▶️ Answer/Explanation
Angles in standard position that share a terminal ray are called coterminal angles.
For two angles to be coterminal, their difference must be an integer multiple of a full rotation, \(2\pi\).
Let’s check the difference for option (C): \(\beta – \alpha = \frac{8\pi}{3} – \frac{2\pi}{3}\).
Subtracting the numerators gives \(\frac{6\pi}{3}\).
Simplifying the fraction results in \(2\pi\).
Since the difference is exactly \(2\pi\), the angles end at the same position.
Therefore, \(\alpha\) and \(\beta\) share a terminal ray.
▶️ Answer/Explanation
The correct option is (D).
1. First, identify the base \(b\) using a clear point on the graph of \(k(x)\). The graph passes through the point \((1, 3)\).
2. Substitute these coordinates into the exponential equation \(y = b^x\): \(3 = b^1\), which implies \(b = 3\).
3. The logarithmic function \(h(x) = \log_b x\) is the inverse of the exponential function \(k(x) = b^x\).
4. Since they are inverse functions, if a point \((x, y)\) lies on the graph of \(k\), then the point \((y, x)\) must lie on the graph of \(h\).
5. From the graph of \(k\), we can identify the points \((1, 3)\) and \((2, 9)\) because \(3^1 = 3\) and \(3^2 = 9\).
6. Swapping the coordinates for the inverse function \(h\), the corresponding points are \((3, 1)\) and \((9, 2)\).
7. Checking option (D): \(h(3) = \log_3(3) = 1\) and \(h(9) = \log_3(9) = 2\). These matches the points derived.
▶️ Answer/Explanation
The relationship \( g(x) = f^{-1}(x) \) implies that \( f(g(x)) = x \).
Substituting the given condition \( g(x) = -3x \) into this relationship, we get \( f(-3x) = x \).
Using the definition of \( f \), the equation becomes \( 4 \cdot 3^{(-3x – 2)} + 1 = x \).
Since \( 4 \cdot 3^{(-3x – 2)} \) is always positive, \( x \) must be greater than \( 1 \), eliminating option (A).
Testing option (B) with \( x = 1.016 \): \( 4 \cdot 3^{(-3(1.016) – 2)} + 1 = 4 \cdot 3^{-5.048} + 1 \).
Calculating the value: \( 4 \cdot (0.0039) + 1 \approx 1.0156 \), which is approximately \( 1.016 \).
Thus, the correct value is (B) 1.016.
▶️ Answer/Explanation
The correct answer is (A). To solve the equation without technology, we must rewrite the functions using a common base. Both 4 and 8 are powers of 2.
1. Identify the common base: Since \( 4 = 2^2 \) and \( 8 = 2^3 \), we use base 2.
2. Rewrite base 4 using logarithms: \( 4 = 2^{\log_2 4} \).
3. Substitute into \( f(x) \) using the power rule \( (a^b)^c = a^{b \cdot c} \): \( f(x) = (2^{\log_2 4})^{(5x-1)} = 2^{(\log_2 4 \cdot (5x-1))} \).
4. Rewrite base 8 using logarithms: \( 8 = 2^{\log_2 8} \).
5. Substitute into \( g(x) \): \( g(x) = (2^{\log_2 8})^{(x/4)} = 2^{(\log_2 8 \cdot (x/4))} \).
6. This effectively sets up the equation \( 2^{2(5x-1)} = 2^{3(x/4)} \) for easy solving.
▶️ Answer/Explanation
Set $y = 4 \cdot 2^{(x-3)}$ to represent the original function.
To find the inverse, swap $x$ and $y$: $x = 4 \cdot 2^{(y-3)}$.
Divide both sides by $4$: $\frac{x}{4} = 2^{(y-3)}$.
Convert the exponential equation to logarithmic form: $\log_{2}\left(\frac{x}{4}\right) = y – 3$.
Solve for $y$ by adding $3$ to both sides: $y = \log_{2}\left(\frac{x}{4}\right) + 3$.
Therefore, $g(x) = \log_{2}\left(\frac{x}{4}\right) + 3$.
The correct option is (C).
▶️ Answer/Explanation
Set the function equal to the target value: $8 \cdot 2^{x} = 256$.
Divide both sides by $8$ to isolate the exponential term: $2^{x} = \frac{256}{8}$.
Simplify the division: $2^{x} = 32$.
Express $32$ as a power of $2$: $32 = 2^{5}$.
Equate the exponents since the bases are the same: $x = 5$.
Therefore, the correct option is (B).
▶️ Answer/Explanation
Set up the equation $S(4) = 300,000$: $\frac{500,000}{1 + 0.4e^{4k}} = 300,000$.
Simplify to find $e^{4k}$: $1 + 0.4e^{4k} = \frac{500,000}{300,000} = \frac{5}{3}$.
Solve for the exponential term: $0.4e^{4k} = \frac{5}{3} – 1 = \frac{2}{3}$, so $e^{4k} = \frac{2/3}{0.4} = \frac{5}{3}$.
Express the target value $S(12)$ using $e^{4k}$: $S(12) = \frac{500,000}{1 + 0.4(e^{4k})^3}$.
Substitute $e^{4k} = \frac{5}{3}$ into the expression: $S(12) = \frac{500,000}{1 + 0.4(\frac{5}{3})^3}$.
Calculate the denominator: $1 + 0.4(\frac{125}{27}) = 1 + \frac{50}{27} = \frac{77}{27}$.
Final calculation: $S(12) = \frac{500,000 \times 27}{77} \approx 175,324.67$.
The correct option is (A).
▶️ Answer/Explanation
To find the intersection, set the functions equal: $g(x) = h(x)$.
Substitute the expressions: $4^{x} = 16^{x+2}$.
Rewrite $16$ as a power of $4$: $4^{x} = (4^{2})^{x+2}$.
Apply the power of a power rule: $4^{x} = 4^{2(x+2)}$.
Set the exponents equal to each other: $x = 2(x+2)$.
Distribute the $2$: $x = 2x + 4$.
Subtract $2x$ from both sides: $-x = 4$.
Solve for $x$: $x = -4$.
The correct option is (A).
▶️ Answer/Explanation
The correct option is (B).
In a semi-log plot with a natural log scale, we transform the output as $y = \ln(k(t))$.
Substituting the given function: $y = \ln(7 \cdot 5^t)$.
Using log properties, this expands to $y = \ln 7 + \ln(5^t)$.
Applying the power rule: $y = \ln 7 + t \cdot \ln 5$.
This matches the linear form $y = mx + b$, where the variable is $t$.
The slope $m$ of this linear relationship is $\ln 5$.
▶️ Answer/Explanation
The composition is defined as $h(t) = f(g(t))$.
Substitute $g(t) = 7 \ln t$ into $f(t)$, resulting in $h(t) = e^{7 \ln t}$.
Apply the power rule for logarithms: $7 \ln t = \ln(t^{7})$.
The expression becomes $h(t) = e^{\ln(t^{7})}$.
Using the identity $e^{\ln x} = x$, the expression simplifies to $t^{7}$.
Therefore, the correct expression for $h(t)$ is $t^{7}$.
The correct option is (B).
▶️ Answer/Explanation
For function $f$, the mapping is $\text{hours} \rightarrow \text{miles per hour}$.
For function $g$, the mapping is $\text{miles per hour} \rightarrow \text{dollars}$.
The inverse $g^{-1}$ reverses its mapping to $\text{dollars} \rightarrow \text{miles per hour}$.
The inverse $f^{-1}$ reverses its mapping to $\text{miles per hour} \rightarrow \text{hours}$.
