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a.
To write:
An equation describing the possible amounts of canned and dry food that you can feed your beagle each day.
a.
![Check Mark](/static/check-mark.png)
Answer to Problem 19E
Our required equation would be
Explanation of Solution
Given:
Your beagle is allowed to eat 800 calories of food each day. You buy canned food containing 40 calories per ounce and dry food containing 100 calories per ounce.
Calculation:
Let x represent ounces of canned food and y represent ounces of dry food.
We have been given that each ounce of canned food contains 40 calories, so calories in x ounces of canned food would be
We are also told that each ounce of dry food contains 100 calories, so calories in y ounces of dry food would be
Since your beagle is allowed to eat 800 calories each day, so the total calories consumed from eating x ounces of canned food and y pounds of dry food should be equal to 800.
We can represent this information in an equation as:
Therefore, our required equation would be
b.
To graph:
The equation from part (a) using the intercepts.
b.
![Check Mark](/static/check-mark.png)
Answer to Problem 19E
The graph of the equation
Explanation of Solution
Given:
Your beagle is allowed to eat 800 calories of food each day. You buy canned food containing 40 calories per ounce and dry food containing 100 calories per ounce.
Calculation:
To find x -intercept, we will substitute
To find y -intercept, we will substitute
Upon graphing our given equation, we will get our required graph as shown below:
c.
To give:
Three possible combinations of canned and dry food that you can feed your beagle.
c.
![Check Mark](/static/check-mark.png)
Answer to Problem 19E
Three possible combinations would be:
1. (5,6)
2. (10,4)
3. (15,2)
Explanation of Solution
Given:
Your beagle is allowed to eat 800 calories of food each day. You buy canned food containing 40 calories per ounce and dry food containing 100 calories per ounce.
Calculation:
To find some possible combinations, we will substitute some values of xin our given equation as shown below:
Therefore, the three possible combinations would be (5,6), (10,4) and (15,2).
Chapter 8 Solutions
ELEMENTARY+INTERMEDIATE ALGEBRA
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