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If you're looking for the cube roots of √(3 + i ), you first have to decide what you mean by the square root √(…), since 3 + i is complex and therefore √(3 + i ) is multi-valued. There are 2 choices, but I'll stick with 1 of them.

First write 3 + i in polar form:

3 + i = √(3² + 1²) exp(i arctan(1/3)) = √10 exp(i arctan(1/3))

Then the 2 possible square roots are

• √(3 + i ) = ∜10 exp(i arctan(1/3)/2)

• √(3 + i ) = ∜10 exp(i (arctan(1/3)/2 + π))

and I'll take the one with the smaller argument,

√(3 + i ) = ∜10 exp(i arctan(1/3)/2)

Then the 3 cube roots of √(3 + i ) are

• ∛(√(3 + i )) = ¹²√10 exp(i arctan(1/3)/6)

• ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + π/3))

• ∛(√(3 + i )) = ¹²√10 exp(i (arctan(1/3)/6 + 2π/3))

On the off-chance you meant to ask about the cube roots of 3 + i, and not √(3 + i ), then these would be

• ∛(3 + i ) = ⁶√10 exp(i arctan(1/3)/3)

• ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/3 + 2π/3))

• ∛(3 + i ) = ⁶√10 exp(i (arctan(1/3)/6 + 4π/3))

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.