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IrinaVladis [17]
3 years ago
12

John bought 3 pineapples and 2 watermelons for $8.50. Sue bought 2 pineapples and 4 watermelons for $13.00. Write and show a sys

tem of equations. You do not have to solve the system. Just show the system.​
Mathematics
1 answer:
AleksAgata [21]3 years ago
5 0
As per your question, total cost of watermelon should end with either 5(for odd quantity) or 0(for even quantity).

If the quantity of watermelon is odd, then the total cost value of pineapple should end with 3 and this is not possible when the cost of pineapple is ₹7.

So let's come to conclusion that the count(quantity) of watermelon should be any one of 0, 2, 4, 6.

If count of watermelon is 6: It will cost ₹30 and for remaining ₹8, we can buy 1 pineapple but still ₹1 will not be utilised. So 1 pineapple is not possible

If count of watermelon is 4: It will cost ₹20 and for remaining ₹18, we can buy 2 pineapple with ₹4 not being utilised. So 2 pineapple is also not possible.

If count of watermelon is 2: It will cost ₹10 and for remaining ₹28, we can buy 4 pineapple with all amount being utilised. We can buy 4 pineapple along with with 2 watermelon for ₹38.

If count of watermelon is 0: It will cost you ₹0 and for remaining ₹38, we can buy 5 pineapple with ₹3 being not utilised. So 5 pineapple is also not possible.

So the answer is 4 pineapple.
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Given that a large mixing tank initially holds 400 gallons of water in which 70 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 4 gal/min, and when the solution is well stirred, it is then pumped out at the same rate.

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Volgvan

<u>Answer</u>

1st solution: a = 1

                   b = 3

2nd solution: a = 2

                      b = 6


<u>Explanation</u>

1st solution

                  <em>         1</em>

<em>                       333</em>

<em>                       333</em>

<em>                       333</em>

<em>                   </em><em><u>  +333</u></em>

<em>               </em><em><u>       1333</u></em>

2nd solution

<em>         2</em>

<em>    666</em>

<em>    666</em>

<em>    666</em>

<em><u> + 666</u></em>

<em><u> 2666</u></em>

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