evaluate the function w(t) = -2t + 1; find w(-7)
is this a function? why or why not?
graph
no, because its not straight
is this a function? why or why not?
| x | y |
| 9 | -18 |
| -20 | 0 |
| -6 | 1 |
| -17 | 16 |
| 9 | 17 |
| 11 | 19 |
explain two ways you can solve 20 = 5(-3 + x)
graph this equation y = -1/3x + 4
graph
graph this equation y - 2 = 2(x + 3)
graph
use these two points to write an equation in point slope form
(3, -7) and (-12, 3)
solve for k -4k + 2(5k - 6) = -3k - 39
what are three things you are probably going to forget for the test:
make a plan to use the quizlet on oncampus to review those things
love you lots! youll be fine

Question

evaluate the function w(t) = -2t + 1; find w(-7)
is this a function? why or why not?
graph
no, because its not straight
is this a function? why or why not?
| x | y |
| 9 | -18 |
| -20 | 0 |
| -6 | 1 |
| -17 | 16 |
| 9 | 17 |
| 11 | 19 |
explain two ways you can solve 20 = 5(-3 + x)
graph this equation y = -1/3x + 4
graph
graph this equation y - 2 = 2(x + 3)
graph
use these two points to write an equation in point slope form
(3, -7) and (-12, 3)
solve for k -4k + 2(5k - 6) = -3k - 39
what are three things you are probably going to forget for the test:
make a plan to use the quizlet on oncampus to review those things
love you lots! youll be fine <3

Accepted Answer

### First Sub - Question: Evaluate $ w(t)=-2t + 1 $; Find $ w(-7) $
# Explanation:
## Step 1: Substitute $ t=-7 $ into the function
We have the function $ w(t)=-2t + 1 $. To find $ w(-7) $, we replace $ t $ with $ -7 $ in the function. So we get $ w(-7)=-2\times(-7)+1 $.
## Step 2: Simplify the expression
First, calculate $ -2\times(-7) $. According to the rule of multiplying two negative numbers, the result is positive, so $ -2\times(-7) = 14 $. Then we add 1 to this result: $ 14 + 1=15 $.
# Answer:
$ 15 $

### Second Sub - Question: Is the graph a function? Why or why not?
# Brief Explanations:
To determine if a graph represents a function, we use the vertical line test. A vertical line test states that if any vertical line drawn through the graph intersects the graph at more than one point, then the graph does not represent a function. Looking at the given graph, we can see that there exists a vertical line that will intersect the graph at two points (since the graph is a curve that fails the vertical line test). So, it is not a function because it fails the vertical line test (a vertical line can intersect it more than once).
# Answer:
No, because it fails the vertical line test (a vertical line can intersect it more than once).

### Third Sub - Question: Is the table a function? Why or why not?
# Brief Explanations:
A relation is a function if each input (x - value) has exactly one output (y - value). In the given table, the x - value $ 9 $ is associated with two different y - values, $ - 18 $ and $ 17 $. Since an input has more than one output, this relation is not a function.
# Answer:
No, because the input $ x = 9 $ has two different outputs ($ y=-18 $ and $ y = 17 $), so it does not satisfy the definition of a function (each input must have exactly one output).

### Fourth Sub - Question: Explain two ways to solve $ 20=5(-3 + x) $
# Explanation:
## Method 1: Using the distributive property first
### Step 1: Distribute the 5
We start with the equation $ 20 = 5(-3+x) $. Using the distributive property $ a(b + c)=ab+ac $, we get $ 20=5\times(-3)+5\times x $, which simplifies to $ 20=-15 + 5x $.
### Step 2: Solve for $ x $
Add 15 to both sides of the equation: $ 20 + 15=-15+15 + 5x $, so $ 35 = 5x $. Then divide both sides by 5: $ \frac{35}{5}=\frac{5x}{5} $, which gives $ x = 7 $.

## Method 2: Dividing both sides by 5 first
### Step 1: Divide both sides by 5
Start with $ 20=5(-3 + x) $. Divide both sides of the equation by 5: $ \frac{20}{5}=\frac{5(-3 + x)}{5} $. This simplifies to $ 4=-3 + x $.
### Step 2: Solve for $ x $
Add 3 to both sides of the equation: $ 4+3=-3 + 3+x $, so $ x = 7 $.
# Answer:
Two ways: 1. Distribute first: $ 20=-15 + 5x\Rightarrow35 = 5x\Rightarrow x = 7 $; 2. Divide by 5 first: $ 4=-3 + x\Rightarrow x = 7 $.

