CLEP Precalculus Practice Test

Explanation

To evaluate the one-sided limit lim x→0+ (x / |x|), we consider x approaching 0 from the right, meaning x is positive. When x is positive, |x| equals x. So, the expression simplifies to x/x, which equals 1 for all x ≠ 0. Therefore, as x approaches 0 from the right, the limit is 1.

Explanation

Using the law of cosines, c^2 = a^2 + b^2 - 2ab * cos(C), where c = 5, a = 3, and b = 4. Substituting these values in gives 5^2 = 3^2 + 4^2 - 2*3*4 * cos(C). This simplifies to 25 = 9 + 16 - 24 * cos(C), which further simplifies to 25 = 25 - 24 * cos(C). Then, 24 * cos(C) = 0, so cos(C) = 0. The angle whose cosine is 0 is 90 degrees.

Explanation

To find the limit of the difference quotient, first calculate f(2) and f(2 + h). f(2) = 2*2 + 1 = 5. f(2 + h) = 2*(2 + h) + 1 = 4 + 2h + 1 = 5 + 2h. Then, the difference quotient is [f(2 + h) - f(2)]/h = [(5 + 2h) - 5]/h = 2h/h = 2. As h approaches 0, the limit of the difference quotient is simply 2.

Explanation

To find the value of t when x = 7, substitute x = 7 into the equation x = 2t + 1. This gives 7 = 2t + 1. Subtracting 1 from both sides yields 6 = 2t. Dividing both sides by 2 gives t = 3.

Explanation

To find the limit of f(x) as x approaches 2, first factor the numerator: f(x) = ((x + 2)(x - 2)) / (x - 2). Cancel (x - 2) from the numerator and denominator to get f(x) = x + 2. Now, substitute x = 2 into the simplified expression: f(2) = 2 + 2 = 4. Thus, the limit of f(x) as x approaches 2 is 4.

Explanation

To convert from polar to rectangular form, we use the relationships $x = rcos( heta)$ and $y = rsin( heta)$. Given $r = 5$, we can express $r$ in terms of $x$ and $y$ as $r^2 = x^2 + y^2$. Substituting $r = 5$ gives us $5^2 = x^2 + y^2$, which simplifies to $25 = x^2 + y^2$. Thus, the correct rectangular form of the equation $r = 5$ is $x^2 + y^2 = 25$.

Explanation

To solve the equation sin(x) = 1/2, recall the unit circle definition of sine. The sine of an angle in a right-angled triangle is the ratio of the length of the side opposite the angle to the length of the hypotenuse. From the unit circle, we find that sin(π/6) = 1/2 and sin(5π/6) = 1/2 in the interval [0, 2π). These are the angles whose sine is 1/2.

Explanation

To find the limit as x approaches infinity, we need to look at the leading terms of the numerator and denominator. The leading term in the numerator is 2x^2 and in the denominator is x^2. Dividing both by x^2, we get (2 + 3/x) / (1 + 1/x^2). As x approaches infinity, the terms 3/x and 1/x^2 approach 0. So, the limit approaches 2 / 1 = 2.

Explanation

To find the limit of the difference quotient, first calculate f(2 + h) and f(2). f(2) = 3(2)^2 = 12. f(2 + h) = 3(2 + h)^2 = 3(4 + 4h + h^2) = 12 + 12h + 3h^2. Then, the difference quotient becomes [f(2 + h) - f(2)]/h = [12 + 12h + 3h^2 - 12]/h = (12h + 3h^2)/h. Factoring out h, we get h(12 + 3h)/h. As h approaches 0, the limit of the difference quotient is 12.

Explanation

To find the coordinates when t = 2, substitute t into both equations. For x: x = 2(2) + 1 = 4 + 1 = 5. For y: y = 2 - 3 = -1. Thus, the coordinates are (5, -1).