ACT ACT Math Practice Test 2025

To find the total sugar needed, multiply $\dfrac{3}{4}$ by $2\dfrac{1}{2}$.

First, convert the mixed number to an improper fraction: $2\dfrac{1}{2} = \dfrac{5}{2}$.

Then multiply: $\dfrac{3}{4} \times \dfrac{5}{2} = \dfrac{15}{8}$.

Convert to a mixed number: $\dfrac{15}{8} = 1\dfrac{7}{8}$.

Therefore, Marcus needs $1\dfrac{7}{8}$ cups of sugar.

Substitute $x = -3$ and $y = 4$ into the expression:

$2x^2 - 3xy + y$ $= 2(-3)^2 - 3(-3)(4) + 4$ $= 2(9) - 3(-12) + 4$ $= 18 + 36 + 4$ $= 58$

Step by step:

• $2(-3)^2 = 2 \times 9 = 18$
• $-3(-3)(4) = -3 \times (-12) = 36$
• $y = 4$
• Sum: $18 + 36 + 4 = 58$

Using the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a = 2$, $b = 5$, $c = -3$:

$x = \dfrac{-5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \dfrac{-5 \pm \sqrt{25 + 24}}{4} = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}$

So $x = \dfrac{-5 + 7}{4} = \dfrac{2}{4} = \dfrac{1}{2}$ or $x = \dfrac{-5 - 7}{4} = \dfrac{-12}{4} = -3$.

The solutions are $x = \dfrac{1}{2}$ and $x = -3$.

Verification: $2(\tfrac{1}{2})^2 + 5(\tfrac{1}{2}) - 3 = \tfrac{1}{2} + \tfrac{5}{2} - 3 = 3 - 3 = 0$. Also $2(-3)^2 + 5(-3) - 3 = 18 - 15 - 3 = 0$.

The slope formula is $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.

Substituting the given points $(-2, 7)$ and $(4, -5)$:

$m = \dfrac{-5 - 7}{4 - (-2)} = \dfrac{-12}{6} = -2$

The slope of the line is $-2$. This means for every 1 unit increase in $x$, $y$ decreases by 2 units.

The area of a triangle is given by the formula:

$A = \dfrac{1}{2} \times \text{base} \times \text{height}$

$A = \dfrac{1}{2} \times 14 \times 9 = \dfrac{126}{2} = 63$

The area of the triangle is $63$ square centimeters.

In a right triangle, $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{5}{13}$.

To find $\cos\theta$, we first need the adjacent side. Using the Pythagorean theorem:

$\text{adjacent} = \sqrt{\text{hypotenuse}^2 - \text{opposite}^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$

Therefore, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{12}{13}$.

Note: This is a 5-12-13 right triangle, a well-known Pythagorean triple.

To find the LCM, first find the prime factorizations:

$12 = 2^2 \times 3$ $18 = 2 \times 3^2$

The LCM is the product of the highest powers of all prime factors:

$\text{LCM} = 2^2 \times 3^2 = 4 \times 9 = 36$

Verification:

• $36 \div 12 = 3$ (whole number)
• $36 \div 18 = 2$ (whole number)
• No smaller positive integer is divisible by both 12 and 18.

Substitute $x = -2$ into the function:

$f(-2) = 3(-2)^2 - 2(-2) + 7 = 3(4) - 2(-2) + 7 = 12 + 4 + 7 = 23$

Step by step:

• $3(-2)^2 = 3 \times 4 = 12$ (the squaring happens before the multiplication by 3)
• $-2(-2) = 4$ (negative times negative is positive)
• Constant term: $7$
• Sum: $12 + 4 + 7 = 23$

Solve the inequality step by step:

$-3(2x - 4) > 18$

Distribute $-3$: $-6x + 12 > 18$

Subtract 12 from both sides: $-6x > 6$

Divide both sides by $-6$ (remember to flip the inequality sign when dividing by a negative number): $x < -1$

Verification: Test $x = -2$ (which should satisfy $x < -1$): $-3(2(-2) - 4) = -3(-4 - 4) = -3(-8) = 24 > 18$. Correct.

Test $x = 0$ (which should NOT satisfy $x < -1$): $-3(2(0) - 4) = -3(-4) = 12$, which is not greater than 18. Correct.

When two objects travel toward each other, their combined rate is the sum of their individual rates.

Combined rate $= 15 + 10 = 25$ miles per hour.

Using the formula $\text{time} = \dfrac{\text{distance}}{\text{rate}}$:

$t = \dfrac{75}{25} = 3$ hours.

Verification:

• In 3 hours, cyclist 1 travels $15 \times 3 = 45$ miles.
• In 3 hours, cyclist 2 travels $10 \times 3 = 30$ miles.
• Total distance covered: $45 + 30 = 75$ miles (equals the distance between the towns).