The diagram below shows a tall beaker with four different liquids and their densities. A 0.89 g/cm B 1.00 g/cm с 1.21 g/cm3 D 2.34 g/cm3 If a ball that has a density of 1.73 g/cm3 is placed in the beaker, where will the ball come to rest? O on top of liquid A O between liquids B and C between liquids C and D O on the bottom of the beaker​

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)

i use my brain for prombles like this maybe you could try it

step-by-step explanation:

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )

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