Digital SAT Math Practice Test 14: Advanced Module

Identify two distinct points from the line to calculate the slope $m$. Let's use the $y$-intercept $(0, 2000)$ and another clear point $(12000, 8000)$.

$$m = \frac{8000 - 2000}{12000 - 0} = \frac{6000}{12000} = 0.5$$

The slope $m$ is $0.5$ dollars per unit.

The fixed cost $b$ is the $y$-intercept of the graph, which is where $x = 0$. From the graph, this value is $2000$.

Answer: $m = 0.5$ USD/unit, $b = 2000$ USD

Identify two points to find the slope $a$. Using the origin $(0, 0)$ and a clear point on the grid like $(8000, 6000)$:

$$a = \frac{6000 - 0}{8000 - 0} = \frac{6}{8} = 0.75$$

Answer: $a = 0.75$ USD/unit

Set the two equations equal to each other to find the break even quantity $x$:

$$ax = mx + b \implies ax - mx = b \implies x(a - m) = b \implies x = \frac{b}{a - m}$$

Now substitute this quantity $x$ back into the revenue equation $R(x) = ax$ to find the revenue at that point:

$$R = a \cdot \left(\frac{b}{a - m}\right) = \frac{ab}{a - m}$$

Answer: $R = \frac{ab}{a - m}$

First, write the main denominator as a single rational expression by finding a common denominator:

$$\frac{1}{x+2} + 4 = \frac{1}{x+2} + \frac{4(x + 2)}{x + 2} = \frac{1 + 4x + 8}{x+2} = \frac{4x + 9}{x+2}$$

Now substitute this back into the original complex fraction:

$$\frac{\frac{x^2+1}{x+2}}{\frac{4x + 9}{x+2}}$$

To divide fractions, multiply by the reciprocal of the denominator:

$$= \frac{x^2+1}{x+2} \cdot \frac{x+2}{4x+9}$$

Cancel out the common $(x+2)$ terms:

$$= \frac{x^2+1}{4x+9}$$

Answer: B

Set the height equation equal to 25 to find the exact times the ball crosses the 25-meter mark:

$$-5t^2 + 30t = 25$$

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

$$-5t^2 + 30t - 25 = 0$$

Divide the entire equation by $-5$ to simplify:

$$t^2 - 6t + 5 = 0$$

Factor the quadratic:

$$(t - 1)(t - 5) = 0$$

The solutions are $t = 1$ and $t = 5$. Since the parabola opens downward (negative $t^2$ coefficient), the ball is above 25 meters between $t = 1$ and $t = 5$.

Duration = $5 - 1 = 4$ seconds.

Answer: 4 seconds

Calculate the height at $t = 1$:

$$h(1) = -5(1)^2 + 30(1) = -5 + 30 = 25 \text{ meters}$$

Calculate the height at $t = 2$:

$$h(2) = -5(2)^2 + 30(2) = -20 + 60 = 40 \text{ meters}$$

Find the difference:

$$\text{Change} = 40 - 25 = 15 \text{ meters}$$

Answer: The height increases by 15 meters.

We want to find $f(x)$, but we are given $f(t+2)$. Let's do a substitution: let $x = t + 2$.

This means $t = x - 2$. Substitute this $t$ into the right side of the equation:

$$f(x) = g(2(x - 2) - 3) = g(2x - 4 - 3) = g(2x - 7)$$

Now, apply the definition of the function $g(u) = 3u + 3$ using $u = 2x - 7$:

$$f(x) = 3(2x - 7) + 3 = 6x - 21 + 3 = 6x - 18$$

Answer: $f(x) = 6x - 18$

First, write the quadratic equation in standard form ($Ax^2 + Bx + C = 0$):

$$2x^2 + 3mx + 2x - 4 = 0 \implies 2x^2 + (3m + 2)x - 4 = 0$$

According to Vieta's formulas, the sum of the roots of a quadratic equation is $-\frac{B}{A}$.

$$\text{Sum} = -\frac{3m + 2}{2}$$

Set up the inequality as stated in the problem (Sum $> 5$):

$$-\frac{3m + 2}{2} > 5$$

Multiply by 2:

$$-(3m + 2) > 10 \implies -3m - 2 > 10 \implies -3m > 12$$

Divide by $-3$ (remember to flip the inequality sign!):

$$m < -4$$

Answer: $m < -4$

This is a proportional reasoning question. Notice that distance is squared in the denominator. Substitute $4d$ into the formula:

$$F_2 = \frac{G m_1 m_2}{(4d)^2} = \frac{G m_1 m_2}{16d^2} = \frac{1}{16} \cdot \left(\frac{G m_1 m_2}{d^2}\right)$$

The force is $\frac{1}{16}$ of the original force.

$$F_2 = \frac{1}{16} \cdot 2000 = 125 \text{ N}$$

Answer: 125 Newtons

Since the total number of students is 200, we can subtract all the known entries from the total to find the missing value.

