1. if ln = 8x - 6, lm = x + 7, and mn = 5x + 11, find x and mn.
3. given the points below, find xy. round to the nearest hundredths. x(3, 2) and y(6, 3)
4. find the midpoint of $overline{ab}$ if a(-2, 11) and b(-6, -7)
5. if m is the midpoint of $overline{ab}$, find the coordinates of b if a(11, -2) and m(5, 2)
6. if line n bisects $overline{ce}$, find x and cd.
7. find the value of x.
8. find the value of x.

Question

Accepted Answer

### 1.
# Explanation:
## Step1: Use segment - addition postulate
Since $LN = LM+MN$, we substitute the given expressions: $8x - 6=(x + 7)+(5x + 11)$.
## Step2: Simplify the right - hand side
$(x + 7)+(5x + 11)=x+5x + 7 + 11=6x+18$. So, $8x - 6=6x + 18$.
## Step3: Solve for $x$
Subtract $6x$ from both sides: $8x-6x - 6=6x-6x + 18$, which gives $2x-6 = 18$. Then add 6 to both sides: $2x-6 + 6=18 + 6$, so $2x=24$. Divide both sides by 2: $x = 12$.
## Step4: Find $MN$
Substitute $x = 12$ into the expression for $MN$: $MN=5x + 11=5\times12+11=60 + 11=71$.
# Answer:
$x = 12$, $MN = 71$

### 2. (Since no graph is provided, we'll skip this one for now. If you can provide the graph details, we can solve it.)

### 3.
# Explanation:
## Step1: Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$
Here, $x_1 = 3,y_1 = 2,x_2 = 6,y_2 = 3$.
## Step2: Substitute the values into the formula
$XY=\sqrt{(6 - 3)^2+(3 - 2)^2}=\sqrt{3^2+1^2}=\sqrt{9 + 1}=\sqrt{10}\approx3.16$.
# Answer:
$XY\approx3.16$

### 4.
# Explanation:
## Step1: Use the mid - point formula $(\frac{x_1+x_2}{2},\frac{y_1 + y_2}{2})$
Here, $x_1=-2,y_1 = 11,x_2=-6,y_2=-7$.
## Step2: Calculate the coordinates of the mid - point
$\frac{-2+( - 6)}{2}=\frac{-2-6}{2}=\frac{-8}{2}=-4$ and $\frac{11+( - 7)}{2}=\frac{11 - 7}{2}=\frac{4}{2}=2$.
# Answer:
$(-4,2)$

### 5.
# Explanation:
## Step1: Use the mid - point formula $(\frac{x_A+x_B}{2},\frac{y_A + y_B}{2})=(x_M,y_M)$
We know $x_A = 11,y_A=-2,x_M = 5,y_M = 2$. For the $x$ - coordinate: $\frac{11+x_B}{2}=5$, multiply both sides by 2: $11+x_B = 10$, then $x_B=-1$. For the $y$ - coordinate: $\frac{-2+y_B}{2}=2$, multiply both sides by 2: $-2+y_B = 4$, then $y_B = 6$.
# Answer:
$(-1,6)$

### 6.
# Explanation:
## Step1: Since line $n$ bisects $\overline{CE}$, then $CD=DE$
So, $x + 3=4x-21$.
## Step2: Solve for $x$
Subtract $x$ from both sides: $x - x+3=4x-x-21$, which gives $3 = 3x-21$. Add 21 to both sides: $3+21=3x-21 + 21$, so $24 = 3x$. Divide both sides by 3: $x = 8$.
## Step3: Find $CD$
Substitute $x = 8$ into the expression for $CD$: $CD=x + 3=8 + 3=11$.
# Answer:
$x = 8$, $CD = 11$

### 7.
# Explanation:
## Step1: Since the angles are vertical angles, they are equal
So, $9x-6=6x - 9$.
## Step2: Solve for $x$
Subtract $6x$ from both sides: $9x-6x-6=6x-6x - 9$, which gives $3x-6=-9$. Add 6 to both sides: $3x-6 + 6=-9 + 6$, so $3x=-3$. Divide both sides by 3: $x=-1$.
# Answer:
$x=-1$

### 8.
# Explanation:
## Step1: Since the angles are vertical angles, they are equal
So, $5x-8=8x-71$.
## Step2: Solve for $x$
Subtract $5x$ from both sides: $5x-5x-8=8x-5x-71$, which gives $-8 = 3x-71$. Add 71 to both sides: $-8+71=3x-71 + 71$, so $63 = 3x$. Divide both sides by 3: $x = 21$.
# Answer:
$x = 21$

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QUESTION IMAGE

Question

Explanation:

1.

Step1: Use segment - addition postulate

Step2: Simplify the right - hand side

Step3: Solve for \(x\)

Step4: Find \(MN\)

Step1: Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

Step2: Substitute the values into the formula

Step1: Use the mid - point formula \((\frac{x_1+x_2}{2},\frac{y_1 + y_2}{2})\)

Step2: Calculate the coordinates of the mid - point

Answer:

2. (Since no graph is provided, we'll skip this one for now. If you can provide the graph details, we can solve it.)

3.