remember
for problems 11–14, evaluate the expression for the given value.
11. $0.4x + 5$ when $x = 0.8$
12. $1.2x - 5$ when $x = 0.3$
13. $0.6x + 4$ when $x = -1.4$
14. $6.2x - 9$ when $x = -0.8$
15. the diagram shows a dilation with center $o$ and scale factor $\frac{7}{3}$.
diagram of dilation with center o, points p, q on a line segment, p, q on a longer line segment
a. place a point $r$ on $overline{pq}$. draw and label the image of point $r$ under a dilation with center $o$ and scale factor $\frac{7}{3}$.
b. how do you know the image of point $r$ is in the correct location?

Question

Accepted Answer

### Problem 11: Evaluate $ 0.4x + 5 $ when $ x = 0.8 $
# Explanation:
## Step 1: Substitute $ x = 0.8 $ into the expression
$ 0.4(0.8) + 5 $
## Step 2: Multiply $ 0.4 $ and $ 0.8 $
$ 0.4 \times 0.8 = 0.32 $
## Step 3: Add the result to 5
$ 0.32 + 5 = 5.32 $
# Answer:
$ 5.32 $

### Problem 12: Evaluate $ 1.2x - 5 $ when $ x = 0.3 $
# Explanation:
## Step 1: Substitute $ x = 0.3 $ into the expression
$ 1.2(0.3) - 5 $
## Step 2: Multiply $ 1.2 $ and $ 0.3 $
$ 1.2 \times 0.3 = 0.36 $
## Step 3: Subtract 5 from the result
$ 0.36 - 5 = -4.64 $
# Answer:
$ -4.64 $

### Problem 13: Evaluate $ 0.6x + 4 $ when $ x = -1.4 $
# Explanation:
## Step 1: Substitute $ x = -1.4 $ into the expression
$ 0.6(-1.4) + 4 $
## Step 2: Multiply $ 0.6 $ and $ -1.4 $
$ 0.6 \times (-1.4) = -0.84 $
## Step 3: Add the result to 4
$ -0.84 + 4 = 3.16 $
# Answer:
$ 3.16 $

### Problem 14: Evaluate $ 6.2x - 9 $ when $ x = -0.8 $
# Explanation:
## Step 1: Substitute $ x = -0.8 $ into the expression
$ 6.2(-0.8) - 9 $
## Step 2: Multiply $ 6.2 $ and $ -0.8 $
$ 6.2 \times (-0.8) = -4.96 $
## Step 3: Subtract 9 from the result
$ -4.96 - 9 = -13.96 $
# Answer:
$ -13.96 $

### Problem 15a: Dilation of Point $ R $ on $ \overline{PQ} $
# Brief Explanations:
To find the image of point $ R $ under dilation with center $ O $ and scale factor $ \frac{7}{3} $:
1. Draw a line from $ O $ through $ R $ (since dilation preserves collinearity with the center).
2. Measure the distance from $ O $ to $ R $, then multiply that distance by $ \frac{7}{3} $ to find the distance from $ O $ to $ R' $ (the image of $ R $).
3. Mark the point $ R' $ on the line $ OR $ at the calculated distance. $ R' $ should lie on the same line as $ O $ and $ R $, and the ratio $ \frac{OR'}{OR} = \frac{7}{3} $.
# Answer:
(Diagrammatic answer: Draw line $ OR $, extend it, and mark $ R' $ such that $ OR' = \frac{7}{3}OR $.)

### Problem 15b: Justification for Location of $ R' $
# Brief Explanations:
By the definition of dilation, a dilation with center $ O $ and scale factor $ k $ maps a point $ R $ to a point $ R' $ such that $ \overrightarrow{OR'} = k \cdot \overrightarrow{OR} $. This means $ R' $ lies on the line through $ O $ and $ R $ (collinearity) and the distance from $ O $ to $ R' $ is $ |k| $ times the distance from $ O $ to $ R $. For scale factor $ \frac{7}{3} $ (positive, so same direction from $ O $ as $ R $), $ R' $ must be on line $ OR $ with $ \frac{OR'}{OR} = \frac{7}{3} $, confirming its location.
# Answer:
Dilation preserves collinearity with the center, so $ R' $ lies on $ \overleftrightarrow{OR} $, and $ \frac{OR'}{OR} = \frac{7}{3} $ (by dilation definition), so its location is correct.

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QUESTION IMAGE

Question

Explanation:

Problem 11: Evaluate \( 0.4x + 5 \) when \( x = 0.8 \)

Step 1: Substitute \( x = 0.8 \) into the expression

Step 2: Multiply \( 0.4 \) and \( 0.8 \)

Step 3: Add the result to 5

Step 1: Substitute \( x = 0.3 \) into the expression

Step 2: Multiply \( 1.2 \) and \( 0.3 \)

Step 3: Subtract 5 from the result

Step 1: Substitute \( x = -1.4 \) into the expression

Step 2: Multiply \( 0.6 \) and \( -1.4 \)

Step 3: Add the result to 4

Answer:

Problem 12: Evaluate \( 1.2x - 5 \) when \( x = 0.3 \)