Holt Mcdougal Larson Pre-algebra: Student Edition 2012
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN: 9780547587776
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Chapter 8.3, Problem 19E

a.

To determine

To write:

An equation describing the possible amounts of canned and dry food that you can feed your beagle each day.

a.

Expert Solution
Check Mark

Answer to Problem 19E

Our required equation would be 40x+100y=800 .

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 40x .

We are also told that each ounce of dry food contains 100 calories, so calories in y ounces of dry food would be 100y .

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:

  40x+100y=800

Therefore, our required equation would be 40x+100y=800 .

b.

To determine

To graph:

The equation from part (a) using the intercepts.

b.

Expert Solution
Check Mark

Answer to Problem 19E

The graph of the equation 40x+100y=800 would look like:

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 8.3, Problem 19E , additional homework tip  1

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 y=0 in our equation as:

  40x+100y=800 (Given equation)40x+100(0)=800 (Substituting y=0)40x=800x=20

To find y -intercept, we will substitute x=0 in our equation as:

  40x+100y=800 (Given equation)40(0)+100y=800 (Substituting x=0)100y=800y=8

Upon graphing our given equation, we will get our required graph as shown below:

Holt Mcdougal Larson Pre-algebra: Student Edition 2012, Chapter 8.3, Problem 19E , additional homework tip  2

c.

To determine

To give:

Three possible combinations of canned and dry food that you can feed your beagle.

c.

Expert Solution
Check Mark

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:

  40x+100y=800 (Given equation)40(5)+100y=800 (Substituting x=5)200+100y=800100y=600 (Subtracting 200 from both sides)y=6 (Dividing both sides by 100)

  40x+100y=800 (Given equation)40(10)+100y=800 (Substituting x=10)400+100y=800100y=400 (Subtracting 400 from both sides)y=4 (Dividing both sides by 100)

  40x+100y=800 (Given equation)40(15)+100y=800 (Substituting x=15)600+100y=800100y=200 (Subtracting 600 from both sides)y=2 (Dividing both sides by 100)

Therefore, the three possible combinations would be (5,6), (10,4) and (15,2).

Chapter 8 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 8.1 - Prob. 7ECh. 8.1 - Prob. 8ECh. 8.1 - Prob. 9ECh. 8.1 - Prob. 10ECh. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.1 - Prob. 33ECh. 8.1 - Prob. 34ECh. 8.1 - Prob. 35ECh. 8.1 - Prob. 36ECh. 8.1 - Prob. 37ECh. 8.2 - Prob. 1CCh. 8.2 - Prob. 2CCh. 8.2 - Prob. 3CCh. 8.2 - Prob. 4CCh. 8.2 - Prob. 5CCh. 8.2 - Prob. 6CCh. 8.2 - Prob. 7CCh. 8.2 - Prob. 8CCh. 8.2 - Prob. 9CCh. 8.2 - Prob. 10CCh. 8.2 - Prob. 1ECh. 8.2 - Prob. 2ECh. 8.2 - Prob. 3ECh. 8.2 - Prob. 4ECh. 8.2 - Prob. 5ECh. 8.2 - Prob. 6ECh. 8.2 - Prob. 7ECh. 8.2 - Prob. 8ECh. 8.2 - Prob. 9ECh. 8.2 - Prob. 10ECh. 8.2 - Prob. 11ECh. 8.2 - Prob. 12ECh. 8.2 - Prob. 13ECh. 8.2 - 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