Digital SAT Math Practice Test 3

First, use the given point \((5, -14)\) to solve for the constant \(k\):

\[ -14 = k(5) + 2k \]

\[ -14 = 7k \Rightarrow k = -2 \]

Now that we know \(k = -2\), we can rewrite the full linear equation:

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

\[ y = -2x - 4 \]

Finally, find the value of \(x\) when \(y = -24\) by substituting it into the equation:

\[ -24 = -2x - 4 \]

Add 4 to both sides: \[ -20 = -2x \]

Divide by -2: \[ x = 10 \]

While you could solve for \(x\) and \(y\) individually, the Digital SAT often rewards looking for a direct linear combination. Try adding the two given equations together:

\[ (3x + 2y) + (5x + 4y) = 5 + 9 \]

Combine the like terms: \[ 8x + 6y = 14 \]

Notice the relationship between our new equation and the target expression \(4x + 3y\). Every coefficient in \(8x + 6y = 14\) is exactly double the coefficients in the target expression. Divide the entire equation by 2:

\[ \dfrac{8x + 6y}{2} = \dfrac{14}{2} \]

\[ 4x + 3y = 7 \]

Multiply both sides by the denominator \((Mx - N)\) to clear the fraction, and distribute the negative sign in the numerator:

\[ -Kx + L = P(Mx - N) \]

Expand the right side of the equation:

\[ -Kx + L = PMx - PN \]

To solve for \(x\), group all terms containing \(x\) on one side and all other terms on the opposite side:

\[ L + PN = PMx + Kx \]

Factor out the common variable \(x\) on the right side:

\[ L + PN = x(PM + K) \]

Finally, divide by the expression \((PM + K)\) to isolate \(x\):

\[ x = \dfrac{L + PN}{PM + K} \]

To compare exponents, both sides of the equation must have the same base. Since \(9 = 3^2\), we can rewrite the right side:

\[ 3^m = (3^2)^{1/n} \]

Apply the power of a power rule by multiplying the exponents on the right side:

\[ 3^m = 3^{2/n} \]

Because the bases are identical, the exponents must be equal to each other:

\[ m = \dfrac{2}{n} \]

Multiply both sides by \(n\) to find the product:

\[ m \times n = 2 \]

Let the two roots be \(p\) and \(q\). We can set up a system of equations based on the information provided:

1) Average is 3: \(\dfrac{p + q}{2} = 3 \Rightarrow p + q = 6\)

2) Positive difference is 2: \(p - q = 2\)

Add the two equations together: \(2p = 8 \Rightarrow p = 4\). Substitute \(p = 4\) back into the first equation: \(4 + q = 6 \Rightarrow q = 2\).

If the roots of the quadratic are \(4\) and \(2\), the factors of the equation must be \((x - 4)\) and \((x - 2)\). Expand the factors to find the standard form:

\[ (x - 4)(x - 2) = x^2 - 2x - 4x + 8 \]

The resulting quadratic equation is: \(x^2 - 6x + 8 = 0\).

Recognize that the expression \(a^2 - b^2\) is a difference of squares, which factors as follows:

\[ (a - b)(a + b) = 30 \]

Substitute the given value for \((a - b)\):

\[ 5(a + b) = 30 \]

Divide by 5 to find the value of the sum: \[ a + b = 6 \]

Now use the two linear equations to solve for \(a\):

1) \(a - b = 5\)

2) \(a + b = 6\)

Add the equations together to eliminate \(b\): \[ 2a = 11 \Rightarrow a = 5.5 \]

First, determine the individual reading rates in pages per minute:

John's rate: \(\dfrac{15}{20} = 0.75\) pages per minute.

Linda's rate: \(\dfrac{20}{16} = 1.25\) pages per minute.

When working together, their rates are additive: \[ 0.75 + 1.25 = 2 \text{ pages per minute.} \]

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

\[ \text{Time} = \dfrac{100 \text{ pages}}{2 \text{ pages/min}} = 50 \text{ minutes.} \]

To find the overall average, we must calculate the total sum of all numbers and divide by the total count.

Sum of the first 4 numbers: \(4 \times 21 = 84\)

Sum of the remaining 2 numbers: \(2 \times 27 = 54\)

Total sum of all 6 numbers: \(84 + 54 = 138\)

Overall average: \[ \dfrac{138}{6} = 23 \]

Let \(x\) represent the original price. A 15% discount means the sale price is 85% of the original (since \(100\% - 15\% = 85\%\)).

Convert the percentage to a decimal and set up the equation:

\[ 0.85x = 680 \]

To solve for \(x\), divide by 0.85:

\[ x = \dfrac{680}{0.85} = 800 \]

The original price of the television was $800.

To find the range of possible areas, we first find the minimum possible circle that passes through both points. This occurs when the segment \(AB\) is the diameter of the circle.

Calculate the distance between \(A(8,0)\) and \(B(0,6)\) using the distance formula:

\[ d = \sqrt{(8 - 0)^2 + (0 - 6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \]

If the minimum diameter is 10, the minimum radius is \(r = 5\). The minimum area is:

\[ \text{Area}_{\text{min}} = \pi(5)^2 = 25\pi \]

Since the circle can be infinitely large while still passing through points \(A\) and \(B\), the area \(a\) can be any value greater than or equal to the minimum: \(a \ge 25\pi\).

A regular hexagon has 6 equal sides. If the perimeter is \(6x\), the length of each side is \(x\).

A regular hexagon is composed of 6 congruent equilateral triangles. The area formula for an equilateral triangle with side length \(s\) is \(\dfrac{\sqrt{3}}{4}s^2\). Here, \(s = x\).

Total Area = \(6 \times \left(\dfrac{\sqrt{3}}{4}x^2\right)\)

Simplify the fraction: \[ \text{Area} = \dfrac{6\sqrt{3}}{4}x^2 = \dfrac{3\sqrt{3}}{2}x^2 \]

Let \(l\) be the length and \(w\) be the width. The perimeter is \(P = 2l + 2w\). We are given \(P = 8w\).

Set up the relationship: \[ 2l + 2w = 8w \Rightarrow 2l = 6w \Rightarrow l = 3w \]

The area is \(l \times w = 48\). Substitute \(3w\) for \(l\):

\[ (3w)(w) = 48 \Rightarrow 3w^2 = 48 \]

Divide by 3: \[ w^2 = 16 \Rightarrow w = 4 \]

Since the width is 4 meters, the perimeter is \(8 \times 4 = 32\) meters.