Alex has 64 cubes with dimensions in feet like the one shown. (1/2 ft). He uses all the cubes to fill a box shaped like a larger rectangular prism. What is the volume, in feet, of the larger rectangular prism?

MARK AS BRAINLIST PLZ:)

A) First, find the volume of each cube:

V = s³

  = (1/2)³

  = 1/8 ft³

We have a total of 64 cubes, therefore the maximum volume will be:

Vmax = 64 · (1/8) = 8 ft³

B) The volume of a rectangular prism is given by:

V = w × l × h

Therefore any combination of w × l × h = Vmax = 8 will be fine.

Examples:

2 × 2 × 2

4 × 2 × 1

Step-by-step explanation:

HOPE IT HELPS BRAINLIST

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

answer: ouch. thats difficult

step-by-step explanation: