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Order the set of numbers from least to greatest: square root 64, 8 and 1 over 7, 8.14 repeating 14, 15 over 2

A 15 over 2, square root 64, 8 and 1 over 7, 8.14 repeating 14
B 8 and 1 over 7, 8.14 repeating 14, square root 64, 15 over 2
C square root 64, 8 and 1 over 7, 8.14 repeating 14, 15 over 2
D 15 over 2, square root 64, 8.14 repeating 14, 8 and 1 over 7

PLEASE HELP

Respuesta :

LammettHash
First, some rewriting:

[tex]\sqrt{64}=8[/tex]

[tex]8.1414\ldots=\dfrac{806}{99}[/tex]

Why? If [tex]x=8.1414\ldots[/tex], then [tex]100x=814.1414\ldots[/tex], which means [tex]100x-x=99x=814-8=806[/tex], and so [tex]x=\dfrac{806}{99}[/tex].

[tex]\dfrac{15}2=7.5[/tex]

So from the four given numbers, it's obvious that [tex]\dfrac{15}2[/tex] is the smallest, followed by [tex]\sqrt{64}[/tex]. So you just need to determine which of [tex]8.1414\ldots[/tex] and [tex]8+\dfrac17[/tex] is smallest.

Note that

[tex]8+\dfrac17=\dfrac{57}7[/tex]

To easily compare this number to [tex]8.1414\ldots=\dfrac{806}{99}[/tex], we need to find a common denominator. The greatest common divisor of 7 and 99 is 1 (they are relatively prime), so [tex]\mathrm{lcm}(7,99)=7\cdot99=693[/tex]. So we write


[tex]\dfrac{57}7\cdot\dfrac{99}{99}=\dfrac{5643}{693}[/tex]
[tex]\dfrac{806}{99}\cdot\dfrac77=\dfrac{5642}{693}[/tex]

So we find that [tex]\dfrac{806}{99}<\dfrac{57}7[/tex], that is, [tex]8.1414\ldots<8+\dfrac17[/tex].

So the proper ordering from least to greatest is

[tex]\dfrac{15}2,\sqrt{64},8.1414\ldots,8+\dfrac17[/tex]

Answer:

D.

Step-by-step explanation:

15/2, √64, 8.14 repeating, 8 1/7

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