Arectangular box is constructed in 3-space with one corner at the origin and other vertices at (1, 0, 0), (0, 4, 0), (0, 0, 2). find the length of the diagonal of the box.

The space diagonal will have length ...

... d = √(1² +4² +2²) = √(1 +16 +4) = √21

This can be found using the Pythagorean theorem. A drawing can help. Find the length of any face diagonal, then use that length as the leg of a right triangle whose hypotenuse is the space diagonal and whose other leg is the edge length not used in the first calculation.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb

false

step-by-step explanation:

the reason i say this is because for a certain length of sides you can pick any two and the length is wither equal or greater than the third side and in this case it's not.

3 + 5 = 8

8 is smaller than 9 which is the third side.

hope this : )