1. decide whether there is enough information to prove that $m \parallel n$. if so, state the theorem you can use.

2. write an equation of the line passing through point $p(-6, 4)$ that is perpendicular to $y + 8 = 2(x - 14)$. graph the equations to check that the lines are perpendicular.

3. graph quadrilateral $abcd$ with vertices $a(-2, 4)$, $b(-1, 6)$, $c(4, 4)$, and $d(2, 3)$ and its image after the translation $(x, y) \to (x + 2, y - 3)$.

4. write an equation for the $n$th term of the arithmetic sequence. then find $a_{10}$.
$7, 3, -1, -5, \dots$

5. solve the equation $50x = 2x^3$.

6. the function $h(x) = -16x^2 + 32x + 4$ models the height $h$ (in feet) of a ball ejected from a ball launcher after $x$ seconds. when is the ball at a height of 15 feet?

Question

1. decide whether there is enough information to prove that $m \parallel n$. if so, state the theorem you can use.

2. write an equation of the line passing through point $p(-6, 4)$ that is perpendicular to $y + 8 = 2(x - 14)$. graph the equations to check that the lines are perpendicular.

3. graph quadrilateral $abcd$ with vertices $a(-2, 4)$, $b(-1, 6)$, $c(4, 4)$, and $d(2, 3)$ and its image after the translation $(x, y) \to (x + 2, y - 3)$.

4. write an equation for the $n$th term of the arithmetic sequence. then find $a_{10}$.
   $7, 3, -1, -5, \dots$

5. solve the equation $50x = 2x^3$.

6. the function $h(x) = -16x^2 + 32x + 4$ models the height $h$ (in feet) of a ball ejected from a ball launcher after $x$ seconds. when is the ball at a height of 15 feet?

Accepted Answer

### Problem 1
# Explanation:
## Step1: Identify the angles
The two angles given are both $138^\circ$ and they are alternate interior angles (or also can be considered as same - side angles in a way that shows parallelism).
## Step2: Apply the theorem
The theorem for proving two lines parallel when alternate interior angles are equal is: If alternate interior angles are congruent, then the two lines are parallel. Since the two $138^\circ$ angles are alternate interior angles (formed by a transversal cutting lines $m$ and $n$) and they are equal, we can prove that $m\parallel n$ using the "Alternate Interior Angles Theorem" (or also, since they are same - side angles and supplementary? Wait, no, $138^\circ+138^\circ = 276^\circ$, no. Wait, actually, the two angles are alternate interior angles. So the theorem is: If alternate interior angles are congruent, then the lines are parallel.

# Answer:
Yes, we can prove $m\parallel n$. The theorem used is the Alternate Interior Angles Theorem (since the alternate interior angles are congruent, each equal to $138^\circ$).

### Problem 2
# Explanation:
## Step1: Find the slope of the given line
The given line is $y + 8=2(x - 14)$. In slope - intercept form $y=mx + b$, where $m$ is the slope, the given line has a slope $m_1 = 2$.
## Step2: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is $- 1$. Let the slope of the perpendicular line be $m_2$. Then $m_1\times m_2=-1$. Substituting $m_1 = 2$, we get $2\times m_2=-1$, so $m_2=-\frac{1}{2}$.
## Step3: Use the point - slope form to find the equation
The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-6,4)$ and $m =-\frac{1}{2}$. Substituting these values, we have $y - 4=-\frac{1}{2}(x + 6)$.
## Step4: Simplify the equation
$y-4=-\frac{1}{2}x-3$, then $y=-\frac{1}{2}x + 1$.

# Answer:
The equation of the line is $y =-\frac{1}{2}x + 1$

### Problem 3
# Explanation:
## Step1: Graph the original quadrilateral
- For point $A(-2,4)$: Move 2 units to the left of the origin on the x - axis and 4 units up on the y - axis.
- For point $B(-1,6)$: Move 1 unit to the left of the origin on the x - axis and 6 units up on the y - axis.
- For point $C(4,4)$: Move 4 units to the right of the origin on the x - axis and 4 units up on the y - axis.
- For point $D(2,3)$: Move 2 units to the right of the origin on the x - axis and 3 units up on the y - axis. Connect the points $A - B - C - D - A$ to get quadrilateral $ABCD$.
## Step2: Apply the translation $(x,y)\to(x + 2,y - 3)$
- For point $A(-2,4)$: New $x=-2 + 2=0$, new $y = 4-3 = 1$, so $A'(0,1)$.
- For point $B(-1,6)$: New $x=-1+2 = 1$, new $y=6 - 3=3$, so $B'(1,3)$.
- For point $C(4,4)$: New $x=4 + 2=6$, new $y=4-3 = 1$, so $C'(6,1)$.
- For point $D(2,3)$: New $x=2+2 = 4$, new $y=3-3 = 0$, so $D'(4,0)$.
## Step3: Graph the image
Plot the points $A'(0,1)$, $B'(1,3)$, $C'(6,1)$, $D'(4,0)$ and connect them to get the image of quadrilateral $ABCD$ after translation.

