Find x and h in this trinomial right angle

Step-by-step explanation:

Set up the Pythagorean Equation for the two triangles

x^2 + h^2 = 4^2  = 16

(x + 6)^2 + h^2 = 8^2 = 64

x^2 +    h^2 = 16                        Subtract. That will get rid of h^2

(x + 6)^2 - x^2 = 64 - 16            Expand the brackets. combine the right.

x^2 + 12x+36 - x^2 = 48           Combine

12x + 36 = 48                            Subtract 36 from both sides

12x +36 -36 = 48-36                 Combine

12x = 12                                      Divide by 12

12x/12 = 12/12

x = 1

x^2 + h^2 = 4^2

1^2 + h^2 = 16

h^2 = 15

h = sqrt(15)  

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

u rich

step-by-step explanation: