What are the solutions to f (x) = x2 + 5x – 24
1 answer:
You might be interested in
Answer:
A. 3 possible combinations
B. 8 4-ounce's bags and 3 3-ounce's bags
C. 2 4-ounce's bags and 11 3-ounce's bags
D. 8 4-ounce's bags and 3 3-ounce's bags
E. All solutions offer the same revenue.
Step-by-step explanation:
You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then
A. Find all integer solutions:
- When x=0, then 3y=41 - impossible, because 41 is not divisible by 3.
- When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.
- When x=2, then 3y=33, y=11 - possible.
- When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.
- When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.
- When x=5, then 3y=21, y=7 - possible.
- When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.
- When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.
- When x=8, then 3y=9, y=3 - possible.
- When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.
- When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.
You get 3 possible combinations.
B. 1. 2 + 11 = 13,
2. 5 + 7 = 12,
3. 8 + 3 = 11.
The minimal number of bags is 11.
C. 1. 2·7+11·5=69 cents
2. 5·7+7·5=70 cents
3. 8·7+3·5=71 cents
The cheapest is 1st solution.
D. 1. 2·6+11·5=67 cents
2. 5·6+7·5=65 cents
3. 8·6+3·5=63 cents
The cheapest is 3rd solution.
E. 1. 2·2+11·1.50=$20.50
2. 5·2+7·1.50=$20.50
3. 8·2+3·1.50=$20.50
All solutions offer the same revenue.
<h2>Answer:</h2>
Shown below
<h2>Step-by-step explanation:</h2>First of all, we have the following points:
To plot these points, we need to graph a coordinate grid. Recall that a coordinate grid is composed of two perpendicular lines, that we call the axes and are labeled like number lines. The horizontal axis is called the x-axis while the vertical axis is called the y-axis, and these two axes intersect at a point called the origin. Next, let's plot each point as follows:
<h2>PART 1. Plot the points on your own coordinates grid and connect in alphabetical order. </h2><h2></h2>
Point a:
Let's stand at the origin and then move three units to the left and three units up. The resulting point is shown in the firs figure below.
Point b:
Let's stand at the origin and then move two units to the right and three units up. The resulting point is shown in the firs figure below.
Point c:
Let's stand at the origin and then move five units to the right and two units down. The resulting point is shown in the firs figure below.
Point d:
Let's stand at the origin and then move three units to down. The resulting point is shown in the firs figure below.
Point e:
Let's stand at the origin and then move three units to the left and two units down. The resulting point is shown in the firs figure below.
Next, let's connect connect the points in alphabetical order, so that the line are connected from a to b, from b to c, from c to d, from d to e and from e to a again. So the resulting graph is shown in the second figure. As you can see, this is an irregular pentagon.
<h2>PART 2. Decompose figure ABCDE into rectangles and triangles.</h2>Every pentagon can be divided up into three triangles. Hence, from the third figure. the triangles are:
ΔABE
ΔBDE
ΔBCD
Since a rectangle is a quadrilateral whose angles are all right angles, then there is no chance to obtain a rectangle from ABCDE. To obtain a rectangle, we'd need other points and this is not what the problem asks.