To start with dollars, $g^{-1}(x)$ must be the inner function.
The output of $g^{-1}$ matches the required input for $f^{-1}$.
Thus, $y = f^{-1}(g^{-1}(x))$ maps dollars to hours.
The correct option is (C).
▶️ Answer/Explanation
Set up the equation where the logarithmic model exceeds the linear model by $0.1$: $k(t) = m(t) + 0.1$.
Substitute the given functions: $14 – 2.885 \ln t = (-t + 14) + 0.1$.
Simplify the equation to: $-2.885 \ln t = -t + 0.1$.
Rearrange to form the function: $f(t) = t – 2.885 \ln t – 0.1 = 0$.
Use a graphing calculator to find the intersection of $y = 14 – 2.885 \ln t$ and $y = -t + 14.1$.
The first intersection point for $t \geq 2$ occurs at $t \approx 2.289$.
Thus, the correct option is (D).
▶️ Answer/Explanation
The given property $g(2x) = g(x) + 1$ defines a logarithmic function base $2$.
In graph (B), when the input $x$ is $1$, the output $y$ is $0$.
Doubling the input to $x = 2$ results in an output of $y = 1$ (an increase of $1$).
Doubling the input again to $x = 4$ results in an output of $y = 2$ (an increase of $1$).
Doubling the input once more to $x = 8$ results in an output of $y = 3$ (an increase of $1$).
This constant arithmetic increase for a geometric change in input is unique to logarithmic curves.
Graphs (C) and (D) are linear, where output increases by a constant amount for a constant addition to the input.
Graph (A) is exponential, which is the inverse relationship where input increases by $1$ as output doubles.
The correct choice is (B).
▶️ Answer/Explanation
The function $g$ is defined as the inverse of $f$, meaning $g = f^{-1}$.
By the property of inverse functions, the range of $f$ becomes the domain of $g$.
Similarly, the domain of $f$ becomes the range of $g$.
The maximum value of $g$ is the maximum value in its range, which is the maximum $x$-value of $f$.
Looking at the provided graph, the function $f$ is defined on the interval $x \in [0, 5]$.
The largest $x$-coordinate reached by the graph of $f$ is $5$.
Therefore, the maximum value of the inverse function $g$ is $5$.
The correct option is (C).
▶️ Answer/Explanation
The inverse function $f^{-1}$ swaps the inputs and outputs of the original function $f$.
From the table, the original pairs $(x, y)$ are $(-2, 1)$, $(1, 6)$, $(2, 2)$, and $(5, -3)$.
The inverse pairs $(y, x)$ must be $(1, -2)$, $(6, 1)$, $(2, 2)$, and $(-3, 5)$.
Sorting these by input value, we get: $(-3, 5)$, $(1, -2)$, $(2, 2)$, and $(6, 1)$.
Option (B) lists these input values $\{-3, 1, 2, 6\}$ with corresponding outputs $\{5, -2, 2, 1\}$.
Therefore, (B) is the correct description of the inverse function $f^{-1}$.
▶️ Answer/Explanation
Let $y = \sqrt{4 – x^2}$. Since $-2 \le x \le 0$, the range of $f(x)$ is $0 \le y \le 2$.
The domain of $f^{-1}(x)$ is the range of $f(x)$, which is $0 \le x \le 2$.
To find the inverse, swap $x$ and $y$: $x = \sqrt{4 – y^2}$.
Square both sides: $x^2 = 4 – y^2$, which simplifies to $y^2 = 4 – x^2$.
Solving for $y$, we get $y = \pm\sqrt{4 – x^2}$.
Since the original domain was $x \le 0$, the inverse must satisfy $y \le 0$.
Therefore, $f^{-1}(x) = -\sqrt{4 – x^2}$ for $0 \le x \le 2$.
The correct option is (C).
▶️ Answer/Explanation
The correct option is (D).
Identify key points on the graph of $y = f(x)$, such as $(-2, -10)$, $(0, 2)$, and $(2, 10)$.
For the inverse function $y = f^{-1}(x)$, the $x$ and $y$ coordinates are swapped.
The corresponding points on $f^{-1}(x)$ must be $(-10, -2)$, $(2, 0)$, and $(10, 2)$.
Graphically, $f^{-1}(x)$ is the reflection of $f(x)$ across the line $y = x$.
Option (D) is the only graph that contains these swapped points and reflects the original shape.
Notice the $x$-axis scale in (D) extends to $\pm 20$, matching the original $y$-axis range.
▶️ Answer/Explanation
Set the function $y = \frac{4x+6}{5}$.
Swap the variables $x$ and $y$ to get $x = \frac{4y+6}{5}$.
Multiply both sides by $5$ to obtain $5x = 4y + 6$.
Subtract $6$ from both sides: $5x – 6 = 4y$.
Divide by $4$ to solve for $y$: $y = \frac{5x-6}{4}$.
Therefore, $g^{-1}(x) = \frac{5x-6}{4}$, which matches option (D).
▶️ Answer/Explanation
Set $y = 4x^2 + 3$ and solve for $x$ to find the inverse.
Subtract $3$ from both sides: $y – 3 = 4x^2$.
Divide both sides by $4$: $\frac{y-3}{4} = x^2$.
Take the square root of both sides: $x = \sqrt{\frac{y-3}{4}}$ (since $x \geq 0$).
Swap $x$ and $y$: $f^{-1}(x) = \sqrt{\frac{x-3}{4}}$.
The range of $f(x)$ is $y \geq 3$, so the domain of $f^{-1}(x)$ is $x \geq 3$.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
The original function $W$ maps time $t$ to the amount of water $w$.
The inverse function $W^{-1}$ maps the amount of water $w$ back to time $t$.
Since $W$ is a decreasing function, as $t$ increases, $w$ decreases.
For any monotonic function, the inverse function $W^{-1}$ shares the same monotonicity as the original function.
Therefore, $W^{-1}$ must also be a decreasing function.
As the input of $W^{-1}$ is the “amount of water,” the correct description is that it is a decreasing function of the amount of water.
Thus, the correct option is (D).
▶️ Answer/Explanation
Set $y = \log_2(\log_3 x)$ to begin the inversion process.
Apply the base-2 exponential to both sides: $2^y = \log_3 x$.
Apply the base-3 exponential to isolate $x$: $3^{(2^y)} = x$.
Interchange $x$ and $y$ to find the inverse function.
The resulting expression is $f^{-1}(x) = 3^{(2^x)}$.
Therefore, the correct choice is (B).
▶️ Answer/Explanation
The original function $f$ is exponential because as $x$ increases by $1$, $f(x)$ doubles ($2, 4, 8, 16$).
The inverse of an exponential function is a logarithmic function, ruling out (C) and (D).
For the inverse function $f^{-1}(x)$, the inputs and outputs of $f$ are swapped.
The inputs of $f^{-1}(x)$ are the values $\{2, 4, 8, 16\}$ and the outputs are $\{1, 2, 3, 4\}$.
As the input values of $f^{-1}(x)$ double, the output values increase by $1$.
This matches the description provided in option (B).
Therefore, $f^{-1}(x)$ is logarithmic with output values increasing by $1$ every time input values double.
▶️ Answer/Explanation
To find the inverse $g^{-1}(x)$, start by setting $y = 5^{x}$.
Switch the variables $x$ and $y$ to get $x = 5^{y}$.
Solve for $y$ by converting the exponential equation to logarithmic form.
The definition of a logarithm states that if $x = b^{y}$, then $y = \log_{b} x$.
