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### Problem 11
# Explanation:
## Step1: Recall parallelogram diagonals property
In a parallelogram, diagonals bisect each other. So, $ WQ=\frac{1}{2}WY $ and $ QX = \frac{1}{2}XZ $, also $ WQ=\frac{1}{2}WY $ can be used to find $ x $, then find $ XZ $ and then $ QX $.

First, find $ x $ using $ WQ=\frac{1}{2}WY $. Given $ WQ = 2x - 6 $ and $ WY=24 $, so:
$ 2x - 6=\frac{1}{2}\times24 $

## Step2: Solve for $ x $
Simplify the right - hand side: $ \frac{1}{2}\times24 = 12 $
So the equation is $ 2x-6 = 12 $
Add 6 to both sides: $ 2x=12 + 6=18 $
Divide both sides by 2: $ x=\frac{18}{2}=9 $

## Step3: Find $ XZ $
Given $ XZ = 8x+16 $, substitute $ x = 9 $:
$ XZ=8\times9 + 16=72+16 = 88 $

## Step4: Find $ QX $
Since in a parallelogram diagonals bisect each other, $ QX=\frac{1}{2}XZ $
$ QX=\frac{1}{2}\times88 = 44 $

# Answer:
$ 44 $

### Problem 12
#### Part (a)
# Explanation:
## Step1: Recall slope formula
The slope of a line passing through two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is given by $ m=\frac{y_2 - y_1}{x_2 - x_1} $

For $ \overline{AB} $, $ A(-13,5) $ and $ B(-1,9) $, so $ x_1=-13,y_1 = 5,x_2=-1,y_2 = 9 $

## Step2: Calculate the slope
$ m_{AB}=\frac{9 - 5}{-1-(-13)}=\frac{4}{-1 + 13}=\frac{4}{12}=\frac{1}{3}\approx0.3333 $

# Answer:
$ \frac{1}{3}$ (or approximately $ 0.3333 $)

#### Part (b)
# Explanation:
## Step1: Use slope formula
For $ \overline{BC} $, $ B(-1,9) $ and $ C(11,5) $, $ x_1=-1,y_1 = 9,x_2 = 11,y_2=5 $

## Step2: Calculate the slope
$ m_{BC}=\frac{5 - 9}{11-(-1)}=\frac{-4}{12}=-\frac{1}{3}\approx - 0.3333 $

# Answer:
$ -\frac{1}{3}$ (or approximately $ - 0.3333 $)

#### Part (c)
# Explanation:
## Step1: Use slope formula
For $ \overline{CD} $, $ C(11,5) $ and $ D(-1,1) $, $ x_1 = 11,y_1=5,x_2=-1,y_2 = 1 $

## Step2: Calculate the slope
$ m_{CD}=\frac{1 - 5}{-1-11}=\frac{-4}{-12}=\frac{1}{3}\approx0.3333 $

# Answer:
$ \frac{1}{3}$ (or approximately $ 0.3333 $)

#### Part (d)
# Explanation:
## Step1: Use slope formula
For $ \overline{AD} $, $ A(-13,5) $ and $ D(-1,1) $, $ x_1=-13,y_1 = 5,x_2=-1,y_2 = 1 $

## Step2: Calculate the slope
$ m_{AD}=\frac{1 - 5}{-1-(-13)}=\frac{-4}{12}=-\frac{1}{3}\approx - 0.3333 $

# Answer:
$ -\frac{1}{3}$ (or approximately $ - 0.3333 $)

#### Part (e)
# Explanation:
## Step1: Recall distance formula
The distance between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is $ d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} $

For $ \overline{AB} $, $ A(-13,5) $ and $ B(-1,9) $, $ x_1=-13,y_1 = 5,x_2=-1,y_2 = 9 $

## Step2: Calculate the distance
$ AB=\sqrt{(-1-(-13))^2+(9 - 5)^2}=\sqrt{(12)^2+(4)^2}=\sqrt{144 + 16}=\sqrt{160}=4\sqrt{10}\approx12.6491 $

