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Oksana_A [137]
3 years ago
13

Suppose the rent for an apartment is $9,000 for the first year. The rent is expected

Mathematics
2 answers:
Maurinko [17]3 years ago
8 0

Answer: $57,000

Step-by-step explanation:

Understand what is being asked. 1st yr is 9000. Yr 2 is 9000+1200= 10200. Yr 3 =10,200+1200=11,400. Yr 4 is 11,400 +1200=12,600 and Yr 5 is 12600+1200=13,800. Over the course of 5 yrs you are paying 9000+10200+11400+12600+13800 which has a total sum of $57,000.

MaRussiya [10]3 years ago
8 0

Answer:

upper 5 lover n=1 (9000+(n-1)1200)

upper 4 lover n=0 (9000+1200n)

Step-by-step explanation:

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2 years ago
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How many multiples of 4 are there in {n; 37< n <1001}?
s2008m [1.1K]
Let's begin by listing the first few multiples of 4:  4, 8, 12, 16, 20, 24, 28, 32, 36, 38, 40, 44.  So, between 1 and 37 there are 9 such multiples:  {4, 8, 12, 16, 20, 24, 28, 32, 36}.  Note that 4 divided into 36 is 9.

Let's experiment by modifying the given problem a bit, for the purpose of discovering any pattern that may exist:

<span>How many multiples of 4 are there in {n; 37< n <101}?  We could list and then count them:  {40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100}; there are 16 such multiples in that particular interval.  Try subtracting 40 from 100; we get 60.  Dividing 60 by 4, we get 15, which is 1 less than 16.  So it seems that if we subtract 40 from 1000 and divide the result by 4, and then add 1, we get the number of multiples of 4 between 37 and 1001:

1000
   -40
-------
 960

Dividing this by 4, we get 240.  Adding 1, we get 241.

Finally, subtract 9 from 241:  We get 232.

There are 232 multiples of 4 between 37 and 1001.

Can you think of a more straightforward method of determining this number? </span>

5 0
3 years ago
A problem states: "There are 9 more children than parents in a room. There are 25 people in the room in all. How many children a
Elza [17]
C = number of parents
c+9 = number of children
2c+9=25
2c=16
c=8
8 parents, 17 children
4 0
3 years ago
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