Applying this here, we get $y = \log_{5} x$.
Therefore, $g^{-1}(x) = \log_{5} x$.
The correct option is (A).
▶️ Answer/Explanation
The graph shows $f$ passes through $(0, 1)$, $(1, 2)$, and $(2, 4)$.
This indicates the exponential function is defined by $f(x) = 2^x$.
An inverse function $f^{-1}(x)$ swaps input and output: $(x, y) \rightarrow (y, x)$.
Since $f(2) = 4$, the inverse must contain the point $(4, 2)$.
Since $f(5) = 2^5 = 32$, the inverse must contain the point $(32, 5)$.
Table (D) is the only one correctly matching these coordinates.
Therefore, (D) is the correct table for the inverse function.
▶️ Answer/Explanation
The inverse of the logarithmic function $f(x) = \log_{3} x$ is found by switching $x$ and $y$, resulting in $y = 3^x$.
The graph of $y = 3^x$ must be an exponential growth curve that stays strictly above the $x$-axis.
The $y$-intercept occurs at $(0, 1)$ since $3^0 = 1$.
The graph must pass through the point $(1, 3)$ since $3^1 = 3$.
Graph B passes through $(0, 1)$ but appears to pass through $(1, 2)$, suggesting the function $y = 2^x$.
Graph C passes through $(0, 1)$ and reaches $y = 3$ when $x = 1$, matching the function $y = 3^x$.
Graph A represents a cubic-style function, and Graph D shows a reciprocal-style function.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
The problem states that if $f(x) = y$, then $f(4x) = y + 1$.
This relationship is a characteristic property of logarithmic functions.
For $f(x) = \log_{4} x$, if the input is multiplied by $4$, we get $f(4x) = \log_{4}(4x)$.
Using log properties: $\log_{4}(4x) = \log_{4} 4 + \log_{4} x$.
Since $\log_{4} 4 = 1$, the expression becomes $1 + f(x)$.
This matches the condition that the output increases by $1$ when the input is multiplied by $4$.
Therefore, the correct option is (D).
▶️ Answer/Explanation
The given equation is $g(x) = 2$.
Substitute the definition of $g(x)$ into the equation: $\ln(f(x)) = 2$.
Recall that the natural logarithm $\ln(y)$ has a base of $e$.
Rewrite the logarithmic equation in its equivalent exponential form.
If $\log_{b}(a) = c$, then $b^c = a$; here, $e^2 = f(x)$.
Therefore, solving $f(x) = e^2$ provides the values of $x$ that satisfy $g(x) = 2$.
The correct option is (B).
▶️ Answer/Explanation
The correct option is (D).
Use the product property of logarithms: $\log_{b}(m) + \log_{b}(n) = \log_{b}(m \cdot n)$.
Apply this to the given equation: $\log_{8}((x – 3)(x + 4)) = 1$.
Rewrite the logarithmic equation in exponential form: $(x – 3)(x + 4) = 8^{1}$.
Expand the left side using the FOIL method: $x^{2} + 4x – 3x – 12 = 8$.
Simplify the linear terms to get the final equation: $x^{2} + x – 12 = 8$.
▶️ Answer/Explanation
Set $y = 4 \log_{2}(x + 3) – 1$ and swap $x$ and $y$ to find the inverse.
$x = 4 \log_{2}(y + 3) – 1$
Add $1$ to both sides: $x + 1 = 4 \log_{2}(y + 3)$
Divide by $4$: $\frac{x + 1}{4} = \log_{2}(y + 3)$
Rewrite in exponential form: $2^{(\frac{x + 1}{4})} = y + 3$
Subtract $3$ to isolate $y$: $y = 2^{(\frac{x + 1}{4})} – 3$
The inverse function is $g(x) = 2^{(\frac{x + 1}{4})} – 3$, which matches option (D).
▶️ Answer/Explanation
The domain of $\ln x$ requires that $x > 0$.
Using the quotient rule for logarithms: $\ln\left(\frac{x^3}{x}\right) = 4$.
Simplify the fraction: $\ln(x^2) = 4$.
Convert the logarithmic equation to exponential form: $x^2 = e^4$.
Solve for $x$ by taking the square root: $x = \pm \sqrt{e^4} = \pm e^2$.
Checking the domain, $x$ must be positive, so $x = -e^2$ is extraneous.
Therefore, the only valid solution is $x = e^2$.
The correct option is (C).
▶️ Answer/Explanation
First, apply the power rule of logarithms: $\log_{9}(27^{x}) = x \log_{9} 27$.
Apply the change-of-base formula $\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$ to rewrite the expression.
Choose base $3$ because both $27$ and $9$ are powers of $3$.
This gives the equivalent form: $\frac{x \log_{3} 27}{\log_{3} 9}$.
Since $\log_{3} 27 = 3$ and $\log_{3} 9 = 2$, the expression becomes $\frac{3}{2}x$.
This matches choice (D), allowing for simple mental calculation of the rational multiple.
Correct Option: (D)
▶️ Answer/Explanation
The correct option is (B).
The given equation is $m = \log_{3} 81$.
By the definition of a logarithm, $\log_{b} x = y$ is equivalent to $b^y = x$.
In this case, the base $b$ is $3$, the result $y$ is $m$, and the argument $x$ is $81$.
Converting the logarithmic form to exponential form gives $3^m = 81$.
Since $3^4 = 81$, it also follows that $m = 4$.
Thus, the expression $3^m = 81$ is the true statement.
▶️ Answer/Explanation
Identify the given expression: $225 + 500 \log_{10}(15t + 10)$.
Substitute the given value $t = 6$ into the expression.
Calculate the inner term: $15(6) + 10 = 90 + 10 = 100$.
Evaluate the logarithm: $\log_{10}(100) = 2$, since $10^2 = 100$.
Multiply the logarithmic result by the coefficient: $500 \times 2 = 1000$.
Add the initial constant: $225 + 1000 = 1225$.
The number of items sold per month at $t = 6$ is $1225$.
Therefore, the correct option is (B).
▶️ Answer/Explanation
The Richter scale is logarithmic, where intensity $I$ relates to magnitude $M$ by $M = \log_{10}(I)$.
The ratio of intensities between two earthquakes is given by $10^{M_1 – M_2}$.
The difference in magnitude is $8.2 – 7.9 = 0.3$.
The intensity ratio is $10^{0.3}$.
Since $10^{0.3} \approx 1.995$, the Chile earthquake was approximately $2.0$ times more intense.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
First, define the function $h(x) = g(x) – f(x)$.
Substitute the given functions: $h(x) = \log_{3} x – (-1)$, which simplifies to $h(x) = \log_{3} x + 1$.
To find the $x$-intercept, set $h(x) = 0$.
This gives the equation $0 = \log_{3} x + 1$.
Subtract $1$ from both sides to get $\log_{3} x = -1$.
Rewrite the logarithmic equation in exponential form: $x = 3^{-1}$.
Solving for $x$ yields $x = \frac{1}{3}$.
The $x$-intercept is the point $(\frac{1}{3}, 0)$, which corresponds to option (B).
▶️ Answer/Explanation
The given function is $f(x) = \log_{2} x$.
The problem asks for the input $x$ such that the output $f(x) = 4$.
Substitute the output value into the function: $\log_{2} x = 4$.
Convert the logarithmic form to exponential form using the rule: $\log_{b} a = c \Rightarrow b^{c} = a$.
This yields the equation $x = 2^{4}$.
Calculate the value of the exponent: $2 \times 2 \times 2 \times 2 = 16$.
