remember
for problems 11–14, evaluate the expression for the given value.
11. $2(x + 4)$ when $x = 6$
12. $5(x + 9)$ when $x = 3$
13. $-4(x + 8)$ when $x = -8$
14. $-7(x - 3)$ when $x = -5$
15. draw the image of figure $ab$ under a dilation with center $o$ and scale factor $2$.

16. a small bottle of soap costs $\$8.00$ for $40$ fluid ounces. a large bottle of soap costs $\$19.20$ for $120$ fluid ounces. what is the unit cost for each bottle size?

Question

remember
for problems 11–14, evaluate the expression for the given value.
11. $2(x + 4)$ when $x = 6$
12. $5(x + 9)$ when $x = 3$
13. $-4(x + 8)$ when $x = -8$
14. $-7(x - 3)$ when $x = -5$
15. draw the image of figure $ab$ under a dilation with center $o$ and scale factor $2$.

16. a small bottle of soap costs $\$8.00$ for $40$ fluid ounces. a large bottle of soap costs $\$19.20$ for $120$ fluid ounces. what is the unit cost for each bottle size?

Accepted Answer

### Problem 11: $ 2(x + 4) $ when $ x = 6 $
# Explanation:
## Step 1: Substitute $ x = 6 $ into the expression
Substitute $ x = 6 $ into $ 2(x + 4) $, we get $ 2(6 + 4) $.
## Step 2: Simplify the expression inside the parentheses
Calculate $ 6 + 4 = 10 $, so the expression becomes $ 2\times10 $.
## Step 3: Multiply
$ 2\times10 = 20 $.
# Answer: 20

### Problem 12: $ 5(x + 9) $ when $ x = 3 $
# Explanation:
## Step 1: Substitute $ x = 3 $ into the expression
Substitute $ x = 3 $ into $ 5(x + 9) $, we get $ 5(3 + 9) $.
## Step 2: Simplify the expression inside the parentheses
Calculate $ 3 + 9 = 12 $, so the expression becomes $ 5\times12 $.
## Step 3: Multiply
$ 5\times12 = 60 $.
# Answer: 60

### Problem 13: $ -4(x + 8) $ when $ x = -8 $
# Explanation:
## Step 1: Substitute $ x = -8 $ into the expression
Substitute $ x = -8 $ into $ -4(x + 8) $, we get $ -4(-8 + 8) $.
## Step 2: Simplify the expression inside the parentheses
Calculate $ -8 + 8 = 0 $, so the expression becomes $ -4\times0 $.
## Step 3: Multiply
$ -4\times0 = 0 $.
# Answer: 0

### Problem 14: $ -7(x - 3) $ when $ x = -5 $
# Explanation:
## Step 1: Substitute $ x = -5 $ into the expression
Substitute $ x = -5 $ into $ -7(x - 3) $, we get $ -7(-5 - 3) $.
## Step 2: Simplify the expression inside the parentheses
Calculate $ -5 - 3 = -8 $, so the expression becomes $ -7\times(-8) $.
## Step 3: Multiply
$ -7\times(-8) = 56 $.
# Answer: 56

### Problem 15: Draw the image of figure $ AB $ under a dilation with center $ O $ and scale factor 2.
# Explanation:
## Step 1: Recall the definition of dilation
Dilation with center $ O $ and scale factor $ k $ means that for any point $ P $ in the figure, the image $ P' $ satisfies $ \overrightarrow{OP'} = k\overrightarrow{OP} $. So we need to find the images of points $ A $ and $ B $ under this dilation.
## Step 2: Find the image of point $ A $
Let the vector from $ O $ to $ A $ be $ \overrightarrow{OA} $. Then the vector from $ O $ to the image of $ A $ (let's call it $ A' $) is $ 2\overrightarrow{OA} $. So we draw a line from $ O $ through $ A $, and extend it so that the length from $ O $ to $ A' $ is twice the length from $ O $ to $ A $.
## Step 3: Find the image of point $ B $
Similarly, for point $ B $, let the vector from $ O $ to $ B $ be $ \overrightarrow{OB} $. The vector from $ O $ to the image of $ B $ (let's call it $ B' $) is $ 2\overrightarrow{OB} $. We draw a line from $ O $ through $ B $, and extend it so that the length from $ O $ to $ B' $ is twice the length from $ O $ to $ B $.
## Step 4: Draw the image of the arc $ AB $
After finding $ A' $ and $ B' $, we draw an arc from $ A' $ to $ B' $ that is similar in shape to the original arc $ AB $ (since dilation preserves the shape of the figure, just scales the size) with center $ O $ and scale factor 2.
# Answer: The image of arc $ AB $ under dilation with center $ O $ and scale factor 2 is an arc $ A'B' $ where $ A' $ and $ B' $ are located such that $ OA' = 2OA $ and $ OB' = 2OB $, and the arc $ A'B' $ is a scaled - up version of arc $ AB $ with the same curvature (since dilation is a similarity transformation).

### Problem 16: Unit cost for each bottle size
# Explanation:
## Step 1: Find the unit cost for the small bottle
The unit cost is calculated by dividing the total cost by the number of fluid ounces. For the small bottle, the cost is $ \$8.00 $ and the volume is 40 fluid ounces. So the unit cost $ c_{small}=\frac{8.00}{40} $.
## Step 2: Calculate the unit cost for the small bottle
$ \frac{8.00}{40}=0.20 $ dollars per fluid ounce.
## Step 3: Find the unit cost for the large bottle
For the large bottle, the cost is $ \$19.20 $ and the volume is 120 fluid ounces. So the unit cost $ c_{large}=\frac{19.20}{120} $.
## Step 4: Calculate the unit cost for the large bottle
$ \frac{19.20}{120} = 0.16 $ dollars per fluid ounce.
# Answer: The unit cost for the small bottle is $ \$0.20 $ per fluid ounce, and the unit cost for the large bottle is $ \$0.16 $ per fluid ounce.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 11: \( 2(x + 4) \) when \( x = 6 \)

Step 1: Substitute \( x = 6 \) into the expression

Step 2: Simplify the expression inside the parentheses

Step 3: Multiply

Step 1: Substitute \( x = 3 \) into the expression

Step 2: Simplify the expression inside the parentheses

Step 3: Multiply

Step 1: Substitute \( x = -8 \) into the expression

Step 2: Simplify the expression inside the parentheses

Step 3: Multiply

Answer:

Problem 12: \( 5(x + 9) \) when \( x = 3 \)