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Diano4ka-milaya [45]
2 years ago
6

In a certain clothing store, 3 shirts and 5 ties cost $60, and 2 shirts and 3 ties cost $39. What is the cost of each shirt? If

s = the cost of a shirt and t = the cost of a tie, which system of equations represents the problem? 3t + 5s = 60 and 2t + 3s = 39 3s + 5t = 60 and 2s + 3t = 39 s + t = 60 and 5s + 8t = 99
Mathematics
2 answers:
katrin [286]2 years ago
8 0

Answer:

3s + 5t = 60 and 2s + 3t = 39

Step-by-step explanation:

STALIN [3.7K]2 years ago
7 0

Answer:

The second set of equations is the one you want.

Step-by-step explanation:

If s is the cost of a shirt and you buy 3 of them, you represent that as 3s.  Likewise with all the other numbers of articles of clothing you are given.  The first situation then is 3s + 5t = 60.  The second situation is 2s + 3t = 39.

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4 0
3 years ago
Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the
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Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees wasx+\frac{1}{4} x=\frac{5}{4} x, and for the next three years we have that

                             Start                                          End

Second year     \frac{5}{4}x --------------   \frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x

Third year    (\frac{5}{4} )^{2}x-------------(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x

Fourth year (\frac{5}{4})^{3}x--------------(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.

So  the formula to calculate the number of trees in the fourth year  is  

(\frac{5}{4} )^{4} x, we know that all of the trees thrived and there were 6250 at the end of 4 year period, then  

6250=(\frac{5}{4} )^{4}x⇒x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.

Therefore the number of trees at the begging of the 4-year period was 2560.  

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