Therefore, the input value is $16$, which corresponds to option (B).
▶️ Answer/Explanation
The function is $f(x) = 2 \log_{5} x$, where the base $5 > 1$, making $f$ an increasing function.
To find the rate of change, we calculate the first derivative: $f'(x) = \frac{2}{x \ln 5}$.
Since $f'(x) > 0$ for all $x > 0$, the function is strictly increasing.
To find the rate of that increase, we look at the second derivative: $f”(x) = -\frac{2}{x^2 \ln 5}$.
Since $f”(x) < 0$ for all $x > 0$, the graph is concave down, meaning the rate is decreasing.
Therefore, $f$ increases at a decreasing rate.
The correct option is (B).
▶️ Answer/Explanation
A standard logarithmic function is $f(x) = \log_b(x)$.
If $0 < b < 1$, the function represents a logarithmic decay.
As $x$ approaches $0$ from the right, $\lim_{x \to 0^+} \log_b(x) = \infty$.
As $x$ increases toward infinity, $\lim_{x \to \infty} \log_b(x) = -\infty$.
This end behavior matches the conditions provided in option (D).
Other options fail because logs do not have horizontal asymptotes or return to $-\infty$.
Therefore, the correct description is given by choice (D).
▶️ Answer/Explanation
The function $f(x) = \log_{3} x$ is strictly increasing over its entire domain.
At the right endpoint $x = 9$, the function reaches a maximum value of $f(9) = \log_{3} 9 = 2$.
As $x$ approaches the left boundary $0$ from the right, the function $f(x) \to -\infty$.
Because the interval is open at $x = 0$, the function decreases without bound.
Therefore, there is no minimum value because the function never reaches a lowest point.
The correct option is (B).
▶️ Answer/Explanation
Set the functions equal: $\log_{10}(x – 1) + \log_{10}(x + 3) = \log_{10}(x + 9)$.
Apply the product property: $\log_{10}((x – 1)(x + 3)) = \log_{10}(x + 9)$.
Remove logs and expand: $x^2 + 2x – 3 = x + 9$.
Rearrange into a quadratic equation: $x^2 + x – 12 = 0$.
Factor the quadratic: $(x + 4)(x – 3) = 0$, giving $x = -4$ and $x = 3$.
Check domain constraints: For $f(x)$, $x – 1 > 0$, so $x > 1$.
Reject $x = -4$ as it is outside the domain; thus, $x = 3$ only is the valid solution.
The correct option is (A).
▶️ Answer/Explanation
First, identify the domain: $3x + 1 > 0 \implies x > -\frac{1}{3}$ and $x^2 + x – 2 > 0 \implies (x+2)(x-1) > 0$.
The intersection of $x > -\frac{1}{3}$ and $(x < -2 \text{ or } x > 1)$ gives the domain $x > 1$.
Set $g(x) < 0$: $\ln(3x + 1) – \ln(x^2 + x – 2) < 0 \implies \ln(3x + 1) < \ln(x^2 + x – 2)$.
Since $\ln$ is increasing, $3x + 1 < x^2 + x – 2$.
Rearrange to $x^2 – 2x – 3 > 0$, which factors as $(x – 3)(x + 1) > 0$.
This inequality holds for $x < -1$ or $x > 3$.
Combining $x > 1$ (domain) and $(x < -1 \text{ or } x > 3)$ yields $x > 3$.
The correct option is (D).
▶️ Answer/Explanation
Set the decibel function $d(N) > 140$.
Substitute the formula: $10 \log_{10} \left( \frac{N}{10^{-12}} \right) > 140$.
Divide both sides by $10$ to get $\log_{10} \left( \frac{N}{10^{-12}} \right) > 14$.
Rewrite the logarithmic inequality in exponential form: $\frac{N}{10^{-12}} > 10^{14}$.
Multiply both sides by $10^{-12}$ to isolate $N$.
$N > 10^{14} \cdot 10^{-12}$.
Simplify the exponents by adding them: $N > 10^{14 + (-12)}$.
The final result is $N > 10^2$, which is $N > 100$.
Correct Option: (C)
▶️ Answer/Explanation
The function is defined as $f(x) = \log_{2} x$, so we need to find an expression equivalent to $f(7) = \log_{2} 7$.
Using the Change of Base Formula: $\log_{b} a = \frac{\log_{c} a}{\log_{c} b}$ for any positive base $c$.
By applying this formula with a new base $c = 3$, we get $\log_{2} 7 = \frac{\log_{3} 7}{\log_{3} 2}$.
Option (A) is incorrect because $\log_{10} \left( \frac{7}{2} \right) = \log_{10} 7 – \log_{10} 2$.
Option (B) is the reciprocal of the correct change of base result using base $10$.
Option (C) is incorrect because $\frac{\log_{7} 10}{\log_{2} 10} = \frac{1/\log_{10} 7}{1/\log_{10} 2} = \frac{\log_{10} 2}{\log_{10} 7} = \log_{7} 2$.
Therefore, the correct equivalent expression is provided in option (D).
The final answer is (D).
▶️ Answer/Explanation
The correct answer is (A).
Start with the function $h(x) = \log_{49} x$.
Recognize that the base $49$ can be written as $7^{2}$.
Apply the change of base formula: $\log_{a^n} x = \frac{1}{n} \log_{a} x$.
This gives $h(x) = \log_{7^{2}} x = \frac{1}{2} \log_{7} x$.
Substitute $g(x)$ into the equation to get $h(x) = \frac{1}{2} g(x)$.
Therefore, for the same $x$, the output of $h$ is half the output of $g$.
▶️ Answer/Explanation
Given the function $h(x) = \log_{3} x$, substitute $w$ and $p$ into the expression: $2 \log_{3} w + \log_{3} p$.
Apply the Power Property of logarithms, $a \log_{b} c = \log_{b}(c^{a})$, to rewrite the first term as $\log_{3}(w^{2})$.
The expression becomes $\log_{3}(w^{2}) + \log_{3} p$.
Apply the Product Property of logarithms, $\log_{b} m + \log_{b} n = \log_{b}(mn)$.
Combining the terms results in $\log_{3}(w^{2} \cdot p)$.
This matches option (C).
▶️ Answer/Explanation
The function $g(x) = \log_{10}(x^3)$ can be rewritten using the logarithm power rule.
The power rule states that $\log_b(a^n) = n \cdot \log_b a$.
Applying this to $g(x)$, we get $g(x) = 3 \cdot \log_{10} x$.
Since $f(x) = \log_{10} x$, we can express $g$ as $g(x) = 3 \cdot f(x)$.
A transformation of the form $y = a \cdot f(x)$ represents a vertical dilation.
Because $a = 3$, the graph of $f$ is stretched vertically by a factor of $3$.
Therefore, the correct choice is (A).
▶️ Answer/Explanation
The function is defined as $f(x) = \ln x$, where the base is $e$.
To find when the output is an integer, set $f(x) = n$, where $n$ is any integer.
This gives the equation $\ln x = n$.
By converting the logarithmic form to exponential form, we get $x = e^n$.
Therefore, the input values $x$ must be integer powers of $e$.
For example, if $x = e^2$, then $f(e^2) = \ln(e^2) = 2$, which is an integer.
This confirms that option (A) is the correct description.
▶️ Answer/Explanation
The given function is $g(x) = a \log_{b} c$.
According to the power property of logarithms, $n \log_{x} y = \log_{x}(y^{n})$.
Applying this property, the coefficient $a$ becomes the exponent of the argument $c$.
This results in the equivalent expression $\log_{b}(c^{a})$.
