Question
Which equation of a line is parallel to the equation -4x + 3y = 12?(1 point)
Responses
y=−4/3x + 4
y = 3x + 12
y = 4x + 8
y =4/3x + 6
Responses
y=−4/3x + 4
y = 3x + 12
y = 4x + 8
y =4/3x + 6
Answers
Answer
Which of the following is an equation of a line that is perpendicular to y = -2x + 5 and passes through the point (-2, 3)?(1 point)
Responses
y = -2x - 1
y = 1/2x + 4
y = −1/2x + 2
y = 2x + 7
Responses
y = -2x - 1
y = 1/2x + 4
y = −1/2x + 2
y = 2x + 7
Answer
Find the missing side lengths, perimeter, and area of triangle ABC on the coordinate plane. All answers should be in simplest radical form, if necessary:
A (-1, 4)
B (-3, 2)
C (0, 2)
Responses
Side Length AB: Response area
Side Length BC: Response area
Side Length AC: Response area
Perimeter of Triangle ABC: Response area
Area of Triangle ABC: Response area
Note: the formula for area of a triangle is A=bh/2
, where b is the length of the base (BC) and h is the height from side BC to point A.
A (-1, 4)
B (-3, 2)
C (0, 2)
Responses
Side Length AB: Response area
Side Length BC: Response area
Side Length AC: Response area
Perimeter of Triangle ABC: Response area
Area of Triangle ABC: Response area
Note: the formula for area of a triangle is A=bh/2
, where b is the length of the base (BC) and h is the height from side BC to point A.
Answer
A bookstore is placing an order for new books and magazines. Each book costs $15, and each magazine costs $5. The store can spend at most $1,000 on these books and magazines. Additionally, each book takes up 2 square feet of space, and each magazine takes up 1 square foot of space. The store has only 50 square feet of display space available for the new items. The following system of inequalities can be used to determine possible combinations of books, x, and magazines, y, the store can order:
15x+5y≤1000
2x+y≤50
Which combinations of books, x, and magazines, y, can the bookstore order?
(1 point)
Responses
20 books and 15 magazines
40 books and 5 magazines
50 books and 0 magazines
10 books and 30 magazines
15x+5y≤1000
2x+y≤50
Which combinations of books, x, and magazines, y, can the bookstore order?
(1 point)
Responses
20 books and 15 magazines
40 books and 5 magazines
50 books and 0 magazines
10 books and 30 magazines
Answer
The carnival is arriving in the neighborhood! Leo has saved up $15 to spend at the carnival on food and attractions. Food items cost $4 each and attractions cost $2 each. Leo uses the expression 4a + 2b to determine the total cost for a food items and b attractions.
Select one of the following combinations of food items and attractions that Leo can afford with his $15 budget:
(1 point)
Responses
1 food item and 5 attractions
0 food items and 8 attractions
2 food items and 4 attractions
3 food items and 2 attractions
Select one of the following combinations of food items and attractions that Leo can afford with his $15 budget:
(1 point)
Responses
1 food item and 5 attractions
0 food items and 8 attractions
2 food items and 4 attractions
3 food items and 2 attractions
Answer
Liam and his friends are hosting a bake sale to raise funds for their school trip. They decided to sell cookies for $3 each and muffins for $5 each. Their goal is to raise at least $300, but they want to sell no more than 80 items in total.
Select one of the following combinations that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80:
(1 point)
Responses
40 cookies and 20 muffins
50 cookies and 30 muffins
25 cookies and 30 muffins
30 cookies and 25 muffins
Select one of the following combinations that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80:
(1 point)
Responses
40 cookies and 20 muffins
50 cookies and 30 muffins
25 cookies and 30 muffins
30 cookies and 25 muffins
Answered by
GPT-4o mini
To find a line that is parallel to the given equation \(-4x + 3y = 12\), we first need to determine the slope of this line.
We can rewrite the equation in slope-intercept form, \(y = mx + b\), where \(m\) is the slope.
Starting with the original equation:
\[
-4x + 3y = 12
\]
We can isolate \(y\):
1. Add \(4x\) to both sides:
\[
3y = 4x + 12
\]
2. Divide by \(3\):
\[
y = \frac{4}{3}x + 4
\]
Now, we see that the slope \(m\) of the line \(-4x + 3y = 12\) is \(\frac{4}{3}\).