### Fifth Sub - Question: Graph $ y=-\frac{1}{3}x + 4 $
# Explanation:
## Step 1: Identify the slope and y - intercept
The equation $ y=-\frac{1}{3}x + 4 $ is in slope - intercept form $ y=mx + b $, where $ m $ is the slope and $ b $ is the y - intercept. Here, the y - intercept $ b = 4 $, so we plot the point $ (0,4) $ on the y - axis. The slope $ m=-\frac{1}{3} $, which means for every 3 units we move to the right (increase in x by 3), we move down 1 unit (decrease in y by 1).
## Step 2: Plot the second point
From the point $ (0,4) $, using the slope, we move 3 units to the right (to $ x = 3 $) and 1 unit down (to $ y=4 - 1=3 $). So we plot the point $ (3,3) $.
## Step 3: Draw the line
Draw a straight line through the points $ (0,4) $ and $ (3,3) $. This line represents the graph of the equation $ y=-\frac{1}{3}x + 4 $.
# Answer:
The graph is a straight line with y - intercept at $ (0,4) $ and passing through $ (3,3) $ (drawn using the slope - intercept method as described).

### Sixth Sub - Question: Graph $ y - 2=2(x + 3) $
# Explanation:
## Step 1: Rewrite in slope - intercept form (optional, but helpful for graphing)
First, we expand the right - hand side: $ y-2=2x+6 $. Then we add 2 to both sides to get $ y=2x + 8 $. Now, the equation is in slope - intercept form $ y=mx + b $, where the y - intercept $ b = 8 $ (so we plot the point $ (0,8) $) and the slope $ m = 2 $ (which means for every 1 unit we move to the right (increase in x by 1), we move up 2 units (increase in y by 2)).
## Step 2: Plot the second point
From the point $ (0,8) $, using the slope, we move 1 unit to the right (to $ x = 1 $) and 2 units up (to $ y=8 + 2=10 $). So we plot the point $ (1,10) $.
## Step 3: Draw the line
Draw a straight line through the points $ (0,8) $ and $ (1,10) $. This line represents the graph of the equation $ y - 2=2(x + 3) $ (or $ y=2x + 8 $).
# Answer:
The graph is a straight line with y - intercept at $ (0,8) $ and passing through $ (1,10) $ (drawn using the slope - intercept method as described).

### Seventh Sub - Question: Use the points $ (3,-7) $ and $ (-12,3) $ to write an equation in point - slope form
# Explanation:
## Step 1: Find the slope $ m $
The formula for the slope between two points $ (x_1,y_1) $ and $ (x_2,y_2) $ is $ m=\frac{y_2 - y_1}{x_2 - x_1} $. Let $ (x_1,y_1)=(3,-7) $ and $ (x_2,y_2)=(-12,3) $. Then $ m=\frac{3-(-7)}{-12 - 3}=\frac{3 + 7}{-15}=\frac{10}{-15}=-\frac{2}{3} $.
## Step 2: Write the point - slope form
The point - slope form of a line is $ y - y_1=m(x - x_1) $. We can use either of the two points. Let's use the point $ (3,-7) $. Substituting $ x_1 = 3 $, $ y_1=-7 $, and $ m=-\frac{2}{3} $ into the point - slope form, we get $ y-(-7)=-\frac{2}{3}(x - 3) $, which simplifies to $ y + 7=-\frac{2}{3}(x - 3) $. If we use the point $ (-12,3) $, we get $ y - 3=-\frac{2}{3}(x+12) $.
# Answer:
Using the point $ (3,-7) $, the equation is $ y + 7=-\frac{2}{3}(x - 3) $; using the point $ (-12,3) $, the equation is $ y - 3=-\frac{2}{3}(x + 12) $ (either form is correct for point - slope form).

### Eighth Sub - Question: Solve for $ k $ in $ -4k+2(5k - 6)=-3k - 39 $
# Explanation:
## Step 1: Expand the left - hand side
First, we use the distributive property to expand $ 2(5k - 6) $. We get $ 2\times5k-2\times6 = 10k-12 $. So the equation becomes $ -4k + 10k-12=-3k - 39 $.
## Step 2: Combine like terms on the left - hand side
Combine $ -4k $ and $ 10k $: $ (-4k+10k)-12=-3k - 39 $, which simplifies to $ 6k-12=-3k - 39 $.
## Step 3: Add $ 3k $ to both sides
Adding $ 3k $ to both sides gives $ 6k+3k-12=-3k + 3k-39 $, so $ 9k-12=-39 $.
## Step 4: Add 12 to both sides
Adding 12 to both sides: $ 9k-12 + 12=-39 + 12 $, which simplifies to $ 9k=-27 $.
## Step 5: Divide both sides by 9
Dividing both sides by 9: $ \frac{9k}{9}=\frac{-27}{9} $, so $ k=-3 $.
# Answer:
$ k=-3 $

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QUESTION IMAGE

Question

Explanation:

First Sub - Question: Evaluate \( w(t)=-2t + 1 \); Find \( w(-7) \)

Step 1: Substitute \( t=-7 \) into the function

Step 2: Simplify the expression

Answer:

Second Sub - Question: Is the graph a function? Why or why not?