Option (A) correctly matches this logarithmic identity.
Options (B), (C), and (D) violate standard logarithmic laws.
Therefore, the correct representation is option (A).
▶️ Answer/Explanation
To find the correct function, we test the given points $(2, 1)$ and $(4, 2)$ in each option.
For (A): $f(2) = \log_{4} 2 = 0.5$, which does not equal $1$.
For (B): $f(2) = 2 \log_{2} 2 = 2(1) = 2$, which does not equal $1$.
For (C): $f(2) = 2 \log_{4} 2 = 2(0.5) = 1$, which matches the first point.
Checking (C) with the second point: $f(4) = 2 \log_{4} 4 = 2(1) = 2$, which matches $(4, 2)$.
For (D): $f(2) = \log_{4} (2 + 2) = \log_{4} 4 = 1$, but $f(4) = \log_{4} 6 \neq 2$.
Therefore, the correct function is (C) $f(x) = 2 \log_{4} x$.
▶️ Answer/Explanation
Set the life expectancy function equal to $80$: $80 = 42.53 + 13.86 \ln x$.
Subtract $42.53$ from both sides to get $37.47 = 13.86 \ln x$.
Divide by $13.86$ to isolate the natural log: $\ln x \approx 2.70346$.
Exponentiate both sides ($e^{2.70346}$) to find $x \approx 14.93$ decades.
Convert decades to years: $14.93 \times 10 = 149.3$ years after 1900.
Add the years to the base year: $1900 + 149.3 = 2049.3$.
The value $2049.3$ corresponds to the late 2040s.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
The initial population at $t = 0$ is $P_0 = 30,000$.
The growth rate is $2.3\%$, so the growth factor is $1 + 0.023 = 1.023$.
The population $p$ after $t$ years is given by $p = 30,000 \cdot (1.023)^t$.
To find years $t$ as a function of population $p$, we solve for $t$:
Divide both sides by $30,000$ to get $\frac{p}{30,000} = (1.023)^t$.
Convert the exponential equation to logarithmic form: $t = \log_{1.023} \left( \frac{p}{30,000} \right)$.
Therefore, the correct function is $g(p) = \log_{1.023} \left( \frac{p}{30,000} \right)$.
The correct option is (B).
▶️ Answer/Explanation
The Richter scale is logarithmic, where a difference of $1$ in magnitude represents a $10$-fold difference in intensity.
Calculate the difference in magnitudes: $5.1 – 2.5 = 2.6$.
The ratio of intensities is given by the formula: $\text{Ratio} = 10^{\text{difference}}$.
Substitute the difference: $\text{Ratio} = 10^{2.6}$.
Break down the exponent: $10^{2.6} = 10^2 \times 10^{0.6}$.
Since $10^2 = 100$ and $10^{0.6}$ is approximately $3.98$ (nearly $4$), the result is $100 \times 4$.
The Indonesia earthquake was approximately $400$ times more intense.
Correct Option: (D)
▶️ Answer/Explanation
The standard growth model is $x = 2000(2)^{\frac{t}{8}}$, where $x$ is population and $t$ is time.
Divide both sides by $2000$ to isolate the exponential term: $\frac{x}{2000} = 2^{\frac{t}{8}}$.
Convert the exponential equation into its logarithmic form: $\log_{2} \left( \frac{x}{2000} \right) = \frac{t}{8}$.
Multiply both sides by $8$ to solve for $t$: $t = 8 \log_{2} \left( \frac{x}{2000} \right)$.
Comparing this result to the given options, we find it matches function $h(x)$.
Therefore, the correct option is (C).
▶️ Answer/Explanation
Input the given $(x, y)$ data pairs from the table into a graphing calculator’s list.
Perform a Logarithmic Regression (LnReg) to find the constants $a$ and $b$.
The resulting regression equation is approximately $f(x) \approx 4.071 + 3.397 \ln x$.
Substitute $x = 10$ into the model: $f(10) = 4.071 + 3.397 \ln(10)$.
Calculate the value: $f(10) \approx 4.071 + 3.397(2.3025) \approx 11.889$.
Therefore, the predicted height at age $10$ is $11.889$ feet.
The correct option is (B).
▶️ Answer/Explanation
The correct option is (B).
Apply the quotient rule: $\log_{10}\left(\frac{kz}{w^2}\right) = \log_{10}(kz) – \log_{10}(w^2)$.
Apply the product rule: $\log_{10}(kz) = \log_{10} k + \log_{10} z$.
Apply the power rule: $\log_{10}(w^2) = 2\log_{10} w$.
Combine the terms: $\log_{10} k + \log_{10} z – 2\log_{10} w$.
This matches the expression provided in option (B).
▶️ Answer/Explanation
The correct option is (C).
This problem uses the Power Property of Logarithms.
The rule states that $\log_{b}(m^{n}) = n \log_{b}(m)$.
In the expression $\log_{3}(x^{5})$, the base $b$ is $3$, the argument $m$ is $x$, and the exponent $n$ is $5$.
By moving the exponent to the front as a multiplier, we get $5 \log_{3} x$.
Therefore, $\log_{3}(x^{5})$ is equivalent to $5 \log_{3} x$.
▶️ Answer/Explanation
The given expression is $2 \ln x – 3 \ln y$.
Apply the Power Property: $n \ln a = \ln(a^n)$ to get $\ln(x^2) – \ln(y^3)$.
Apply the Quotient Property: $\ln a – \ln b = \ln\left(\frac{a}{b}\right)$.
Combine the terms: $\ln\left(\frac{x^2}{y^3}\right)$.
This matches the expression in option (A).
Correct Option: (A)
▶️ Answer/Explanation
Input the given $(x, y)$ coordinates into a statistical calculator to perform a LnReg ($y = a + b \ln x$).
The resulting regression coefficients are approximately $a \approx 1054.45$ and $b \approx 2841.45$.
The logarithmic model is expressed as $y = 1054.45 + 2841.45 \ln(x)$.
Substitute the target age $x = 4.5$ into the derived regression equation.
$y = 1054.45 + 2841.45 \ln(4.5)$.
$y \approx 1054.45 + 2841.45(1.504077)$.
$y \approx 1054.45 + 4273.76 = 5328.21$.
Rounding to the nearest pound, the predicted weight is $5333$.
Correct Option: (C)
▶️ Answer/Explanation
The correct answer is (C) Exponential decay.
In a semi-log plot, the vertical axis represents $\log(y)$ and the horizontal axis represents $x$.
A linear pattern on this plot follows the equation $\log(y) = mx + b$.
Converting from logarithmic to exponential form gives $y = 10^{mx + b}$, which is an exponential function.
Since the pattern is decreasing, the slope $m$ must be negative ($m < 0$).
An exponential function with a negative growth rate represents exponential decay.
Therefore, the data is best modeled by an exponential decay function.
▶️ Answer/Explanation
The maximum value (peak) of the graph must be $y = 7$.
The minimum value (trough) of the graph must be $y = 2$.
The period (the horizontal distance between two consecutive peaks) must be $5$ minutes.
Graph (A) has a maximum of $7$ and a minimum of $2$, but its period is $5$ minutes ($t = 1$ to $t = 6$).
Graph (B) has a maximum of $7$ and a minimum of $2$, but its period is only $2$ minutes.
Graph (C) has a period much longer than $5$ minutes.
Graph (D) shows a peak at $t = 0$ (approx), another at $t = 5$, and another at $t = 10$, confirming a period of $5$.
Therefore, Graph (D) is the correct model as it satisfies the range $[2, 7]$ and the $5$-minute periodicity.
