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7nadin3 [17]
3 years ago
7

How do I solve this math problem 53x6=6x53=

Mathematics
1 answer:
AlladinOne [14]3 years ago
5 0
You only need to solve one part of the problem, because the second problem is the exact same as the first, so you would do 53 x 6= 318, and 6 x 53 also equals 318
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Plz help me plzz i beg of u
Stells [14]

Answer:

2.0 x 10^11

Step-by-step explanation:

2.6 divide it by 1.3 and get 2.0. 9+2 = 11. Together with the, it would be 10^11. That how I got 2.0 x 10^11

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3 years ago
??????? Which exspression represents the total cost of the red shirts?
Vlad1618 [11]

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A

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3 years ago
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pantera1 [17]
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Ratling [72]
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8 0
3 years ago
A metallurgist has one alloy containing 34% copper another containing 48% copper. How many pounds of each alloy must he use to m
Zielflug [23.3K]

Answer:

<h2>36.14 pounds of 34% copper alloy and 9.86 pounds of 48% copper alloy</h2>

Step-by-step explanation:

        First alloy contains 34% copper and the second alloy contains 48% alloy.

We wish to make 46 pounds of a third alloy containing 37% copper.

       Let the weight of first alloy used be x in pounds and the weight of second alloy used be y in pounds.

       Total weight = 46\text{ }pounds=x+y        -(i)

      Total weight of copper = 37\%\text{ of 46 pounds = }34\%\text{ of }x\text{ pounds + }48\%\text{ of }y\text{ pounds }

       \dfrac{37\times 46}{100}=\dfrac{34x}{100}+\dfrac{48y}{100}\\\\ 34x+48y=1702        -(ii)

       Subtracting 34 times first equation from second equation,

34x+48y-34x-34y=1702-34\times46\\14y=138\\y=9.857\text{ }pounds \\x=36.143\text{ }pounds

∴ 36.14 pounds of first alloy and 9.86 pounds of second alloy were used.

5 0
3 years ago
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