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Hunter-Best [27]
3 years ago
9

Is 9.373 a repeating? is it rational?

Mathematics
1 answer:
kow [346]3 years ago
7 0

Decimal numbers are repeating if the part after the decimal separator never ends, but has a cicle. For example, 14.020202020202.... is a repeating decimal number, because after the decimal separator you can see a pattern: 02 repeated indefinitely.

In your case, the part after the decimal separator is finite: there is no repeating part. Every number with a finite decimal part is rational. In your case, the number 9.373 can be written as the fraction

\dfrac{9373}{1000}

which proves that the number is rational.

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Sever21 [200]

Answer:

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Step-by-step explanation:

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amm1812

9514 1404 393

Answer:

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  • stock: $238,500
  • mutual fund: $228,500

Step-by-step explanation:

We can let x and y represent the amounts invested in the stock and mutual fund, respectively. Then the amount invested in the money market account was (468000 -x -y). Two relations can be written.

  x = y + 10000 . . . . . 10,000 more was invested in stock than in mutual funds

  0.035(468000-x-y) +0.04(x) -0.03(y) = 2720 . . . . first-year return

Collecting terms in the second equation gives ...

  0.005x -0.065y +16380 = 2720

  -x + 13y = 2732000 . . . . . . subtract 16380, multiply by -200

  12y -10000 = 2732000 . . . . substitute for x using the first equation

  y = 2742000/12 = 228,500 . . . . solve for y

  x = 228,500 +10,000 = 238,500 . . . . . find x from the first equation

  468,000 -x -y = 468,000 -467,000 = 1,000 . . . . find the money market amount

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Pierre invested $1000 in the money market account, $238,500 in the stock, and $228,500 in the mutual fund.

6 0
3 years ago
How many positive integers less than 100 have at least one digit that is a 9? (Let T be the set of positive integers less than 1
Ymorist [56]

Answer:

19

Step-by-step explanation:

Let T be the set of positive integers less than 100 with a 9 in the ten's place.

T = {90, 91, 92, 93, 94, 95, 96, 97, 98, 99}

n(T) = 10

Let O be the set of positive integers less than 100 with a 9 in the one's place.

O = {9, 19, 29, 39, 49, 59, 69, 79, 89, 99}

n(O) = 10

The common element is 99.

T\cap O={99}

n(T\cap O)=1

The union of both sets is

n(T\cup O)=n(T)+n(O)+n(T\cap O)

n(T\cup O)=10+10-1

n(T\cup O)=19

Therefore, 19 positive integers are less than 100 those have at least one 9.

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Answer:

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3 years ago
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Answer:

2x+30 = 90

Step-by-step explanation:

The two angles are complementary so they add to 90 degrees

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