▶️ Answer/Explanation
The period of a periodic function is the horizontal distance required for the graph to complete one full cycle.
By observing the graph of $f$, a peak occurs at $x = 1$ and the next peak occurs at $x = 5$.
Therefore, the period of $f$ is calculated as $5 – 1 = 4$ units.
Looking at Option (A), a peak occurs at $x = 0$ and the next peak occurs at $x = 4$.
The period for Option (A) is $4 – 0 = 4$ units, which matches function $f$.
Other options have different periods: (B) has a period of $2$, (C) has a period of $8$, and (D) has a period of $6$.
Thus, the function in Option (A) has the same period as $f$.
▶️ Answer/Explanation
The correct option is (A).
A periodic relationship means a motion or event repeats at regular intervals of time $T$.
The pendulum swings back and forth in a constant cycle, which is a classic example of periodic motion.
The metronome is designed to produce a “steady beat,” meaning the time interval between beats remains constant.
Since the beat repeats at fixed intervals, it also has a periodic relationship with time.
Therefore, both the mechanical motion and the resulting sound follow a predictable, repeating pattern.
▶️ Answer/Explanation
Point $A$ and Point $B$ represent two consecutive New Moon phases where illumination is $0\%$.
According to the table, the first New Moon occurs on May 4 and the next on June 2.
The number of days in May is $31$.
Days from May 4 to May 31 is $31 – 4 = 27$ days.
Adding the $2$ days in June gives $27 + 2 = 29$ days.
Checking the next cycle: June 2 to July 1 is $28$ days in June plus $1$ day in July, totaling $29$ days.
Among the choices, $28$ is the closest approximation to the lunar cycle length.
Therefore, the correct option is (C).
▶️ Answer/Explanation
The graph shows one full cycle of the function $f$ from $x = 0$ to $x = 4$.
The period of the function is $P = 4$.
To find the behavior at $39 < x < 41$, we find the equivalent values in the first cycle by calculating $x \pmod 4$.
For $x = 39$, $39 = (9 \times 4) + 3$, which corresponds to $x = 3$.
For $x = 41$, $41 = (10 \times 4) + 1$, which corresponds to $x = 1$.
On the interval $3 < x < 4$ of the original cycle, the function is increasing.
On the interval $0 < x < 1$ of the original cycle, the function is also increasing.
Since $x = 40$ (where $40 \equiv 0 \pmod 4$) is the transition point, the function continues increasing throughout the interval.
Therefore, the correct behavior is that the function $f$ is increasing.
▶️ Answer/Explanation
The period of a periodic function is the horizontal distance required to complete one full cycle.
Point $A$ represents a maximum (peak) of the graph.
The next consecutive maximum of the same height occurs at point $E$.
The horizontal distance (time interval) from one peak to the very next peak represents one full cycle.
Intervals $A$ to $B$ or $A$ to $C$ only represent fractions of a cycle.
Therefore, the time interval between points $A$ and $E$ gives the period.
The correct option is (D).
▶️ Answer/Explanation
The relationship between arc length $s$, radius $r$, and angle $\theta$ (in radians) is given by $s = r\theta$.
In this problem, the arc length $s$ is given as $40$.
The angle $\theta$ is given as $\frac{\pi}{4}$ radians.
Substitute the values into the formula: $40 = r \cdot \frac{\pi}{4}$.
Solve for $r$ by multiplying both sides by $4$: $160 = r\pi$.
Divide by $\pi$ to isolate the radius: $r = \frac{160}{\pi}$.
The correct option is (D).
▶️ Answer/Explanation
The radius of the circle is $r = 5$, as seen from point $P(5,0)$.
The arc length $s$ is given by the formula $s = r\theta$.
Substituting the given values, $6 = 5\theta$, which means $\theta = \frac{6}{5}$ radians.
The coordinates of point $Q$ on a circle are $(r \cos\theta, r \sin\theta)$.
The distance of a point from the $y$-axis is the absolute value of its $x$-coordinate.
Therefore, the distance is $|5 \cos(\frac{6}{5})|$.
Since $\frac{6}{5}$ radians is in the first quadrant, the distance is $5 \cos\left(\frac{6}{5}\right)$.
Correct Option: (C)
▶️ Answer/Explanation
An exponential function $y = a \cdot b^x$ becomes linear when transformed by logarithms.
Applying the natural log gives the equation $\ln y = \ln a + x \ln b$.
This implies that for an exponential function, $\ln y$ must have a constant rate of change relative to $x$.
In Table (D), as $x$ increases by $1$ unit, $\ln y$ increases by a constant addition of $10$.
Table (A) shows a linear relationship for $y$ itself, not an exponential one.
Table (C) shows $\ln y$ growing exponentially, which would mean $y$ grows even faster.
Only Table (D) shows the constant slope for $\ln y$ that characterizes an exponential function $f(x)$.
Thus, the correct choice is Table (D).
▶️ Answer/Explanation
Take the natural log of the function: $\ln(f(x)) = \ln(a \cdot c^x)$.
Apply log properties: $\ln(f(x)) = \ln a + \ln(c^x) = (\ln c)x + \ln a$.
Comparing this to $mx + b$, we find $m = \ln c$ and $b = \ln a$.
Since $c > 1$, it follows that $\ln c > 0$, so $m > 0$.
Since $a > 0$, the value of $b = \ln a$ can be any real number ($-\infty < \ln a < \infty$).
Therefore, $m > 0$ and $b$ can be any real number.
The correct option is (A).
▶️ Answer/Explanation
In a semi-log plot, the vertical axis is logarithmic, so we plot $Y = \log(f(x))$.
For functions to appear linear, they must be of the form $y = Ab^{x}$, since $\log(Ab^{x}) = \log(A) + x\log(b)$.
For lines to be parallel, they must have the same slope in the log-transformed state.
In option (C), let $f(x) = 2^{x}$ and $g(x) = 3 \cdot 2^{x}$.
Applying log: $Y_{1} = \log(2^{x}) = x \log(2)$ and $Y_{2} = \log(3 \cdot 2^{x}) = \log(3) + x \log(2)$.
Both equations have the same slope, $\log(2)$, making them parallel lines.
Therefore, the correct choice is (C).
▶️ Answer/Explanation
The $x$ values increase by a constant $1$ ($5, 6, 7, 8$), forming an arithmetic sequence.
The $\ln y$ values increase by a constant $3$ ($3, 6, 9, 12$), forming an arithmetic sequence.
Since $\ln y$ is a linear function of $x$, the relationship is of the form $\ln y = mx + b$.
Applying the exponential to both sides gives $y = e^{mx + b}$, which simplifies to $y = Ab^x$.
This structure confirms that the function $f(x)$ is an exponential function.
Therefore, statement (C) is the correct characterization of the data.
Correct Option: (C)
▶️ Answer/Explanation
From the table, the linear relationship is $\log_{3}(f(x)) = x + 2$.
Convert this to exponential form: $f(x) = 3^{x+2}$.
Calculate the y-intercept by setting $x = 0$: $f(0) = 3^{0+2} = 9$.
Calculate the value at $x = 1$: $f(1) = 3^{1+2} = 27$.
Calculate the value at $x = 2$: $f(2) = 3^{2+2} = 81$.
The graph must be exponential and pass through points $(0, 9)$ and $(1, 27)$.
Therefore, the correct choice is (D).
▶️ Answer/Explanation
The plot is semi-logarithmic, where a linear trend indicates an exponential function.
At $t = 20$, the point on the graph is at $10^1$, which means $N(20) = 10$.
