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klasskru [66]
3 years ago
6

What is the fractional equivalent of the repeating decimal n=0.2727…?

Mathematics
1 answer:
erastova [34]3 years ago
7 0

Answer: 3/11 would be the answer

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Which ordered pairs are a solution to the equation:
denpristay [2]
The first and the second
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3 years ago
Someone do this got me rq
skelet666 [1.2K]

Answer:

The last listed functional expression:

\left \{ {x+1\,\,\,\,{x\geq 2} \atop {x+2 \,\,\,\,x

Step-by-step explanation:

It is important to notice that the two linear expressions that render such graph are parallel lines (same slope), and that the one valid for the left part of the domain, crosses the y-axis at the point (0,2), that is y = 2 when x = 0. On the other hand, if you prolong the line that describes the right hand side of the domain, that line will cross the y axis at a lower position than the previous one (0,1), that is y=1 when x = 0. This info gives us what the y-intercepts of the equations should be (the constant number that adds to the term in x in the equations: in the left section of the graph, the equation should have "x+2", while for the right section of the graph, the equation should have x+1.

It is also important to understand that the "solid" dot that is located in the region where the domain changes, (x=2) belongs to the domain on the right hand side of the graph, So, we are looking for a function definition that contains x+1 for the function, for the domain: x\geq 2.

Such definition is the one given last (bottom right) in your answer options.

\left \{ {x+1\,\,\,\,{x\geq 2} \atop {x+2 \,\,\,\,x

7 0
3 years ago
Compute: 5 ∙ (2 + 3i)
dimulka [17.4K]

Answer:

It will be equal to 40

Step-by-step explanation:

We have to compute 5\times (2+3!)

Let first find 3 !

For this we have to use factorial concept

So 3! will be equal to 3! = 3×2×1 = 6

Now According to 6+2 = 8

And now after solving bracket we have to multiply with 5

So 5×8 = 40

So after computation 5\times (2+3!)=40

So the final answer will be 40

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5(2j+3+j)
= 5(2j) + 5(3) + 5(j) we distribute 5 to all of the expression
= 10j + 15 + 5j connect like terms j
= 15j + 15
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