, this corresponds to line L.

Answer: line L

Solve for $y$ to convert the equation into standard vertex form:

$$2y = 2(x - 2)^2 + 2 \implies y = (x - 2)^2 + 1$$

This equation represents a standard upward-facing parabola shifted $2$ units to the right and $1$ unit up. This corresponds to graph M.

Answer: Parabola M

Find the slopes of both lines by isolating $y$:

Line 1 ($Ax + By = C$): Slope $m_1 = -\frac{A}{B}$

Line 2 ($Mx + Ny = P$): Slope $m_2 = -\frac{M}{N}$

We are given $\frac{M}{N} = 5$, so $m_2 = -5$.

Perpendicular lines have slopes that multiply to $-1$ ($m_1 m_2 = -1$):

$$\left(-\frac{A}{B}\right)(-5) = -1 \implies 5\frac{A}{B} = -1 \implies \frac{A}{B} = -\frac{1}{5}$$

Answer: $-\frac{1}{5}$

First, find the x-intercept of the line $y = x - 2$ by setting $y = 0$:

$$0 = x - 2 \implies x = 2$$

Since the parabola and line intersect at this exact point $(2, 0)$ on the x-axis, the point must also satisfy the parabola's equation.

Substitute $(2, 0)$ into $f(x) = -x^2 + a$:

$$0 = -(2)^2 + a \implies 0 = -4 + a \implies a = 4$$

Answer: $a = 4$