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givi [52]
3 years ago
13

You have 22 coins, dimes and nickels. If the number of dimes and nickels were reversed, you would have $0.40 less than you actua

lly have. How many dimes and nickels do you have?
Mathematics
1 answer:
fenix001 [56]3 years ago
4 0
The coins differ in value by 5¢, so swapping the numbers of them will change the value by 5¢ for each unit difference in the numbers of coins. Since
40¢ = 8 * 5¢
there must be 8 more dimes than nickels.

There are (22 +8)/2 = 15 dimes and 7 nickels.

_____
You could write some equations for this problem. Let n, d represent numbers of nickels and dimes.
.. n +d = 22
.. 10d +5n - (10n +5d) = 40 . . . . . . . cents
.. 5(d -n) = 40 . . . . . . . . . . . . . . . . . . reversing the coin count changes the total by 5 cents for each unit of difference (d -n), as stated above.
.. d -n = 8 . . . . . . . . . . . . . . . . . . . . . . divide the preceding equation by 5
Adding this last equation to the first gives
.. 2d = 22 +8
.. d = 30/2 = 15
.. n = 22 -15 = 7
There are 15 dimes and 7 nickels.
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Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks​
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Let

P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

P(1):\ 1 = \dfrac{1\cdot 2}{2}=1

So, the base case is ok. Now, we need to assume P(n) and prove P(n+1).

P(n+1) states that

P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}

Since we're assuming P(n), we can substitute the sum of the first n terms with their expression:

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4 0
3 years ago
J/2--12=15 <br> j/2 is a fraction
Natasha_Volkova [10]

Answer:

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Step-by-step explanation:

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