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Aloiza [94]
3 years ago
10

An alloy of copper is 10% copper and weighs 25 pounds. A second alloy is 18% copper. How much (to the nearest tenth lb.) of the

second alloy must be added to the first alloy to get a 13% mixture.
Mathematics
1 answer:
Alisiya [41]3 years ago
4 0
  let  the weight  of the  second  alloy  be  y
convert the  percentage  to  fraction, that  is  10/100=0.1,  18/100=0.18,  13/100=0.13
 Total weight  if  the  mixture is  0.1  x 25+0.18y=0.13(25+y)
2.5+0.18y=3.25+0.13y
  like  terms  together
0.18y +0.13y =  3.25-2.5
 0.05y=0.75
divide both  side  by  0.05
 Y  is  therefore  15 pound  of 18%  copper is needed
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Answer:

a = -5

Step-by-step explanation:

15 - a = 20

Move the constant of 15 to the other side by subtracting it.

-a = 5

Divide by -1 to make a positive.

a = -5

Check.

15 - (-5) = 20

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calculates the sum of the first 8 terms of an arithmetic progression starting with 3/5 and ending with 1/4
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We conclude that the sum of the first 8 terms of the arithmetic sequence is 17/5.

<h3>How to get the sum of the first 8 terms?</h3>

In an arithmetic sequence, the difference between any two consecutive terms is a constant.

Here we know that:

a_1 = 3/5\\a_8 = 1/4

There are 7 times the common difference between these two values, so if d is the common difference:

a_1 + 7*d = a_8\\\\3/5 + 7*d = 1/4\\\\7*d = 1/4 - 3/5 = (5 - 12)/20 = -7/20\\\\d = -1/20

Then the sum of the first 8 terms is given by:

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So we conclude that the sum of the first 8 terms of the arithmetic sequence is 17/5.

If you want to learn more about arithmetic sequences:

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olga_2 [115]

Answer:

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Step-by-step explanation:

3(17x – 6.5) = 108

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Divide each side by 17

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