# Answer:
The original quadrilateral $ABCD$ has vertices $A(-2,4)$, $B(-1,6)$, $C(4,4)$, $D(2,3)$. The image after translation $(x,y)\to(x + 2,y - 3)$ has vertices $A'(0,1)$, $B'(1,3)$, $C'(6,1)$, $D'(4,0)$. (Graphing is a visual task, but the coordinates of the image are as above)

### Problem 4
# Explanation:
## Step1: Recall the formula for the nth term of an arithmetic sequence
The formula for the $n$th term of an arithmetic sequence is $a_n=a_1+(n - 1)d$, where $a_1$ is the first term and $d$ is the common difference.
## Step2: Find $a_1$ and $d$
For the sequence $7,3,-1,-5,\cdots$, $a_1 = 7$. The common difference $d=3 - 7=-4$ (or $d=-1 - 3=-4$, etc.).
## Step3: Write the nth term formula
Substitute $a_1 = 7$ and $d=-4$ into the formula: $a_n=7+(n - 1)(-4)=7-4n + 4=11-4n$.
## Step4: Find $a_{10}$
Substitute $n = 10$ into $a_n=11-4n$: $a_{10}=11-4\times10=11 - 40=-29$.

# Answer:
The equation for the $n$th term is $a_n = 11-4n$ and $a_{10}=-29$

### Problem 5
# Explanation:
## Step1: Rearrange the equation
Start with $50x=2x^3$. Subtract $50x$ from both sides to get $2x^3-50x = 0$.
## Step2: Factor out the common factor
Factor out $2x$ from the left - hand side: $2x(x^2 - 25)=0$.
## Step3: Factor the difference of squares
We know that $x^2 - 25=(x - 5)(x + 5)$ (since $a^2-b^2=(a - b)(a + b)$ with $a = x$ and $b = 5$). So the equation becomes $2x(x - 5)(x + 5)=0$.
## Step4: Solve for $x$
Set each factor equal to zero:
    - $2x=0$ gives $x = 0$.
    - $x - 5=0$ gives $x = 5$.
    - $x+5=0$ gives $x=-5$.

# Answer:
The solutions of the equation $50x = 2x^3$ are $x = 0$, $x = 5$ and $x=-5$

### Problem 6
# Explanation:
## Step1: Set up the equation
We know that $h(x)=-16x^2 + 32x+4$ and we want to find $x$ when $h(x)=15$. So we set up the equation $-16x^2+32x + 4=15$.
## Step2: Rearrange the equation
Subtract 15 from both sides: $-16x^2+32x + 4-15 = 0$, which simplifies to $-16x^2+32x-11 = 0$. Multiply both sides by $-1$ to get $16x^2-32x + 11 = 0$.
## Step3: Use the quadratic formula
The quadratic formula is $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for the quadratic equation $ax^2+bx + c = 0$. Here, $a = 16$, $b=-32$, $c = 11$.
First, calculate the discriminant $\Delta=b^2-4ac=(-32)^2-4\times16\times11=1024 - 704 = 320$.
Then $x=\frac{32\pm\sqrt{320}}{32}=\frac{32\pm8\sqrt{5}}{32}=\frac{4\pm\sqrt{5}}{4}=1\pm\frac{\sqrt{5}}{4}$.
$\sqrt{5}\approx2.24$, so $x = 1+\frac{2.24}{4}=1 + 0.56 = 1.56$ and $x=1-\frac{2.24}{4}=1 - 0.56 = 0.44$ (approximate values).

# Answer:
The ball is at a height of 15 feet at $x = 1+\frac{\sqrt{5}}{4}$ seconds (or approximately $1.56$ seconds) and $x = 1-\frac{\sqrt{5}}{4}$ seconds (or approximately $0.44$ seconds).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Identify the angles

Step2: Apply the theorem

Step1: Find the slope of the given line

Step2: Find the slope of the perpendicular line

Step3: Use the point - slope form to find the equation

Step4: Simplify the equation

Step1: Graph the original quadrilateral

Step2: Apply the translation \((x,y)\to(x + 2,y - 3)\)

Step3: Graph the image

Answer:

Problem 2