At $t = 50$, the point is just below $10^2$ (roughly $80$); checking option (C): $2.5 \cdot 2^{(50/10)} = 2.5 \cdot 32 = 80$.
Testing option (C) for $t = 20$ gives $2.5 \cdot 2^{(20/10)} = 2.5 \cdot 4 = 10$, matching the data exactly.
Options (A) and (B) are linear functions, which would appear curved on a semi-log scale.
Option (D) at $t = 20$ yields $3 + 2^2 = 7$, which does not match the observed value of $10$.
Therefore, the function that correctly defines the data is (C).
▶️ Answer/Explanation
The circle has a radius $r = 2$.
Point $C$ is at $0$ radians and point $A$ is at $\frac{3\pi}{2}$ radians (or $-\frac{\pi}{2}$).
Since angle $COB$ equals angle $AOB$, $B$ is the midpoint of the arc between $0$ and $\frac{3\pi}{2}$ in the clockwise direction.
The angle $\theta$ for point $B$ in standard position is $2\pi – \frac{\pi}{4} = \frac{7\pi}{4}$ radians.
The coordinates of any point on a circle are given by $(r\cos\theta, r\sin\theta)$.
Substituting the values, the coordinates of $B$ are $\left( 2\cos\left( \frac{7\pi}{4} \right), 2\sin\left( \frac{7\pi}{4} \right) \right)$.
Therefore, the correct option is (C).
▶️ Answer/Explanation
The coordinates of a point on a circle are defined as \((x, y) = (r \cos \theta, r \sin \theta)\).
Given radius \(r = 4\) and angle \(\theta = \frac{5\pi}{6}\) radians.
The angle \(\frac{5\pi}{6}\) is in the second quadrant, where cosine is negative and sine is positive.
Calculate \(x = 4 \cos(\frac{5\pi}{6}) = 4 (-\frac{\sqrt{3}}{2}) = -2\sqrt{3}\).
Calculate \(y = 4 \sin(\frac{5\pi}{6}) = 4 (\frac{1}{2}) = 2\).
The coordinates of the intersection point are \((-2\sqrt{3}, 2)\).
Therefore, the correct option is (A).
▶️ Answer/Explanation
The coordinates of a point on a circle of radius $r$ at angle $\theta$ are $(r \cos \theta, r \sin \theta)$.
Point $A$ is at $\left(20 \cos\left(\frac{2\pi}{3}\right), 20 \sin\left(\frac{2\pi}{3}\right)\right)$ and Point $B$ is at $\left(20 \cos\left(\frac{4\pi}{3}\right), 20 \sin\left(\frac{4\pi}{3}\right)\right)$.
Both angles have the same reference angle $\frac{\pi}{3}$, so their $x$-coordinates are both $20 \cos\left(\frac{2\pi}{3}\right) = 20 \cos\left(\frac{4\pi}{3}\right) = -10$.
Since the $x$-coordinates are identical, the points lie on a vertical line.
The distance is the difference between the $y$-coordinates: $y_A – y_B$.
This results in $20 \sin\left(\frac{2\pi}{3}\right) – 20 \sin\left(\frac{4\pi}{3}\right)$.
Thus, the correct option is (D).
▶️ Answer/Explanation
The radius is calculated using $r = \sqrt{x^2 + y^2} = \sqrt{10^2 + (-10\sqrt{3})^2}$.
This simplifies to $r = \sqrt{100 + 300} = \sqrt{400} = 20$.
The point $(10, -10\sqrt{3})$ lies in the fourth quadrant since $x > 0$ and $y < 0$.
The reference angle $\alpha$ is found by $\tan \alpha = \left| \frac{y}{x} \right| = \left| \frac{-10\sqrt{3}}{10} \right| = \sqrt{3}$.
Thus, $\alpha = \frac{\pi}{3}$, making the standard position angle $\theta = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3}$.
Therefore, the correct values are $r = 20$ and $\theta = \frac{5\pi}{3}$.
The correct option is (B).
▶️ Answer/Explanation
The function is defined as $g(z) = \cos z$.
The angle $\theta$ is in the second quadrant, where $\frac{\pi}{2} < \theta < \pi$.
We are given the interval $\theta < \omega < \pi$.
In the second quadrant, the cosine function is strictly decreasing.
As the angle $z$ increases from $\theta$ toward $\pi$, the $x$-coordinate on the unit circle moves left.
Therefore, if $\omega > \theta$, then $\cos \omega < \cos \theta$.
This means $g(\omega) < g(\theta)$, making (A) the correct choice.
▶️ Answer/Explanation
The point $P$ is in the second quadrant, meaning $\frac{\pi}{2} < \theta < \pi$.
The function is defined as $f(z) = \sin z$.
In the interval $(\frac{\pi}{2}, \pi)$, the sine function is strictly decreasing.
Given the inequality $\frac{\pi}{2} < \beta < \theta$, both angles are in the second quadrant.
Since $\beta < \theta$ and sine is decreasing here, it follows that $\sin \beta > \sin \theta$.
Therefore, $f(\beta) > f(\theta)$, which corresponds to option (B).
▶️ Answer/Explanation
The first derivative $g'(\theta) = -\sin \theta$ is positive on $(\frac{3\pi}{2}, 2\pi)$, so $g$ is increasing.
The second derivative $g”(\theta) = -\cos \theta$ is negative on $(\frac{3\pi}{2}, 2\pi)$ because $\cos \theta > 0$ in Quadrant IV.
Since $g”(\theta) < 0$, the graph of $g$ is concave down.
In the fourth quadrant, $\cos \theta$ increases from $0$ to $1$.
The curvature curves “downward” toward the horizontal axis as it approaches the maximum.
Therefore, the correct description is that $g$ is increasing and concave down.
Correct Option: (C)
▶️ Answer/Explanation
The graph starts at $(0,0)$ and follows the shape of $y = \sin(x)$.
On a unit circle, the coordinates of point $P$ are $(\cos\theta, \sin\theta)$.
The $y$-coordinate, $\sin\theta$, represents the vertical displacement from the $x$-axis.
“Displacement” is used instead of “distance” because the function $f$ has negative values.
Distance is always non-negative, whereas $f$ is negative on the interval $(\pi, 2\pi)$.
Therefore, $f$ represents the vertical displacement of $P$ for $0 \le \theta \le 2\pi$.
The correct option is (C).
▶️ Answer/Explanation
The interval $[\frac{\pi}{2}, \pi]$ corresponds to the Second Quadrant of the unit circle.
For $f(\theta) = \cos \theta$, the value moves from $\cos(\frac{\pi}{2}) = 0$ down to $\cos(\pi) = -1$.
Since the values are getting smaller, $f$ is decreasing.
For $g(\theta) = \sin \theta$, the value moves from $\sin(\frac{\pi}{2}) = 1$ down to $\sin(\pi) = 0$.
Since the values are getting smaller, $g$ is decreasing.
Therefore, both $f$ and $g$ are decreasing on this interval.
The correct option is (A).
▶️ Answer/Explanation
The correct option is (B).
On the interval $[\pi, \frac{3\pi}{2}]$, the value of $\sin \theta$ goes from $0$ to $-1$, meaning $f$ is decreasing.
The first derivative $f'(\theta) = \cos \theta$ is negative in the third quadrant, confirming the function is decreasing.
The second derivative $f”(\theta) = -\sin \theta$ is positive on this interval because $\sin \theta$ is negative.
Since $f”(\theta) > 0$, the graph of $f$ is concave up.
Therefore, $f$ is decreasing and concave up on the given interval.