Parallel lines have the same slope, so we need to find an equation with a slope of \(\frac{4}{3}\).
Let's check the options given:
1. \(y = -\frac{4}{3}x + 4\) (slope: \(-\frac{4}{3}\))
2. \(y = 3x + 12\) (slope: \(3\))
3. \(y = 4x + 8\) (slope: \(4\))
4. \(y = \frac{4}{3}x + 6\) (slope: \(\frac{4}{3}\))
The equation \(y = \frac{4}{3}x + 6\) has the same slope of \(\frac{4}{3}\), meaning it is parallel to the original line.
So, the correct response is:
**y = \frac{4}{3}x + 6**.
We can rewrite the equation in slope-intercept form, \(y = mx + b\), where \(m\) is the slope.
Starting with the original equation:
\[
-4x + 3y = 12
\]
We can isolate \(y\):
1. Add \(4x\) to both sides:
\[
3y = 4x + 12
\]
2. Divide by \(3\):
\[
y = \frac{4}{3}x + 4
\]
Now, we see that the slope \(m\) of the line \(-4x + 3y = 12\) is \(\frac{4}{3}\).
Parallel lines have the same slope, so we need to find an equation with a slope of \(\frac{4}{3}\).
Let's check the options given:
1. \(y = -\frac{4}{3}x + 4\) (slope: \(-\frac{4}{3}\))
2. \(y = 3x + 12\) (slope: \(3\))
3. \(y = 4x + 8\) (slope: \(4\))
4. \(y = \frac{4}{3}x + 6\) (slope: \(\frac{4}{3}\))
The equation \(y = \frac{4}{3}x + 6\) has the same slope of \(\frac{4}{3}\), meaning it is parallel to the original line.
So, the correct response is:
**y = \frac{4}{3}x + 6**.
Answered by
GPT-4o mini
To find an equation of a line that is perpendicular to the line given by \(y = -2x + 5\) and passes through the point \((-2, 3)\), we need to follow these steps:
1. **Determine the slope of the original line**:
The slope of the line \(y = -2x + 5\) is \(-2\).
2. **Find the slope of the perpendicular line**:
The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope.
The negative reciprocal of \(-2\) is \(\frac{1}{2}\).
3. **Use the point-slope form to find the equation of the new line**:
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is the point the line passes through, and \(m\) is the slope.
Here, \((x_1, y_1) = (-2, 3)\) and \(m = \frac{1}{2}\).
Plugging in the values:
\[
y - 3 = \frac{1}{2}(x + 2)
\]
4. **Simplify the equation**:
\[
y - 3 = \frac{1}{2}x + 1 \\
y = \frac{1}{2}x + 4
\]
Now we can check the provided options:
1. \(y = -2x - 1\) (slope: \(-2\))
2. \(y = \frac{1}{2}x + 4\) (slope: \(\frac{1}{2}\))
3. \(y = -\frac{1}{2}x + 2\) (slope: \(-\frac{1}{2}\))
4. \(y = 2x + 7\) (slope: \(2\))
The only equation that fits both criteria (perpendicular and passing through \((-2, 3)\)) is:
**y = \frac{1}{2}x + 4**.
1. **Determine the slope of the original line**:
The slope of the line \(y = -2x + 5\) is \(-2\).
2. **Find the slope of the perpendicular line**:
The slope of a line that is perpendicular to another line is the negative reciprocal of the original slope.
The negative reciprocal of \(-2\) is \(\frac{1}{2}\).
3. **Use the point-slope form to find the equation of the new line**:
The point-slope form of a line is given by:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is the point the line passes through, and \(m\) is the slope.
Here, \((x_1, y_1) = (-2, 3)\) and \(m = \frac{1}{2}\).
Plugging in the values:
\[
y - 3 = \frac{1}{2}(x + 2)
\]
4. **Simplify the equation**:
\[
y - 3 = \frac{1}{2}x + 1 \\
y = \frac{1}{2}x + 4
\]
Now we can check the provided options:
1. \(y = -2x - 1\) (slope: \(-2\))
2. \(y = \frac{1}{2}x + 4\) (slope: \(\frac{1}{2}\))
3. \(y = -\frac{1}{2}x + 2\) (slope: \(-\frac{1}{2}\))
4. \(y = 2x + 7\) (slope: \(2\))
The only equation that fits both criteria (perpendicular and passing through \((-2, 3)\)) is:
**y = \frac{1}{2}x + 4**.
Answered by
GPT-4o mini
To find the missing side lengths, perimeter, and area of triangle ABC, we will first calculate the lengths of the sides AB, BC, and AC using the distance formula. Then we will find the perimeter and area based on the dimensions of the triangle.
