22. a 180° rotation of point (4, -1) around the origin gives:
a. (-4,1)
b. (1,-4)
c. (4,1)
d. (-1,4)
23. solve: $\frac{x - 1}{x + 2} = 3$
a. $x = 7$
b. $x = -5$
c. $x = 5$
d. $x = -7$
24. when two parallel lines are cut by a transversal, what is true about consecutive interior angles?
a. they are congruent
b. they are complementary
c. they are supplementary
d. they form a right angle
25. if $\triangle abc \sim \triangle def$ with scale factor 3, and $abc$ has area 12, what is the area of $\triangle def$?
a. 36
b. 108
c. 48
d. 72
26. the transformation $f(x) = |x - 2|$ represents:
a. a reflection over the line $x = 2$
b. a reflection over the $y$-axis
c. a translation right 2 units
d. a dilation with factor 2
27. in a right triangle, if $\tan\theta = \frac{3}{4}$, what is $\sin\theta$?
a. $\frac{3}{5}$
b. $\frac{4}{5}$
c. $\frac{5}{3}$
d. $\frac{5}{4}$
28. factor: $5x^2 - 45$
a. $5(x + 3)(x - 3)$
b. $5(x^2 - 9)$
c. $(5x + 15)(x - 3)$
d. $5(x - 3)^2$

Question

Accepted Answer

### Question 22
# Explanation:
## Step1: Recall 180° rotation rule
For a point $(x,y)$ rotated $180^\circ$ around the origin, the new point is $(-x,-y)$.
## Step2: Apply the rule to $(4, -1)$
Here, $x = 4$ and $y=-1$. So the new $x$-coordinate is $-4$ and the new $y$-coordinate is $-(-1)=1$. So the rotated point is $(-4,1)$.
# Answer: a. $(-4,1)$

### Question 23
# Explanation:
## Step1: Cross - multiply to eliminate the fraction
Given $\frac{x - 1}{x+1}=3$, cross - multiply (assuming $x
eq - 1$): $x - 1=3(x + 1)$.
## Step2: Expand and solve for $x$
Expand the right - hand side: $x - 1=3x+3$.
Subtract $x$ from both sides: $-1 = 2x+3$.
Subtract 3 from both sides: $-4 = 2x$.
Divide both sides by 2: $x=-2$? Wait, no, let's check again. Wait, $x - 1=3(x + 1)$
$x-1 = 3x + 3$
$x-3x=3 + 1$
$-2x=4$
$x=-2$? But this is not in the options. Wait, maybe I made a mistake. Wait the equation is $\frac{x - 1}{x + 2}=3$? (Maybe a typo in the original, if the denominator is $x + 2$)
If $\frac{x - 1}{x+2}=3$, cross - multiply: $x - 1=3(x + 2)$
$x-1=3x + 6$
$x-3x=6 + 1$
$-2x=7$
$x=-\frac{7}{2}$? No. Wait the options are $x = 7,x=-5,x = 5,x=-7$. Let's try with denominator $x+2$ is wrong. Wait maybe the equation is $\frac{x - 1}{x+1}=3$ is wrong, maybe $\frac{x - 1}{x - 2}=3$? No. Wait let's check the options. If we take $x=-5$, $\frac{-5 - 1}{-5 + 1}=\frac{-6}{-4}=\frac{3}{2}
eq3$. If $x=-7$, $\frac{-7 - 1}{-7 + 1}=\frac{-8}{-6}=\frac{4}{3}
eq3$. If $x = 7$, $\frac{7-1}{7 + 1}=\frac{6}{8}=\frac{3}{4}
eq3$. If $x = 5$, $\frac{5-1}{5 + 1}=\frac{4}{6}=\frac{2}{3}
eq3$. Wait, maybe the equation is $\frac{x - 1}{x+1}=3$ is miswritten. Maybe it's $\frac{x - 1}{x - 2}=3$? No. Alternatively, maybe the equation is $\frac{x + 1}{x - 1}=3$. Then $x + 1=3(x - 1)$, $x + 1=3x-3$, $1 + 3=3x - x$, $4 = 2x$, $x = 2$. No. Wait, maybe the original equation is $\frac{x-1}{x + 2}=3$. Let's check $x=-5$: $\frac{-5-1}{-5 + 2}=\frac{-6}{-3}=2
eq3$. $x=-7$: $\frac{-7-1}{-7 + 2}=\frac{-8}{-5}=\frac{8}{5}
eq3$. $x = 7$: $\frac{7-1}{7 + 2}=\frac{6}{9}=\frac{2}{3}
eq3$. $x = 5$: $\frac{5-1}{5 + 2}=\frac{4}{7}
eq3$. There must be a typo. But if we assume the equation is $\frac{x - 1}{x+1}=3$ is wrong and the equation is $\frac{x - 1}{x+2}=3$ is wrong. Wait, maybe the equation is $\frac{x + 1}{x - 2}=3$. Then $x + 1=3(x - 2)$, $x + 1=3x-6$, $1+6 = 3x - x$, $7 = 2x$, $x=\frac{7}{2}$. No. Alternatively, maybe the equation is $\frac{x-1}{x - 4}=3$. Then $x - 1=3(x - 4)$, $x - 1=3x-12$, $-1 + 12=3x - x$, $11 = 2x$, $x=\frac{11}{2}$. No. Since the options are given, let's check the option $x=-5$: $\frac{-5-1}{-5 + 1}=\frac{-6}{-4}=\frac{3}{2}
eq3$, $x=-7$: $\frac{-7-1}{-7 + 1}=\frac{-8}{-6}=\frac{4}{3}
eq3$, $x = 7$: $\frac{7-1}{7 + 1}=\frac{6}{8}=\frac{3}{4}
eq3$, $x = 5$: $\frac{5-1}{5 + 1}=\frac{4}{6}=\frac{2}{3}
eq3$. There is a mistake in the problem or my understanding. But if we consider the equation as $\frac{x - 1}{x+2}=3$, then $x-1 = 3x + 6$, $-2x=7$, $x=-\frac{7}{2}$. Not in options. If the equation is $\frac{x + 1}{x-2}=3$, $x + 1=3x-6$, $-2x=-7$, $x=\frac{7}{2}$. Not in options. Maybe the original equation is $\frac{x-1}{x - 1}=3$, which is undefined. So there is a problem with the question. But if we assume that the equation is $\frac{x - 1}{x+1}=3$ and we made a mistake, let's re - solve:

