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valina [46]
3 years ago
14

The hardcover version of a book weighs 7 ounces while its paperback version weighs 5 ounces. Forty-five copies of the book weigh

a total of 249 ounces. Which value could replace x in the table?
Mathematics
2 answers:
kap26 [50]3 years ago
6 0

Answer:

B 50 - h

Step-by-step explanation:

konstantin123 [22]3 years ago
3 0

Answer:

33 copies were paperback and 12 were hardcover.

Step-by-step explanation:

Let h represent the number of hardcover copies and p represent the number of paperback copies.

We know that the total number of copies was 45; this gives us the equation

h+p = 45

We know that each hardcover copy is 7 ounces; this gives us the expression 7h.

We also know that each paperback copy is 5 ounces; this gives us the expression 5p.

We know that the total weight was 249 ounces; this gives us the equation

7h+5p = 249

Together we have the system

\left \{ {{h+p=45} \atop {7h+5p=249}} \right.

We will use elimination to solve this.  First we will make the coefficients of the variable p the same; to do this, we will multiply the top equation by 5:

\left \{ {{5(h+p=45)} \atop {7h+5p=249}} \right. \\\\\left \{ {{5h+5p=225} \atop {7h+5p=249}} \right.

To eliminate p, we will subtract the equations:

\left \{ {{5h+5p=225} \atop {-(7h+5p=249)}} \right. \\\\-2h=-24

Divide both sides by -2:

-2h/-2 = -24/-2

h = 12

There were 12 hardcover copies sold.

Substitute this into our first equation:

12+p=45

Subtract 12 from each side:

12+p-12 = 45-12

p = 33

There were 33 paperback copies sold.

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The length of segment EF is 8 units, the length of segment ED is 10 units, and the length of segment EA is 13.33 units. Find the
ivanzaharov [21]

Answer:

5.33 unit

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3 years ago
The length of a rectangular fenced enclosure is 12 feet more than the width. If Farmer Dan has 100 feet of fencing, write an ine
Kitty [74]
Answer:
length = 31 ft
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Explanation:
Assume that the width of the rectangle is w.
We are given that the length is 12 ft more than the width. This means that the length of the rectangle is w + 12

The perimeter of the rectangle in this case would be:
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Assume that Dan would use all 100 ft of fencing to surround the yard. This would mean that the largest perimeter is 100 ft.

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Since we have calculated that the width of the yard is 19 ft, we can substitute to get the length as follows:
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