Accuplacer Advanced Algebra & Functions Practice Test

The correct choice is (D). Since the length of the rectangle is four more than the rectangle’s width, we can write $l$ as $w + 4$.

Substituting in $w + 4$ for $l$ and $12$ for $A$ into the equation $A = l × w$, we get $12 = w(w + 4)$.

Simplifying, we get $12 = w^2 + 4w$.

Subtracting $12$ from both sides, we get $0 = w^2 + 4w - 12$, which can be factored as $(w + 6)(w - 2) = 0$.

Since $w$ must be positive, we equate $w - 2$ to $0$ to get $w = 2$. Since $l = w + 4$, we find that $l = 6$.

The correct choice is (A). We first find the equation of Chandan’s taxi service. First we find the slope, which is equal to:

$\dfrac{7 - 0 \, \text{ dollars}}{2 - 0 \, \text{ kilometers}}$ $= 3.5 \dfrac{\$}{\text{km}}$ $\text{or } \$3.5 \text{ per km}$

Next, we find the equation of Taylor’s taxi service. We find the slope, which is equal to:

$\dfrac{7 - 0 \, \text{ dollars}}{4 - 0 \, \text{ kilometers}}$ $= 1.75 \dfrac{\$}{\text{km}}$ $\text{or } \$1.75 \text{ per km}$

To find how much more Chandan’s taxi service costs than Taylor’s taxi service, we find $3.5 - 1.75 = \$1.75$ per kilometer.

The correct choice is (A). To find $f(x − 3)$, we need to substitute $(x − 3)$ for $x$.

$f(x) = 5x(x + 3)$

$f(x − 3) = 5(x − 3)((x − 3) + 3)$

$f(x − 3) = 5(x − 3)(x − 3 + 3)$

$f(x − 3) = 5(x − 3)(x)$

$f(x − 3) = (5x − 15)(x)$

$f(x − 3) = 5x^2 − 15x$

The correct choice is (C). The equation $(8^x)^{3x} = 24$ can be simplified to $8^{3x^2} = 24$.

We can rewrite this as:

$\require{cancel} \cancel{\log_8}(\cancel{8}^{3x^2}) = \log_8⁡ (24)$

$3x^2 = \log_8⁡ 24$

Solving for $x$, we get: $x = \sqrt{\dfrac{\log_8 24}{3}}$

The correct choice is (B). Since square $ABCD$ has an area of $16 \text{m}^2$, it has a side length of $4\text{m}$. Therefore:

$AD + DC + CB + BE$ $= 4 × 4 = 16$

Next, we find the perimeter of the quarter-circle which is equivalent to:

$\dfrac{2π(4)}{4} = 2π$

Summing up, the perimeter of the entire figure is $2π + 16$.

The correct choice is (B). When factoring quadratics, begin by factoring out any common factor of the coefficients:

$5(x^2 − 3x − 10)$

Then factor the "simplified" quadratic:

$5(x − 5)(x + 2)$

The correct choice is (D). As a general rule, when dealing with rational functions, begin by factoring the numerator and denominator.

$\dfrac{(x + 5)(x − 3)}{(x − 3)(x + 2)}$ $= \require{cancel} \dfrac{(x + 5) \cancel{(x − 3)}}{(\cancel{x − 3})(x + 2)}$ $= \dfrac{x + 5}{x + 2}$

The correct choice is (A). Using the zero product property, we set $(2x + \sqrt{2})$ equal to $0$ and $(5x + \sqrt{19})$ equal to $0$.

Solving $2x + \sqrt{2} = 0$, we get: $x = \dfrac{-\sqrt{2}}{2}$

Solving $5x + \sqrt{19} = 0$, we get: $x = \dfrac{-\sqrt{19}}{5}$

Therefore, the zeroes of the equation are:

$x = \dfrac{-\sqrt{2}}{2}$ $\text{ and }$ $\dfrac{-\sqrt{19}}{5}$

The correct choice is (A). Since the tangent of a triangle can be found by dividing the length of the side opposite to the angle $θ$ by the length of the side adjacent to the angle $θ$. The problem states that $\tan⁡ θ = \frac{8}{15}$, which means that $AB = 8$ and $BC = 15$.

The problem also states that the problem is a right triangle, which means that $AC = \sqrt{8^2 + 15^2} = 17$ via the Pythagorean theorem.

Next, we find:

$\sin⁡ θ =$ $\dfrac{ \text{Length of side opposite to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{8}{17}$

$\cos⁡ θ =$ $\dfrac{\text{Length of side adjacent to } θ}{\text{Length of hypotenuse}}$ $= \dfrac{15}{17}$

Adding $\sin ⁡θ$ to $\cos ⁡θ$, we get: $\dfrac{8}{17} + \dfrac{15}{17} = \dfrac{23}{17}$

The correct choice is (D). First, we know that $∠BCA ≅ ∠DCE$ since they are vertical angles. We are also given that $AC ≅ EC$.

