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AlexFokin [52]
3 years ago
5

Find three consecutive odd integers whose sum is 13 more than twice the largest of the three intergers

Mathematics
1 answer:
Yuki888 [10]3 years ago
4 0

15,17,19 
Suppose the three odd integers are #n#, #n+2# and #n+4# 
Their sum is: 
#n + (n+2) + (n+4) = 3n+6# 
#13# more than twice the largest of the three is: 
#2(n+4)+13 = 2n+21# 
From what we are told these two are equal: 
#3n+6 = 2n+21# 
Subtract #2n+6# from both sides to get: 
#n = 15# 
So the three integers are: 
#15, 17, 19#
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Carl collects coins compulsively, but he only likes quarters ($0.25) and dimes ($0.10). Carl currently has 125 coins for a total
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Carl has 35 dimes and 90 quarters

Step-by-step explanation:

Let the number of quarters be q and the number of dimes be d

Total number of coins is 125;

Hence;

q + d = 125 •••••••••(i)

The total value of quarters present = q * 0.25 = 0.25q

The total value of dimes present = d * 0.1 = 0.1d

Adding both gives the total

0.25q + 0.1d =26 ••••••••(ii)

So we need to solve both equations simultaneously;

From i,

q = 125 - d

Substitute this into ii

0.25(125-d) + 0.1d = 26

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d = 5.25/0.15

d = 35

Recall; q = 125 - d = 125 -35 = 90

7 0
3 years ago
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Step-by-step explanation:


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