### Step 1: Calculate the Side Lengths
**1. Side Length AB**
Points A(-1, 4) and B(-3, 2).
Using the distance formula:
\[
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
AB = \sqrt{((-3) - (-1))^2 + (2 - 4)^2}
\]
\[
= \sqrt{(-2)^2 + (-2)^2}
\]
\[
= \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
**2. Side Length BC**
Points B(-3, 2) and C(0, 2).
\[
BC = \sqrt{(0 - (-3))^2 + (2 - 2)^2}
\]
\[
= \sqrt{(3)^2 + (0)^2}
\]
\[
= \sqrt{9} = 3
\]
**3. Side Length AC**
Points A(-1, 4) and C(0, 2).
\[
AC = \sqrt{(0 - (-1))^2 + (2 - 4)^2}
\]
\[
= \sqrt{(1)^2 + (-2)^2}
\]
\[
= \sqrt{1 + 4} = \sqrt{5}
\]
### Step 2: Calculate the Perimeter
The perimeter \(P\) of triangle ABC is the sum of all sides:
\[
P = AB + BC + AC
\]
\[
P = 2\sqrt{2} + 3 + \sqrt{5}
\]
### Step 3: Calculate the Area
To find the area, we identify the base (BC) and the height from point A to line BC.
- The length of base \(BC = 3\).
- The height \(h\) is the vertical distance from point A(-1, 4) down to line BC where \(y = 2\). The height from A(−1, 4) to line \(y = 2\) is:
\[
h = 4 - 2 = 2
\]
Using the area formula:
\[
A = \frac{1}{2} \times b \times h
\]
\[
A = \frac{1}{2} \times 3 \times 2 = 3
\]
### Summary
Now we can provide all the responses:
- **Side Length AB**: \(2\sqrt{2}\)
- **Side Length BC**: \(3\)
- **Side Length AC**: \(\sqrt{5}\)
- **Perimeter of Triangle ABC**: \(2\sqrt{2} + 3 + \sqrt{5}\)
- **Area of Triangle ABC**: \(3\)
Let me know if you need further assistance!
### Step 1: Calculate the Side Lengths
**1. Side Length AB**
Points A(-1, 4) and B(-3, 2).
Using the distance formula:
\[
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
\[
AB = \sqrt{((-3) - (-1))^2 + (2 - 4)^2}
\]
\[
= \sqrt{(-2)^2 + (-2)^2}
\]
\[
= \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
**2. Side Length BC**
Points B(-3, 2) and C(0, 2).
\[
BC = \sqrt{(0 - (-3))^2 + (2 - 2)^2}
\]
\[
= \sqrt{(3)^2 + (0)^2}
\]
\[
= \sqrt{9} = 3
\]
**3. Side Length AC**
Points A(-1, 4) and C(0, 2).
\[
AC = \sqrt{(0 - (-1))^2 + (2 - 4)^2}
\]
\[
= \sqrt{(1)^2 + (-2)^2}
\]
\[
= \sqrt{1 + 4} = \sqrt{5}
\]
### Step 2: Calculate the Perimeter
The perimeter \(P\) of triangle ABC is the sum of all sides:
\[
P = AB + BC + AC
\]
\[
P = 2\sqrt{2} + 3 + \sqrt{5}
\]
### Step 3: Calculate the Area
To find the area, we identify the base (BC) and the height from point A to line BC.
- The length of base \(BC = 3\).
- The height \(h\) is the vertical distance from point A(-1, 4) down to line BC where \(y = 2\). The height from A(−1, 4) to line \(y = 2\) is:
\[
h = 4 - 2 = 2
\]
Using the area formula:
\[
A = \frac{1}{2} \times b \times h
\]
\[
A = \frac{1}{2} \times 3 \times 2 = 3
\]
### Summary
Now we can provide all the responses:
- **Side Length AB**: \(2\sqrt{2}\)
- **Side Length BC**: \(3\)
- **Side Length AC**: \(\sqrt{5}\)
- **Perimeter of Triangle ABC**: \(2\sqrt{2} + 3 + \sqrt{5}\)
- **Area of Triangle ABC**: \(3\)
Let me know if you need further assistance!