Looking at choice (D), we see that if $BC ≅ DC$, triangles $ABC$ and $EDC$ would be congruent using the Side-Angle-Side postulate.

The correct choice is (C).

The expression $\dfrac{\sqrt{x} + \sqrt{y}}{xy}$ can be separated into: $\dfrac{\sqrt{x}}{xy} + \dfrac{\sqrt{y}}{xy}$

This expression can be rewritten as:

$\dfrac{x^{\frac{1}{2}}}{xy} + \dfrac{y^{\frac{1}{2}}}{xy}$ $= \dfrac{1}{(x^{1 − \frac{1}{2}})(y)}$ $+ \dfrac{1}{(x)(y^{1 − \frac{1}{2}})}$

$= \dfrac{1}{(x^{\frac{1}{2}})(y)}$ $+ \dfrac{1}{(x)(y^{\frac{1}{2}})}$ $= \dfrac{1}{y\sqrt{x}} + \dfrac{1}{x\sqrt{y}}$

An alternate approach:

$\dfrac{\sqrt{x}}{xy} + \dfrac{\sqrt{y}}{xy}$ $= \dfrac{\cancel{\sqrt{x}}}{\cancel{\sqrt{x}} \sqrt{x} y} + \dfrac{\cancel{\sqrt{y}}}{x \cancel{\sqrt{y}}\sqrt{y}}$

$= \dfrac{1}{\sqrt{x} y} + \dfrac{1}{x\sqrt{y}}$

The correct choice is (C). A line parallel to one with the equation of $y = \frac{1}{2}(x + 3)$ has a slope of $\frac{1}{2}$.

Applying the point-slope formula to find the equation of the line that passes through $(3, 16)$, we get $y - 16 = \frac{1}{2}(x - 3)$, which can be simplified to $y = \frac{1}{2}x + \frac{29}{2}$.

The correct choice is (A). We can apply the quadratic formula to $3x^2 + 9x + 12 = 0$.

Plugging in 3, 9, and 12 for $a$,$b$, and $c$ respectively into:

$\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$, we get:

$\dfrac{-9 ± \sqrt{-63}}{6}$ $= \dfrac{-9 ± \sqrt{-1 × 9 × 7}}{6}$

$= \dfrac{-9 ± \sqrt{9 × -1 × 7}}{6}$ $= \dfrac{-9 ± \sqrt{9} \sqrt{-1} \sqrt{7}}{6}$

$= \dfrac{-9 ± 3i \sqrt{7}}{6}$ $= -\dfrac{9}{6} ± \dfrac{3i\sqrt{7}}{6}$

$= -\dfrac{3}{2} ± \dfrac{i\sqrt{7}}{2}$

The correct choice is (B). Based on the graph, we see that initially (t = 0) the bike has a value of \$3,000. After 5 years (t = 5), the bike's value is approximately \$1,000. As time progresses, the bike’s value approaches \$0.

The correct choice is (C). Examining the function above, we know that its degree is odd and it has a negative leading coefficient.

In functions that have odd degrees and negative leading coefficients, as $x$ approaches ∞, $y$ approaches -∞, and as $x$ approaches -∞, $y$ approaches ∞. Thus, the only choice that correctly identifies the degree and the end behavior of the graph is choice C.

The correct choice is (D). To be a function, the graph of the relation must pass the vertical line test. Graph (B) fails the vertical line test, but (A), (C), and (D) pass.

To be a one-to-one function, the graph of the function must pass the horizontal line test. Only graph (D) passes.

The correct choice is (C). First, we find the value of $z(-8)$ which is $4$. Next, we plug $4$ back into the function. We are left to find $z(4)$ which is equal to $2$.

The correct choice is (D). $\sin(x)$ and $\cos(x)$ can never be greater than $1$, regardless of the angle.

$\tan(x)$ will exceed $1$ for acute angles exceeding $45°$, but will be less than $1$ for acute angles less than $45°$.

$\cos(x)$ will be less than $1$ for all acute angles. This means that $\dfrac{1}{\cos(x)}$ is guaranteed to be greater than $1$ for the same angles.

For example, the $\cos(30°) = \dfrac{1}{2}$, therefore $\dfrac{1}{\cos(30°)} = \dfrac{1}{\frac{1}{2}}$ $= 1 \cdot \dfrac{2}{1} = 2$.

The correct choice is (B). The domain represents all values of $x$, the input, for which a value of $y$, the output, exists.

We cannot have a negative under a square root and we cannot divide by $0$.

This means $x + 2 > 0$ and, when cancelling out the $2$ by subtracting it from both sides, $x > −2$.

The correct choice is (A). We are given that the two zeroes of the graph are $(-3, 0)$ and $(5, 0)$, which means that the minimum point of the function has an $x$ value of $\dfrac{-3 + 5}{2} = 1$.

The equation of this parabola is $y = (x + 3)(x - 5)$ from the given roots. This means the $y$-value of the parabola at $x = 1$ is $(1 + 3)(1 - 5) = -16$. Therefore, the correct answer is $(1, -16)$.