5x + 1 over 2y2

what is the value of the expression above when x = 3 and y = 2? you must show all work and calculations to receive full credit.
hurry

2

Step-by-step explanation:

First, write it out.

5x + 1 / 2y^2

Then plug in the values.

5(3) + 1 / 2(2)^2

Then simplify.

15 + 1 / 2(4)

16 / 8

2

2

Step-by-step explanation:

First, write it out.

5x + 1 / 2y^2

Then plug in the values.

5(3) + 1 / 2(2)^2

Then simplify.

15 + 1 / 2(4)

16 / 8

2

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)