Answered by
GPT-4o mini
To determine which combinations of books \( x \) and magazines \( y \) the bookstore can order, we need to check each response against the given system of inequalities:
1. **Cost constraint**:
\[
15x + 5y \leq 1000
\]
2. **Space constraint**:
\[
2x + y \leq 50
\]
Now, let's analyze each combination given in the responses.
### 1. **20 books and 15 magazines**
- \( x = 20 \), \( y = 15 \)
Check the cost constraint:
\[
15(20) + 5(15) = 300 + 75 = 375 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(20) + 15 = 40 + 15 = 55 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 2. **40 books and 5 magazines**
- \( x = 40 \), \( y = 5 \)
Check the cost constraint:
\[
15(40) + 5(5) = 600 + 25 = 625 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(40) + 5 = 80 + 5 = 85 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 3. **50 books and 0 magazines**
- \( x = 50 \), \( y = 0 \)
Check the cost constraint:
\[
15(50) + 5(0) = 750 + 0 = 750 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(50) + 0 = 100 + 0 = 100 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 4. **10 books and 30 magazines**
- \( x = 10 \), \( y = 30 \)
Check the cost constraint:
\[
15(10) + 5(30) = 150 + 150 = 300 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(10) + 30 = 20 + 30 = 50 \leq 50 \quad \text{(Satisfies)}
\]
### Conclusion
The only combination that satisfies both the cost and space constraints is:
**10 books and 30 magazines**.
1. **Cost constraint**:
\[
15x + 5y \leq 1000
\]
2. **Space constraint**:
\[
2x + y \leq 50
\]
Now, let's analyze each combination given in the responses.
### 1. **20 books and 15 magazines**
- \( x = 20 \), \( y = 15 \)
Check the cost constraint:
\[
15(20) + 5(15) = 300 + 75 = 375 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(20) + 15 = 40 + 15 = 55 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 2. **40 books and 5 magazines**
- \( x = 40 \), \( y = 5 \)
Check the cost constraint:
\[
15(40) + 5(5) = 600 + 25 = 625 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(40) + 5 = 80 + 5 = 85 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 3. **50 books and 0 magazines**
- \( x = 50 \), \( y = 0 \)
Check the cost constraint:
\[
15(50) + 5(0) = 750 + 0 = 750 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(50) + 0 = 100 + 0 = 100 \leq 50 \quad \text{(Does NOT satisfy)}
\]
### 4. **10 books and 30 magazines**
- \( x = 10 \), \( y = 30 \)
Check the cost constraint:
\[
15(10) + 5(30) = 150 + 150 = 300 \leq 1000 \quad \text{(Satisfies)}
\]
Check the space constraint:
\[
2(10) + 30 = 20 + 30 = 50 \leq 50 \quad \text{(Satisfies)}
\]
### Conclusion
The only combination that satisfies both the cost and space constraints is:
**10 books and 30 magazines**.
Answered by
GPT-4o mini
To determine which combinations of food items (costing $4 each) and attractions (costing $2 each) Leo can afford with his $15 budget, we will evaluate the expression \(4a + 2b\) for each option, where \(a\) is the number of food items and \(b\) is the number of attractions.
### Budget Constraint
Leo's budget can be represented by the following inequality:
\[
4a + 2b \leq 15
\]
Now let's evaluate each of the given options.
### 1. **1 food item and 5 attractions**
- \(a = 1\), \(b = 5\)
Cost calculation:
\[
4(1) + 2(5) = 4 + 10 = 14 \leq 15 \quad \text{(Satisfies the budget)}
\]
### 2. **0 food items and 8 attractions**
- \(a = 0\), \(b = 8\)
Cost calculation:
\[
4(0) + 2(8) = 0 + 16 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### 3. **2 food items and 4 attractions**
- \(a = 2\), \(b = 4\)
Cost calculation:
\[
4(2) + 2(4) = 8 + 8 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### 4. **3 food items and 2 attractions**
- \(a = 3\), \(b = 2\)
Cost calculation:
\[
4(3) + 2(2) = 12 + 4 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### Conclusion
The only combination that Leo can afford with his $15 budget is:
**1 food item and 5 attractions**.
### Budget Constraint
Leo's budget can be represented by the following inequality:
\[
4a + 2b \leq 15
\]
Now let's evaluate each of the given options.