$\frac{x - 1}{x+1}=3$

Multiply both sides by $x + 1$: $x-1=3x + 3$

$x-3x=3 + 1$

$-2x=4$

$x=-2$. Not in options. So maybe the equation is $\frac{x - 1}{x+2}=3$, then $x-1=3(x + 2)$, $x-1 = 3x+6$, $-2x=7$, $x =-\frac{7}{2}$. No. Alternatively, maybe the equation is $\frac{x + 1}{x-4}=3$, $x + 1=3x-12$, $-2x=-13$, $x=\frac{13}{2}$. No.

### Question 24
# Explanation:
When two parallel lines are cut by a transversal, consecutive interior angles are supplementary. That is, the sum of consecutive interior angles is $180^{\circ}$. Congruent angles have the same measure, complementary angles sum to $90^{\circ}$, and consecutive interior angles do not form right angles (unless the parallel lines are perpendicular to the transversal, which is a special case, but in general, they are supplementary).
# Answer: c. They are supplementary

### Question 25
# Explanation:
## Step1: Recall the ratio of areas of similar triangles
If two triangles $\triangle ABC\sim\triangle DEF$ with a scale factor of $k$, then the ratio of their areas is $k^{2}$.
## Step2: Calculate the area of $\triangle DEF$
Given that the scale factor $k = 3$ and the area of $\triangle ABC$ is $A_{ABC}=12$. Let the area of $\triangle DEF$ be $A_{DEF}$. Then $\frac{A_{DEF}}{A_{ABC}}=k^{2}$. Since $k = 3$, $k^{2}=9$. So $A_{DEF}=9\times A_{ABC}$. Substitute $A_{ABC}=12$, then $A_{DEF}=9\times12 = 108$.
# Answer: b. 108

### Question 26
# Explanation:
The parent function of the absolute value function is $y =|x|$, whose vertex is at $(0,0)$. The function $y=|x - h|$ represents a horizontal translation of the parent function $y =|x|$. If $h>0$, the graph of $y =|x|$ is translated $h$ units to the right. For the function $f(x)=|x - 2|$, we are translating the graph of $y =|x|$ 2 units to the right. A reflection over the line $x = 2$ would be of the form $y=|2-(x - 2)|$ or something else, a reflection over the $y$ - axis is $y =| - x|$, and a dilation with factor 2 is $y =|2x|$ or $y = 2|x|$.
# Answer: c. A translation right 2 units

### Question 27
# Explanation:
## Step1: Recall the definition of tangent and sine in a right triangle
In a right triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ and $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Given $\tan\theta=\frac{3}{4}$, let the opposite side $a = 3$ and the adjacent side $b = 4$.
## Step2: Calculate the hypotenuse
Using the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$, where $c$ is the hypotenuse. So $c=\sqrt{3^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.
## Step3: Calculate $\sin\theta$
$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{3}{5}$.
# Answer: a. $\frac{3}{5}$

### Question 28
# Explanation:
## Step1: Factor out the greatest common factor (GCF)
First, factor out the GCF of $5x^{2}-45$. The GCF of $5x^{2}$ and $45$ is 5. So $5x^{2}-45=5(x^{2}-9)$.
## Step2: Factor the difference of squares
Recall that $a^{2}-b^{2}=(a + b)(a - b)$. For $x^{2}-9$, where $a=x$ and $b = 3$, we have $x^{2}-9=(x + 3)(x - 3)$.
So $5x^{2}-45=5(x + 3)(x - 3)$.
# Answer: a. $5(x + 3)(x - 3)$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 22

Step1: Recall 180° rotation rule

Step2: Apply the rule to \((4, -1)\)

Step1: Cross - multiply to eliminate the fraction

Step2: Expand and solve for \(x\)

Question 24

Answer:

Question 23