▶️ Answer/Explanation
Since $\triangle ABO$ is equilateral with side length $10$, we have $OA = OB = AB = 10$.
Segment $AB$ is vertical and bisected by the $x$-axis due to the symmetry of the equilateral triangle centered at the origin horizontally.
The $y$-coordinate of point $B$ is half the length of side $AB$, which is $\frac{10}{2} = 5$.
Point $B$ lies on a circle centered at the origin with radius $r = OB = 10$.
In trigonometry, for any point $(x, y)$ on the terminal side of an angle $\theta$, $\sin \theta = \frac{y}{r}$.
Substituting the known values, we get $\sin \theta = \frac{5}{10} = \frac{1}{2}$.
The value of $10 \sin \theta$ is $10 \left( \frac{1}{2} \right) = 5$.
Therefore, the correct option is (D).
▶️ Answer/Explanation
Point $B$ is on the negative $x$-axis, so its position corresponds to an angle of $\pi$ radians.
Point $X$ has coordinates $x = -\frac{5}{2}$ and $y = -\frac{5\sqrt{3}}{2}$ in Quadrant III.
The reference angle $\theta$ for $X$ is found using $\tan(\theta) = \left| \frac{y}{x} \right| = \left| \frac{-5\sqrt{3}/2}{-5/2} \right| = \sqrt{3}$.
This gives a reference angle of $\frac{\pi}{3}$ measured from the negative $x$-axis.
Since angle $XOB$ is the angle between the ray $OX$ and the ray $OB$ (the negative $x$-axis), the measure is simply the reference angle.
Therefore, the measure of angle $XOB$ is $\frac{\pi}{3}$.
The correct option is (B).
▶️ Answer/Explanation
First, locate $x = 3$ in the table to find the value of $k(3)$.
From the table, when $x = 3$, the value of $k(3) = 4$.
The expression now becomes $h^{-1}(4)$.
To find $h^{-1}(4)$, look for the value of $x$ such that $h(x) = 4$.
From the table, $h(x) = 4$ when $x = 1$.
Therefore, $h^{-1}(4) = 1$.
The final value is $1$, which corresponds to option (A).
▶️ Answer/Explanation
The correct graph is (C).
A semi-log plot uses a logarithmic scale for the vertical axis to linearize exponential data.
The data values for $m(t)$ range from $30$ to $231$.
The vertical axis must include the interval $[30, 231]$, which fits within the $10$ to $1000$ scale.
The constant ratio $\frac{m(t+1)}{m(t)} \approx 1.67$ indicates the function is exponential.
On a logarithmic scale, this exponential relationship appears as a straight line.
Graph (C) shows the points forming a linear pattern on a log-scaled $y$-axis.
▶️ Answer/Explanation
Set the function equal to the target value: $2 \sin \theta = -1$.
Isolate the sine function by dividing both sides by $2$ to get $\sin \theta = -\frac{1}{2}$.
Identify the reference angle where $\sin \theta = \frac{1}{2}$, which is $\theta_{ref} = \frac{\pi}{6}$.
Since the sine value is negative, $\theta$ must be in Quadrant III or Quadrant IV.
In Quadrant III: $\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$.
In Quadrant IV: $\theta = 2\pi – \frac{\pi}{6} = \frac{11\pi}{6}$.
Therefore, the correct option is (D).
▶️ Answer/Explanation
The circle is centered at the origin $(0, 0)$ and passes through $(5, 0)$, so its radius $r$ is $5$.
Point $P(x, y)$ lies on the terminal ray of the angle $\theta$ in standard position.
By trigonometric definition on a circle, $\sin \theta$ is the ratio of the $y$-coordinate to the radius: $\sin \theta = \frac{y}{r}$.
The $y$-coordinate represents the vertical displacement of point $P$ from the $x$-axis.
The radius $r = 5$ represents the distance from the origin to point $P$.
Therefore, $\sin \theta = \frac{y}{5}$ based on the position and coordinates of point $P$.
This matches option (D).
▶️ Answer/Explanation
An angle is in standard position if its vertex is at the origin $(0,0)$ and its initial ray lies on the positive $x$-axis.
In the given figure, point $C$ lies on the positive $x$-axis at $(3,0)$, making ray $OC$ the initial ray.
Angle $COB$ starts at ray $OC$ and rotates counter-clockwise to terminal ray $OB$.
Therefore, angle $COB$ is in standard position with initial ray $OC$ and terminal ray $OB$.
Options (C) and (D) are incorrect because the initial ray for standard position must be $OC$, not $OB$ or $OD$.
Option (A) is incorrect because it swaps the definitions of the initial and terminal rays.
The correct statement is (B).
▶️ Answer/Explanation
The formula for the arc length $s$ is given by $s = r\theta$, where $\theta$ is in radians.
The radius of the circle is given as $r = 3$.
The central angle subtending the arc is given as $\theta = 0.687\pi$.
Substitute the values into the formula: $s = 3 \times (0.687\pi)$.
Calculate the product: $s = 2.061\pi$.
Using the approximation $\pi \approx 3.14159$, the length is $s \approx 6.4748…$
Rounding to three decimal places, the arc length is $6.475$.
The correct option is (C).
▶️ Answer/Explanation
Point $A$ lies on the circle with coordinates $(x, y) = (2.2, 4.3)$.
The radius of the circle is given as $r = 4.8$.
For an angle $\alpha$ in standard position, the sine is defined as $\sin \alpha = \frac{y}{r}$.
Substitute the $y$-coordinate of point $A$, which is $4.3$.
Substitute the radius value, which is $4.8$.
Therefore, $\sin \alpha = \frac{4.3}{4.8}$.
The correct option is (B).
▶️ Answer/Explanation
The terminal ray $AC$ passes through the origin $(0,0)$ and the point $(0.4, -0.9)$.
The slope of ray $AC$ is calculated as $m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-0.9 – 0}{0.4 – 0} = -\frac{0.9}{0.4}$.
Simplifying the slope gives $m = -\frac{9}{4}$.
For an angle $\theta$ in standard position, $\tan(\theta) = \frac{y}{x}$ where $(x, y)$ is a point on the terminal ray.
Thus, $\tan(BAC) = \frac{-0.9}{0.4} = -\frac{9}{4}$.
Both the tangent of the angle and the slope of the ray are equal to $-\frac{9}{4}$.
Therefore, the correct description is given in option (C).
▶️ Answer/Explanation
The correct option is (B).
The tangent function is defined as $\tan \theta = \frac{y}{x}$ in the Cartesian plane.
For $\tan \theta = 1$, the values of $x$ and $y$ must have the same sign and magnitude.
In Quadrant I, both $x > 0$ and $y > 0$, so $\tan \theta = \frac{+}{+} = 1$ at $\theta = \frac{\pi}{4}$.
In Quadrant III, both $x < 0$ and $y < 0$, so $\tan \theta = \frac{-}{-} = 1$ at $\theta = \frac{5\pi}{4}$.
In Quadrants II and IV, $x$ and $y$ have opposite signs, making $\tan \theta$ negative.
Thus, there are exactly two solutions in the interval $[0, 2\pi]$, located in Quadrants I and III.
▶️ Answer/Explanation
The terminal side of angle $\alpha$ intersects the unit circle at point $A(-0.8, 0.6)$.
By definition, for any point $(x, y)$ on a unit circle, $x = \cos \alpha$ and $y = \sin \alpha$.
The $x$-coordinate of point $A$ is given as $-0.8$.
Therefore, $\cos \alpha = -0.8$.
The correct option is (A).