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2 years ago

Which number (s) below represents a repeating decimal? -2/5, -7, 3/7, 11/12

Mathematics

2 answers:

Ray Of Light [21]2 years ago

6 0

Step-by-step explanation: Arrange the terms in ascending order: -7, -2/5, 3/9, 11/12.

Hope this helps! :D

-TanqR

yulyashka [42]2 years ago

4 0

Answer:

3/9

Step-by-step explanation:

A repeating decimal is a decimal in which the last number is the same forever. (Ex. .83333333333333333)

So...

2/5 = .4 Not Repeating

-7 = -7 Just a regular integer so nothing can actually repeat

3/9 = .33333 Yes. the 3 is repeating.

11/12 = 0.916666667 Repeated BUT it ends in 7, it doesn't keep continuing.

Hope this helped :)

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blagie [28]

2(3x-4) = 5x+6
First, expand the bracket.
6x-8=5x+6
Next, move the variables to the same side.
x-8=6
Move the -8 to the other side, change to +
x=6+8
x=14
Charlie’s solution is incorrect. When expanding the bracket, he changed the - to a +.

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1 year ago

The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:

deff fn [24]

Step-by-step explanation:

Given equation of circle is,

${x}^{2} + {y}^{2} - 6y - 12 = 0 \\ by \: compairing \: it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f= - 6 \\ or \: f= - 3 \\ c = - 12$

again the another form of circle is,

${x}^{2} + {(y - k)}^{2} = 21 \\ or \: {x}^{2} + {y}^{2} - 2ky + {k}^{2} - 21 = 0 \\ by \: compairing \:it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\ g = 0 \\f = - k \\ c = {k}^{2} - 21$

now equating the values of f in both equations,

-k=-3

i.e. k=3

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