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Alex777 [14]
2 years ago
8

Which fraction converts to a repeating decimal number?

Mathematics
2 answers:
nekit [7.7K]2 years ago
8 0

Answer:

Step-by-step explanation:

your answer is A, 1/12

musickatia [10]2 years ago
5 0
<h3>Answer: A) 1/12</h3>

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

Explanation:

Using a calculator shows that

  • 1/12 = 0.0833333.... where the 3's go on forever
  • 7/8 = 0.875
  • 14/25 = 0.56
  • 17/20 = 0.85
  • 6/10 = 0.6

We see that nearly all of the decimal values are terminating decimals. This means the decimals don't go on forever, so the decimals stop at some point.

Only 0.083333... goes on forever, so that's why 1/12 converts to a repeating decimal number.

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

A way you can determine this without a calculator is to use long division. Or you could use the idea that if the denominator has prime factors other than 2 and 5, then the decimal will repeat forever.

Note how the denominators of the fractions for choices B through E all have prime factors of either 2s only, 5s only, or a mix of 2s and 5s. This fact is why we have terminating decimals here.

In contrast, 12 has prime factorization 2*2*3. That 3 as a prime factor is why the decimal repeats forever.

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Use compatible numbers to estimate the quotient <br> 252÷3
ICE Princess25 [194]

Answer:

The value of the given expression is 84.

Step-by-step explanation:

The given expression is

\frac{252}{3}

The number 252 can be written as 240+12, because 24 can easily divisible by 3.

The compatible numbers are 240 and 12

\frac{252}{3}=\frac{240+12}{3}

\frac{252}{3}=\frac{240}{3}+\frac{12}{3}

\frac{252}{3}=80+4

\frac{252}{3}=84

Therefore the value of the given expression is 84.

4 0
2 years ago
Write a polynomial function of least degree with integral coefficients that has the
frutty [35]

A polynomial function of least degree with integral coefficients that has the

given zeros  f(x)=x^4+x^3+9x^2+9x

Given

Given zeros are 3i, -1  and 0

complex zeros occurs in pairs. 3i is one of the zero

-3i is the other zero

So zeros are 3i, -3i, 0 and -1

Now we write the zeros in factor form

If 'a' is a zero then (x-a) is a factor

the factor form of given zeros

\:\left(x-3i\right)\left(x-\left(-3i\right)\right)\left(x-0\right)\left(x-\left(-1\right)\right)\\\left(x-3i\right)\left(x+3i\right)\left(x-0\right)\left(x+1\right)

Now we multiply it to get the polynomial

x\left(x-3i\right)\left(x+3i\right)\left(x+1\right)\\x\left(x^2+9\right)\left(x+1\right)\\x\left(x^3+x^2+9x+9\right)\\x^4+x^3+9x^2+9x

polynomial function of least degree with integral coefficients that has the

given zeros  f(x)=x^4+x^3+9x^2+9x

Learn more : brainly.com/question/7619478

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Answer:

Step-by-step explanation:

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