How is using the pythagorean theorem in a rectangular prism similar to using it in a rectangle

ok

Step-by-step explanation:

well first off a rectangle does not have a diameter...  

it has a width and a length  

u could use pythag to solve for a side of the rectangle, given that there is also a triangle inscribed into the rectangle  

lets say ur given the width of a rectangle and u have to solve for the length  

u are also given the diagonal which stretches from one corner to the other forming a triangle with the width and length of the rectangle  

say the diagonal was 5cm long and the width was 3cm long, and u have to find the length, x  

5^2 = 3^2 + x^2  

25 = 9 + x^2  

x^2 = 16  

x=4

question

what is the value of the third quartile of the data set represented by this box plot? a. 19 b. 21 c. 26 d. 29

answer

c

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1