# Answer:
$ 4\sqrt{10}$ (or approximately $ 12.6491 $)

#### Part (f)
# Explanation:
## Step1: Use distance formula
For $ \overline{BC} $, $ B(-1,9) $ and $ C(11,5) $, $ x_1=-1,y_1 = 9,x_2 = 11,y_2=5 $

## Step2: Calculate the distance
$ BC=\sqrt{(11-(-1))^2+(5 - 9)^2}=\sqrt{(12)^2+(-4)^2}=\sqrt{144 + 16}=\sqrt{160}=4\sqrt{10}\approx12.6491 $

# Answer:
$ 4\sqrt{10}$ (or approximately $ 12.6491 $)

#### Part (g)
# Explanation:
## Step1: Use distance formula
For $ \overline{CD} $, $ C(11,5) $ and $ D(-1,1) $, $ x_1 = 11,y_1=5,x_2=-1,y_2 = 1 $

## Step2: Calculate the distance
$ CD=\sqrt{(-1 - 11)^2+(1 - 5)^2}=\sqrt{(-12)^2+(-4)^2}=\sqrt{144 + 16}=\sqrt{160}=4\sqrt{10}\approx12.6491 $

# Answer:
$ 4\sqrt{10}$ (or approximately $ 12.6491 $)

#### Part (h)
# Explanation:
## Step1: Use distance formula
For $ \overline{AD} $, $ A(-13,5) $ and $ D(-1,1) $, $ x_1=-13,y_1 = 5,x_2=-1,y_2 = 1 $

## Step2: Calculate the distance
$ AD=\sqrt{(-1-(-13))^2+(1 - 5)^2}=\sqrt{(12)^2+(-4)^2}=\sqrt{144 + 16}=\sqrt{160}=4\sqrt{10}\approx12.6491 $

# Answer:
$ 4\sqrt{10}$ (or approximately $ 12.6491 $)

#### Part (i)
# Explanation:
- From the slope calculations: $ m_{AB}=m_{CD}=\frac{1}{3} $ and $ m_{BC}=m_{AD}=-\frac{1}{3} $, so opposite sides are parallel.
- From the distance calculations: $ AB = BC=CD = AD=4\sqrt{10} $, so all sides are equal.

A quadrilateral with all sides equal and opposite sides parallel is a rhombus. Also, since the slopes of adjacent sides are negative reciprocals? Wait, no, $ m_{AB}=\frac{1}{3} $ and $ m_{BC}=-\frac{1}{3} $, the product of slopes of adjacent sides $ \frac{1}{3}\times(-\frac{1}{3})=-\frac{1}{9}
eq - 1 $, but wait, let's check the angles. Wait, actually, since all sides are equal and opposite sides are parallel, it is a rhombus. Also, let's check the diagonals. The mid - point of $ AC $: $ A(-13,5) $, $ C(11,5) $, mid - point is $ (\frac{-13 + 11}{2},\frac{5+5}{2})=(-1,5) $. The mid - point of $ BD $: $ B(-1,9) $, $ D(-1,1) $, mid - point is $ (-1,\frac{9 + 1}{2})=(-1,5) $. So diagonals bisect each other, which is consistent with parallelogram. And since all sides are equal, it is a rhombus. Also, let's check the length of diagonals. Length of $ AC $: distance between $ A(-13,5) $ and $ C(11,5) $ is $ |11-(-13)| = 24 $. Length of $ BD $: distance between $ B(-1,9) $ and $ D(-1,1) $ is $ |9 - 1|=8 $. The diagonals are not equal (24≠8), so it is not a square. So the most precise name is a rhombus.

# Answer:
Rhombus

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QUESTION IMAGE

Question

Explanation:

Problem 11

Step1: Recall parallelogram diagonals property

Step2: Solve for \( x \)

Step3: Find \( XZ \)

Step4: Find \( QX \)

Step1: Recall slope formula

Step2: Calculate the slope

Step1: Use slope formula

Step2: Calculate the slope

Answer:

Problem 12

Part (a)