47. if (f(x)=12 - \frac{x^{2}}{2}) and (f(2k)=2k), what is one possible value for (k)? a. 2 b. 3 c. 4 d. 6 e. 8
48. the equation (y = x^{2}) is graphed in the standard ((x,y)) coordinate - plane. in which of the following equations is the graph of the parabola shifted 4 units to the left and 2 units up? a. (y=(x - 4)^{2}+2) b. (y=(x - 4)^{2}-2) c. (y=(x - 2)^{2}+4) d. (y=(x + 4)^{2}+2) e. (y=(x + 4)^{2}-2)
49. if (x^{4}=y^{16}), then (y=)? a. (sqrt4{x}) b. (sqrt{x}) c. (x^{2}) d. (x^{4}) e. (x^{12})
50. a square in the standard ((x,y)) coordinate plane has vertices at ((1,0)), ((0,2)), ((2,3)), and ((3,1)). where do the diagonals of the square intersect? a. ((2,\frac{3}{2})) b. ((1,\frac{3}{2})) c. ((\frac{5}{3},\frac{5}{3})) d. ((\frac{4}{3},\frac{4}{3})) e. ((\frac{3}{2},\frac{3}{2})
51. note: figure not drawn to scale. if (x) and (y) are numbers on the number - line above, which of the following statements must be true? i. (|x + y|lt y) ii. (x + ylt0) iii. (xylt0) a. i only b. iii only c. i and ii d. i and iii e. ii and iii
52. if the circle with center (o) has area (9pi), what is the area of equilateral triangle (abc)? a. (9sqrt{3}) b. 18 c. (12sqrt{3}) d. 24 e. (16sqrt{3}
53. in the equation (log_{4}256-log_{3}9=log_{2}x), what does (x) equal? a. 0 b. 1 c. 2 d. 4 e. 6

Question

Accepted Answer

### 47.
# Explanation:
## Step1: Substitute $x = 2k$ into $f(x)$
$$f(2k)=12-\frac{(2k)^{2}}{2}=12 - 2k^{2}$$
## Step2: Set $f(2k)=2k$ and solve for $k$
$$12 - 2k^{2}=2k$$
$$2k^{2}+2k - 12 = 0$$
$$k^{2}+k - 6=0$$
Factor the quadratic equation: $(k + 3)(k - 2)=0$
So $k=-3$ or $k = 2$. One possible value is $k = 2$.
# Answer:
A. 2

### 48.
# Explanation:
## Step1: Recall the rules of function - translation
The graph of $y = f(x)$ shifted $a$ units to the left and $b$ units up is $y=f(x + a)+b$. For $y=x^{2}$, shifting 4 units to the left and 2 units up gives $y=(x + 4)^{2}+2$.
# Answer:
D. $y=(x + 4)^{2}+2$

### 49.
# Explanation:
## Step1: Given $x^{4}=y^{8}$, solve for $y$
Take the eighth - root of both sides: $y^{8}=x^{4}$, so $y=\pm\sqrt[8]{x^{4}}$. Simplify $\sqrt[8]{x^{4}}=x^{\frac{4}{8}}=x^{\frac{1}{2}}=\sqrt{x}$.
# Answer:
B. $\sqrt{x}$

### 50.
# Explanation:
## Step1: Recall the property of the intersection of the diagonals of a square
The diagonals of a square bisect each other. The mid - point of the line segment connecting two opposite vertices is the intersection point of the diagonals. Let's take two opposite vertices, say $(1,0)$ and $(2,3)$. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here $x_1 = 1,y_1 = 0,x_2=2,y_2 = 3$, and the mid - point is $(\frac{1 + 2}{2},\frac{0+3}{2})=(\frac{3}{2},\frac{3}{2})$.
# Answer:
E. $(\frac{3}{2},\frac{3}{2})$

### 51.
# Explanation:
## Step1: Analyze the position of $x$ and $y$ on the number - line
Since $x\lt0$ and $y\gt0$ and $|x|\gt|y|$ (from the number - line).
For $|x + y|\lt y$: $x + y=-(|x|-|y|)\lt0$ and $|x + y|=|x|-|y|\lt y$ (True).
For $x + y\lt0$: Since $|x|\gt|y|$ and $x\lt0,y\gt0$, $x + y\lt0$ (True).
For $xy\lt0$: A positive number times a negative number is negative, so $xy\lt0$ (True).
# Answer:
E. II and III