Sum of known entries: $20 + 30 + 45 + 15 + 35 = 145$

Boys in chemistry = $Total - 145 = 200 - 145 = 55$

Answer: 55

First, calculate the total number of students in both Geology and Geography (combining boys and girls):

$$\text{Geology} = 20 + 30 = 50$$

$$\text{Geography} = 15 + 35 = 50$$

Total students in these two subjects = $50 + 50 = 100$.

The probability $p$ is the target group divided by the total population (200):

$$p = \frac{100}{200} = 0.5$$

Answer: $0.5$

Substitute the given values into the formula:

$$m = \frac{1}{\sqrt{1 - \frac{(0.9c)^2}{c^2}}} = \frac{1}{\sqrt{1 - \frac{0.81c^2}{c^2}}}$$

The $c^2$ terms cancel out:

$$m = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}}$$

Using a calculator: $m \approx 2.294$ grams.

Answer: $\approx 2.3$ grams

First, find the unit rate of reading for each person (pages per minute).

John's rate = $\frac{15}{20} = \frac{3}{4}$ pages/min.

Linda's rate = $\frac{20}{16} = \frac{5}{4}$ pages/min.

When working together, their rates add up:

Combined rate = $\frac{3}{4} + \frac{5}{4} = \frac{8}{4} = 2$ pages/min.

To find the time for 100 pages, divide the total pages by the combined rate:

$$\text{Time} = \frac{100 \text{ pages}}{2 \text{ pages/min}} = 50 \text{ minutes}$$

Answer: 50 minutes

According to Thales's Theorem, an angle inscribed across a circle's diameter is always a right angle ($90^\circ$). Since $OB$ is a diameter, $\angle OCB = 90^\circ$.

This means Line $L$ and Line $M$ are perpendicular. Perpendicular lines have negative reciprocal slopes.

Slope of $L = \frac{1}{3}$, so the slope of $M = -3$.

We know Line $M$ passes through point $B(4, 0)$. Use point-slope form:

$$y - y_1 = m(x - x_1) \implies y - 0 = -3(x - 4) \implies y = -3x + 12$$

Answer: $y = -3x + 12$

Let $C = (a, b)$. The base $AB$ lies on the horizontal line $y=1$. The length of $AB$ is the distance between $(1,1)$ and $(5,1)$, which is $4$.

Since the triangle is equilateral, sides $AC$ and $BC$ must also equal $4$. Use the distance formula for $AC$ and $BC$ squared:

1) $AC^2: (a-1)^2 + (b-1)^2 = 16$

2) $BC^2: (a-5)^2 + (b-1)^2 = 16$

Subtracting the two equations gives: $(a-1)^2 = (a-5)^2 \implies a = 3$. (This makes sense, as the peak of an equilateral triangle lies halfway between the base's $x$-coordinates).

Substitute $a=3$ into the first equation to find $b$:

$$(3-1)^2 + (b-1)^2 = 16 \implies 4 + (b-1)^2 = 16 \implies (b-1)^2 = 12$$

$$b - 1 = \pm \sqrt{12} = \pm 2\sqrt{3} \implies b = 1 \pm 2\sqrt{3}$$

Since the triangle points upwards into the first quadrant, we take the positive value: $b = 1 + 2\sqrt{3}$.

Answer: $(3, 1 + 2\sqrt{3})$

The radius of the circle is $AO = 5$. Since $AB + BO = 5$, and $AB = \frac{1}{3}BO$, we can set up an equation:

Let $BO = x$. Then $\frac{1}{3}x + x = 5 \implies \frac{4}{3}x = 5 \implies x = \frac{15}{4}$.

Draw a radius from $O$ to $M$. This creates a right triangle $\triangle MBO$ with hypotenuse $OM = 5$ (radius) and leg $OB = \frac{15}{4}$.

