اختر إجابة لكل سؤال لتظهر النتيجة والتفسير فورًا، وتظهر درجتك بعد الانتهاء. الأسئلة من إعداد المدرس، أما الإجابات فحُدّدت بالذكاء الاصطناعي وقد تحتمل الخطأ؛ يمكنك الإبلاغ عبر أيقونة العلم بجانب السؤال.
Which statement describes the graph of g(x)? g(x) = 5 × 3^x - 2
Explanation
For an exponential function g(x) = a × b^x + k, the range is y > k if a > 0. Here k = -2, so the range is y > -2. The y-intercept occurs at x = 0: g(0) = 5 × 30 - 2 = 5(1) - 2 = 3. Thus, the y-intercept is (0, 3).
Question 2
2
Points: 1
Solve the inequality. $4^{x+1} \le (\frac{1}{8})^{x-2}$
Explanation
Rewrite both sides with base 2: $(2^2)^{x+1} \le (2^{-3})^{x-2} \Rightarrow 2(x+1) \le -3(x-2) \Rightarrow 2x + 2 \le -3x + 6$. Adding 3x and subtracting 2 gives $5x \le 4$, so $x \le \frac{4}{5}$.
Question 3
3
Points: 1
Which logarithmic equation is the logarithmic form of the exponential equation? 26 = 64
Explanation
The exponential form b^x = y is equivalent to the logarithmic form $\log_b y = x$. Applying this to 26 = 64 gives $\log_2 64 = 6$.
Use the product rule of logarithms: $\log_2((x+2)(x-2)) = 5 \Rightarrow \log_2(x^2 - 4) = 5$. Converting to exponential form: x2 - 4 = 25 = 32. Thus x2 = 36, so $x = \pm 6$. Since x must be greater than 2 for the logs to be defined, only x = 6 is a valid solution.
Question 5
5
Points: 1
Use the Change of Base Formula to evaluate the expression. Round to the nearest ten-thousandth. $\log_5 18$
Explanation
The change of base formula is $\log_a b = \frac{\log b}{\log a}$. Calculating $\frac{\log 18}{\log 5} \approx 1.7958889$, which rounds to 1.7959.
Question 6
6
Points: 1
Evaluate the expression. 3eln 5 - 2 ln e4
Explanation
Using inverse properties eln x = x and ln e^x = x, the expression simplifies to 3(5) - 2(4) = 15 - 8 = 7.
Question 7
7
Points: 1
Simplify the expression. $\frac{x^2 - 25}{x^2 - 2x - 15}$
Explanation
Factor the numerator and denominator: $\frac{(x - 5)(x + 5)}{(x - 5)(x + 3)}$. Canceling the common factor (x - 5) leaves $\frac{x + 5}{x + 3}$.
Question 8
8
Points: 1
Simplify the complex fraction. $\frac{\frac{1}{x} + \frac{1}{4}}{\frac{1}{x^2} - \frac{1}{16}}$
Explanation
The numerator is $\frac{4 + x}{4x}$. The denominator is $\frac{16 - x^2}{16x^2} = \frac{(4 - x)(4 + x)}{16x^2}$. Dividing numerator by denominator: $\frac{4 + x}{4x} \cdot \frac{16x^2}{(4 - x)(4 + x)} = \frac{16x^2}{4x(4 - x)} = \frac{4x}{4 - x}$.
Question 9
9
Points: 1
What are the equations of the asymptotes of the graph of g(x)? $g(x) = \frac{2}{x - 3} + 4$
Explanation
For a rational function in the form $y = \frac{a}{x - h} + k$, the vertical asymptote is x = h and the horizontal asymptote is y = k. Here h = 3 and k = 4, so the asymptotes are x = 3 and y = 4.
Multiply the entire equation by (x + 3): $2 + (x + 3) = 6 \Rightarrow x + 5 = 6 \Rightarrow x = 1$. Since x = 1 does not make the denominator zero, it is a valid solution.
Question 11
11
Points: 1
A coin is tossed 50 times, and heads is the result 28 times. Find the experimental probability of heads, and compare it with the theoretical probability.
Explanation
The experimental probability is $\frac{28}{50} = \frac{14}{25}$, which is 0.56. The theoretical probability of a coin toss resulting in heads is $\frac{1}{2} = 0.5$. Since 0.56 > 0.5, the experimental probability is greater.
Question 12
12
Points: 1
Describe the shape of the distribution of the data, and choose the appropriate measures of center and spread. 1, 2, 2, 3, 3, 3, 4, 4, 5, 12
Explanation
The outlier 12 pulls the mean (3.9) higher than the median (3), indicating a positive skew. The median is 3. Q1 = 2 and Q3 = 4, so IQR = 4 - 2 = 2. When a distribution is skewed, median and IQR are appropriate measures.
Question 13
13
Points: 1
Find one positive angle and one negative angle that are coterminal with the given angle. $520^\circ$
Explanation
Coterminal angles are found by adding or subtracting multiples of $360^\circ$. $520^\circ - 360^\circ = 160^\circ$. Subtracting another $360^\circ$: $160^\circ - 360^\circ = -200^\circ$.
Question 14
14
Points: 1
In a \right triangle, the legs are 9 and 40. Angle $\theta$ is opposite the leg of length 40. Find $\cos \theta$.
Explanation
First, find the hypotenuse $c = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41$. Since $\theta$ is opposite the leg of length 40, the adjacent leg is 9. Thus, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{9}{41}$.
Question 15
15
Points: 1
Find the exact value of the expression. $\sin(-210^\circ)$
Explanation
$\sin(-210^\circ)$ is coterminal with $-210^\circ + 360^\circ = 150^\circ$. $\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin(30^\circ) = \frac{1}{2}$.
Question 16
16
Points: 1
The water depth at the end of a dock is modeled by h(t), where h is in feet and t is in hours. Find the minimum depth and the time for one full cycle. $h(t) = 6 \sin(\frac{\pi}{6}t) + 10$
Explanation
The minimum value of a sine function y = A sin(Bx) + k is $k - |A|$. Here 10 - 6 = 4 ft. The period (time for one full cycle) is $\frac{2\pi}{B} = \frac{2\pi}{\pi/6} = 12$ h.
Question 17
17
Points: 1
Find the period of the function. $y = 5 \csc \frac{\theta}{3}$
Explanation
The period of the cosecant function $y = a \csc(b\theta)$ is $\frac{2\pi}{|b|}$. Here $b = \frac{1}{3}$, so the period is $\frac{2\pi}{1/3} = 6\pi$.
Question 18
18
Points: 1
Find the value of $\theta$ in degrees. $\theta = \tan^{-1}(-\frac{\sqrt{3}}{3})$
Explanation
The range of the inverse tangent function is $(-90^\circ, 90^\circ)$. Since $\tan(30^\circ) = \frac{\sqrt{3}}{3}$, then $\tan(-30^\circ) = -\frac{\sqrt{3}}{3}$. Thus $\theta = -30^\circ$.
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