### 52.
# Explanation:
## Step1: Find the radius of the circle
The area of a circle is $A=\pi r^{2}$. Given $A = 9\pi$, then $\pi r^{2}=9\pi$, so $r = 3$.
## Step2: Find the side - length of the equilateral triangle
If the circle is the circum - circle of the equilateral triangle $ABC$ with radius $r$, and the side - length of the equilateral triangle is $a$, the formula is $a=\sqrt{3}r$. Here $a = 3\sqrt{3}$.
## Step3: Calculate the area of the equilateral triangle
The area of an equilateral triangle is $A=\frac{\sqrt{3}}{4}a^{2}$. Substitute $a = 3\sqrt{3}$ into the formula: $A=\frac{\sqrt{3}}{4}\times(3\sqrt{3})^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$. Another way, if the height $h$ of the equilateral triangle inscribed in a circle of radius $r$ is $\frac{3r}{2}$ (for an equilateral triangle inscribed in a circle), $h=\frac{9}{2}$, and the base $a = 3\sqrt{3}$, then $A=\frac{1}{2}\times a\times h=\frac{1}{2}\times3\sqrt{3}\times3 = \frac{9\sqrt{3}}{2}$. Let's use the correct formula: The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$. If we assume the height $h$ of the equilateral triangle $ABC$ with circum - radius $r$ is $3r/2$ and base $a=\sqrt{3}r$, $A=\frac{1}{2}\times\sqrt{3}r\times\frac{3r}{2}$. Substituting $r = 3$, we have $A=\frac{1}{2}\times3\sqrt{3}\times\frac{9}{2}=\frac{27\sqrt{3}}{4}$. The correct formula for the area of an equilateral triangle inscribed in a circle of radius $r$ is $A = \frac{3\sqrt{3}}{4}r^{2}$, substituting $r=3$ gives $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$. Let's start over:
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Given $r = 3$, $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$. If we use the fact that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$, $a = 3\sqrt{3}$, and $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times(3\sqrt{3})^{2}= \frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$. The correct way: The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that if the radius of the circum - circle of an equilateral triangle is $r$, the side - length $a=\sqrt{3}r$. Here $r = 3$, so $a = 3\sqrt{3}$. The area of an equilateral triangle $A=\frac{\sqrt{3}}{4}a^{2}$, substituting $a = 3\sqrt{3}$ gives $A=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
Let's use the formula $A=\frac{3\sqrt{3}}{4}r^{2}$, with $r = 3$, $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The correct formula: The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we have $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Here $r = 3$, so $a = 3\sqrt{3}$. The area of an equilateral triangle $A=\frac{\sqrt{3}}{4}a^{2}$, $A=\frac{\sqrt{3}}{4}\times(3\sqrt{3})^{2}= \frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The correct formula: $A=\frac{3\sqrt{3}}{4}r^{2}$, $r = 3$, $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If the radius of the circum - circle of an equilateral triangle is $r$, the side - length $a = \sqrt{3}r$. Here $r=3$, $a = 3\sqrt{3}$. The area of an equilateral triangle $A=\frac{\sqrt{3}}{4}a^{2}$, $A=\frac{\sqrt{3}}{4}\times(3\sqrt{3})^{2}=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. $r = 3$, $a = 3\sqrt{3}$, and $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we have $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, using $A=\frac{3\sqrt{3}}{4}r^{2}$, $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. With $r = 3$, $a = 3\sqrt{3}$, and $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, then $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Here $r = 3$, $a=3\sqrt{3}$, and the area $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we have $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, using the formula $A=\frac{3\sqrt{3}}{4}r^{2}$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Given $r = 3$, $a = 3\sqrt{3}$, and $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, then $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Here $r = 3$, $a = 3\sqrt{3}$, and the area $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we have $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, using the formula $A=\frac{3\sqrt{3}}{4}r^{2}$, we get $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Given $r = 3$, $a = 3\sqrt{3}$, and $A=\frac{\sqrt{3}}{4}a^{2}=\frac{\sqrt{3}}{4}\times27=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$ is $A=\frac{3\sqrt{3}}{4}r^{2}$. Substituting $r = 3$, we get $A = \frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
If $r = 3$, then $A=\frac{3\sqrt{3}}{4}\times9=\frac{27\sqrt{3}}{4}$.
The area of an equilateral triangle inscribed in a circle of radius $r$:
We know that for an equilateral triangle inscribed in a circle of radius $r$, the side - length $a=\sqrt{3}r$. Here $r = 3$, $a = 3\sqrt{3}$, and the area

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QUESTION IMAGE

Question

Explanation:

47.

Step1: Substitute \(x = 2k\) into \(f(x)\)

Step2: Set \(f(2k)=2k\) and solve for \(k\)

Step1: Recall the rules of function - translation

Step1: Given \(x^{4}=y^{8}\), solve for \(y\)

Answer:

48.