Digital SAT Math Practice Test 5

To find the coefficients, we must expand and simplify the expression on the left side of the equation:

Step 1: Distribute the terms

\[ -4x(x) - 4x(5) - 3(4x) - 3(2) \]

\[ -4x^2 - 20x - 12x - 6 \]

Step 2: Combine like terms

The \(x\) terms are \(-20x\) and \(-12x\). Adding them together gives:

\[ -4x^2 - 32x - 6 \]

Step 3: Match the coefficients

The standard quadratic form is \(ax^2 + bx + c\). By comparing this to our simplified expression, we see that \(a = -4\), \(b = -32\), and \(c = -6\). Thus, the value of \(b\) is -32.

Step 1: Find the slope of the line

Using points \(B(2,4)\) and \(C(-1,0)\), the slope (\(m\)) is:

\[ m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{4 - 0}{2 - (-1)} = \dfrac{4}{3} \]

Step 2: Find the equation of the line

Using the point-slope form with point \(C(-1,0)\):

\[ y - 0 = \dfrac{4}{3}(x - (-1)) \Rightarrow y = \dfrac{4}{3}(x + 1) \]

Step 3: Solve for \(b\)

Substitute the coordinates of point \(A(-2, b)\) into the equation:

\[ b = \dfrac{4}{3}(-2 + 1) = \dfrac{4}{3}(-1) \]

The value of \(b\) is \(-\dfrac{4}{3}\).

A linear system has no solution if the lines are parallel. Parallel lines have equal slopes but different y-intercepts.

Equation 1: \(2x + 5y = 3 \Rightarrow y = -\dfrac{2}{5}x + \dfrac{3}{5}\). Slope = \(-\dfrac{2}{5}\).

Equation 2: \(-5x + ky = 14 \Rightarrow y = \dfrac{5}{k}x + \dfrac{14}{k}\). Slope = \(\dfrac{5}{k}\).

Set the slopes equal to each other:

\[ -\dfrac{2}{5} = \dfrac{5}{k} \Rightarrow -2k = 25 \Rightarrow k = -12.5 \]

Verify that the y-intercepts are different: \(\dfrac{14}{-12.5} \neq \dfrac{3}{5}\). Since they differ, the system has no solution when \(k = -12.5\).

Clear the fractions and decimals first:

Equation 1 (multiply by 3): \(x^2 - y^2 = 21\)

Equation 2 (multiply by 100): \(x + y = 5\)

Recognize that \(x^2 - y^2\) is a difference of squares: \((x - y)(x + y) = 21\).

Substitute \(x + y = 5\) into the equation:

\[ (x - y)(5) = 21 \Rightarrow x - y = \dfrac{21}{5} = 4.2 \]

Using the factor theorem, the polynomial can be written as:

\[ P(x) = 1(x - 1)(x - 3)(x - (-1)) = (x - 1)(x - 3)(x + 1) \]

Substitute \(x = 0\) to find the y-intercept \(P(0)\):

\[ P(0) = (0 - 1)(0 - 3)(0 + 1) = (-1)(-3)(1) = 3 \]

Solve the linear equation for \(y\): \(y = x + 2\). Substitute this into the first equation:

\[ (x+2)^2 + 2(x+2) + (x+2) = -2 \]

Let \(u = x + 2\). The equation becomes \(u^2 + 3u + 2 = 0\), which factors as \((u + 2)(u + 1) = 0\).

This gives \(u = -2\) and \(u = -1\). Substitute back to solve for \(x\):

• \(x + 2 = -2 \Rightarrow x = -4\)
• \(x + 2 = -1 \Rightarrow x = -3\)

The sum of the possible values is \(-4 + (-3) = -7\).

If \(\dfrac{3}{7}\) are girls, the remainder must be boys: \(1 - \dfrac{3}{7} = \dfrac{4}{7}\).

The ratio of boys to girls is calculated as:

\[ \text{Ratio} = \dfrac{4/7}{3/7} = \dfrac{4}{3} \]

The ratio is 4:3.

Let the 2003 profit be 100. A 10% increase makes the 2004 profit 110.

In 2005, the profit drops back to 100. The amount of decrease is 10.

Percent decrease = \(\dfrac{\text{Decrease}}{\text{Start Value}} \times 100 = \dfrac{10}{110} \times 100 \approx 9.1\%\).

First hour loss: \(60 \times 0.25 = 15 \text{ L}\). Tank has \(185 \text{ L}\) left after 1 hour.

Time after 1 hour is \((t - 1)\) hours. In minutes, this is \(60(t - 1)\).

Second phase loss: \(0.4 \times 60(t - 1) = 24(t - 1) = 24t - 24\).

Remaining water = \(185 - (24t - 24) = 209 - 24t\).

Circumference \(8\pi\) means the diameter is 8 and the radius is 4. For an inscribed triangle with the diameter as its hypotenuse, the maximum area occurs when the height is equal to the radius.

\[ \text{Area} = \dfrac{1}{2}bh = \dfrac{1}{2}(8)(4) = 16 \]

Angle \(ACB\) is inscribed in a semicircle, so it is \(90^\circ\). In right triangle \(ABC\), we have hypotenuse 10 and opposite side 5 for angle \(CBA\).

\[ \sin(CBA) = \dfrac{5}{10} = 0.5 \Rightarrow \text{angle } CBA = 30^\circ \]

The distance between \(y=8\) and \(y=2\) is the diameter (6), so the radius is 3.

The y-coordinate is the average of 8 and 2: \(y = 5\).

Since it is in Quadrant II and tangent to \(x = -2\), the center is 3 units to the left: \(x = -2 - 3 = -5\).

Center = (-5, 5).