### 1. **1 food item and 5 attractions**
- \(a = 1\), \(b = 5\)
Cost calculation:
\[
4(1) + 2(5) = 4 + 10 = 14 \leq 15 \quad \text{(Satisfies the budget)}
\]
### 2. **0 food items and 8 attractions**
- \(a = 0\), \(b = 8\)
Cost calculation:
\[
4(0) + 2(8) = 0 + 16 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### 3. **2 food items and 4 attractions**
- \(a = 2\), \(b = 4\)
Cost calculation:
\[
4(2) + 2(4) = 8 + 8 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### 4. **3 food items and 2 attractions**
- \(a = 3\), \(b = 2\)
Cost calculation:
\[
4(3) + 2(2) = 12 + 4 = 16 \quad \text{(Does NOT satisfy the budget)}
\]
### Conclusion
The only combination that Leo can afford with his $15 budget is:
**1 food item and 5 attractions**.
Answered by
GPT-4o mini
To determine which combination of cookies and muffins Liam can sell to raise at least $300 while keeping the total items sold to a maximum of 80, we will evaluate each option using the following equations:
1. **Total cost constraint**:
\[
3c + 5m \geq 300
\]
where \(c\) is the number of cookies and \(m\) is the number of muffins.
2. **Total items constraint**:
\[
c + m \leq 80
\]
Now let's evaluate each option.
### 1. **40 cookies and 20 muffins**
- \(c = 40\), \(m = 20\)
**Cost calculation**:
\[
3(40) + 5(20) = 120 + 100 = 220 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
40 + 20 = 60 \quad \text{(Satisfies the items constraint)}
\]
### 2. **50 cookies and 30 muffins**
- \(c = 50\), \(m = 30\)
**Cost calculation**:
\[
3(50) + 5(30) = 150 + 150 = 300 \quad \text{(Satisfies the cost constraint)}
\]
**Total items calculation**:
\[
50 + 30 = 80 \quad \text{(Satisfies the items constraint)}
\]
### 3. **25 cookies and 30 muffins**
- \(c = 25\), \(m = 30\)
**Cost calculation**:
\[
3(25) + 5(30) = 75 + 150 = 225 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
25 + 30 = 55 \quad \text{(Satisfies the items constraint)}
\]
### 4. **30 cookies and 25 muffins**
- \(c = 30\), \(m = 25\)
**Cost calculation**:
\[
3(30) + 5(25) = 90 + 125 = 215 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
30 + 25 = 55 \quad \text{(Satisfies the items constraint)}
\]
### Conclusion
The only combination that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80 is:
**50 cookies and 30 muffins**.
1. **Total cost constraint**:
\[
3c + 5m \geq 300
\]
where \(c\) is the number of cookies and \(m\) is the number of muffins.
2. **Total items constraint**:
\[
c + m \leq 80
\]
Now let's evaluate each option.
### 1. **40 cookies and 20 muffins**
- \(c = 40\), \(m = 20\)
**Cost calculation**:
\[
3(40) + 5(20) = 120 + 100 = 220 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
40 + 20 = 60 \quad \text{(Satisfies the items constraint)}
\]
### 2. **50 cookies and 30 muffins**
- \(c = 50\), \(m = 30\)
**Cost calculation**:
\[
3(50) + 5(30) = 150 + 150 = 300 \quad \text{(Satisfies the cost constraint)}
\]
**Total items calculation**:
\[
50 + 30 = 80 \quad \text{(Satisfies the items constraint)}
\]
### 3. **25 cookies and 30 muffins**
- \(c = 25\), \(m = 30\)
**Cost calculation**:
\[
3(25) + 5(30) = 75 + 150 = 225 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
25 + 30 = 55 \quad \text{(Satisfies the items constraint)}
\]
### 4. **30 cookies and 25 muffins**
- \(c = 30\), \(m = 25\)
**Cost calculation**:
\[
3(30) + 5(25) = 90 + 125 = 215 \quad \text{(Does NOT satisfy the cost constraint)}
\]
**Total items calculation**:
\[
30 + 25 = 55 \quad \text{(Satisfies the items constraint)}
\]
### Conclusion
The only combination that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80 is:
**50 cookies and 30 muffins**.
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