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topjm [15]
2 years ago
13

There are 31 coins of dimes and quarters. If the number of dimes is 5 more than the number of quarters, how many quarters are th

ere? How many are dimes? Hint: let x=dimes and y=quarters. Write your answer in an ordered pair ( , ).
Mathematics
1 answer:
emmasim [6.3K]2 years ago
5 0

Answer:

<u>13 Quarters and 18 Dimes</u>

Step-by-step explanation:

Let D and Q stand for the numbers of Dimes(D) and Quarters(Q).  Give x and y a break, and make it easier to remember which is which.

D + Q = 31

D = Q + 5

Therefore:

D + Q = 31

(Q+5) + Q = 31

2Q + 5 = 31

2Q = 26

<u>Q = 13</u>

<u>Then D = (13+5) or 18</u>

================

Check:

D + Q = 31 ?

18 + 13 = 31  <u>YES</u>

----------------------

D = Q + 5 ?

18 = 13 + 5   <u>YES</u>

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The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given that:

a_n=1 + (-\frac{4}{9} )^n\\\\When\ n=1:a_1=1 + (-\frac{4}{9} )^1=0.5556\\\\When\ n=2:a_2=1 + (-\frac{4}{9} )^2=1.1975\\\\When\ n=3:a_3=1 + (-\frac{4}{9} )^3=0.9122\\\\When\ n=4:a_4=1 + (-\frac{4}{9} )^4=1.039\\\\When\ n=5:a_5=1 + (-\frac{4}{9} )^5=0.9827\\\\When\ n=6:a_6=1 + (-\frac{4}{9} )^6=1.0077\\\\When\ n=7:a_7=1 + (-\frac{4}{9} )^7=0.9966\\\\When\ n=8:a_8=1 + (-\frac{4}{9} )^8=1.0015\\\\When\ n=9:a_9=1 + (-\frac{4}{9} )^9=0.9993\\\\

When\ n=10:a_{10}=1 + (-\frac{4}{9} )^{10}=1.0003\\

The first ten terms of the sequence a_n=1 + (-\frac{4}{9} )^n is<em> 0.5556, 1.1975, 0.9122, 1.039, 0.9827, 1.0077, 0.9966, 1.0015, 0.9993 and 1.0003</em>

Find out more on equation at: brainly.com/question/2972832

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