Use the Pythagorean theorem to find the other leg $MB$ (which is half the chord):

$$MB^2 + \left(\frac{15}{4}\right)^2 = 5^2 \implies MB^2 + \frac{225}{16} = 25$$

$$MB^2 = \frac{400}{16} - \frac{225}{16} = \frac{175}{16}$$

$$MB = \frac{\sqrt{175}}{4} = \frac{5\sqrt{7}}{4}$$

The full chord $MN$ is twice $MB$:

$$MN = 2 \cdot \frac{5\sqrt{7}}{4} = \frac{5\sqrt{7}}{2}$$

Answer: $\frac{5\sqrt{7}}{2}$ cm

Since $AB$ is tangent to the circle at $B$, and $BC$ is the diameter, $\angle ABC$ is $90^\circ$. By Thales's Theorem, angle $BDC$ is inscribed on the diameter $BC$, so $\angle BDC$ is also $90^\circ$.

Because $\angle BDA$ and $\angle BDC$ are supplementary, $\triangle BDA$ is also a right triangle.

We have similar triangles: $\triangle ABC \sim \triangle BDA$ (they share angle $A$ and a $90^\circ$ angle).

First, find the hypotenuse $AC$ of the large triangle:

$$AC = \sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13}$$

Using the similarity ratio to find altitude $BD$:

$$\frac{AC}{AB} = \frac{BC}{BD} \implies \frac{2\sqrt{13}}{6} = \frac{4}{BD} \implies BD = \frac{24}{2\sqrt{13}} = \frac{12}{\sqrt{13}}$$

Now, use the Pythagorean theorem on the small $\triangle BDA$ to find $AD$:

$$AD = \sqrt{AB^2 - BD^2} = \sqrt{36 - \frac{144}{13}} = \sqrt{\frac{468 - 144}{13}} = \sqrt{\frac{324}{13}} = \frac{18}{\sqrt{13}}$$

Answer: $\frac{18}{\sqrt{13}}$

Look at the top intersection on line $m$. The angle adjacent to the $116^\circ$ angle forms a straight line ($180^\circ$). Let's call it $\alpha$:

$$\alpha = 180^\circ - 116^\circ = 64^\circ$$

Because lines $m$ and $n$ are parallel, alternate interior angles are equal. This means the angle inside the triangle at the bottom intersection is also $64^\circ$ ($\beta = 64^\circ$).

Now, look at the triangle formed at intersection $O$. The angles in a triangle must add up to $180^\circ$:

$$x + 64^\circ + 48^\circ = 180^\circ \implies x + 112^\circ = 180^\circ \implies x = 68^\circ$$

Answer: $68^\circ$

The inscribed angle $\angle CAB$ is $30^\circ$. The central angle subtending the same arc $CB$ is exactly double: $\angle COB = 2 \times 30^\circ = 60^\circ$.

Since $AB$ passes through the center $O$, it is a diameter, meaning it's a straight line ($180^\circ$). The adjacent central angle is $\angle AOC = 180^\circ - 60^\circ = 120^\circ$.

The shaded area is equal to the area of the top semi-circle minus the area of the two triangles ($\triangle AOC$ and $\triangle COB$).

We use the triangle area formula $A = \frac{1}{2}ab\sin(\theta)$, where $a$ and $b$ are the radius ($R=5$):

$$\text{Area } \triangle AOC = \frac{1}{2}(5)(5)\sin(120^\circ) = \frac{25}{2}\left(\frac{\sqrt{3}}{2}\right) = \frac{25\sqrt{3}}{4}$$

$$\text{Area } \triangle COB = \frac{1}{2}(5)(5)\sin(60^\circ) = \frac{25}{2}\left(\frac{\sqrt{3}}{2}\right) = \frac{25\sqrt{3}}{4}$$

Now, calculate the final shaded area:

$$\text{Shaded Area} = \frac{1}{2} \pi R^2 - (\text{Area } \triangle AOC + \text{Area } \triangle COB)$$

$$= \frac{25\pi}{2} - \left(\frac{25\sqrt{3}}{4} + \frac{25\sqrt{3}}{4}\right) = \frac{25\pi}{2} - \frac{25\sqrt{3}}{2} \approx 39.27 - 21.65 \approx 17.6 \text{ cm}^2$$

(Note: Calculations yield approximately $17.6 \text{ cm}^2$)

Answer: $\approx 17.6 \text{ cm}^2$