Ok, this is a ratio problem; the ratio of the length to width is constant (and therefore equal): 4 /6 = 15 / x Now, with a ratio, we may do any allowable algebra operation: cross-multiply, invert both sides, multiply or divide both sides by the same amount, etc. Let's cross-multiply: 4x = (15)(6) x = 90/4<span> x = 22.5 in. </span>
Label the 3 distinct sides of the box. I arbitrarily chose the letters a, b and c.
Use the info about areas as follows:
ab=54 in^2
ac=90 in^2
bc=60 in^2
Here you have 3 equations in 3 unknowns (a, b and c), which is enough info to use to determine a, b and c. Then the volume of the box is a*b*c.
Example: bc = 90, but c = 60/b. You could subst. 60/b for c in the 2nd and 3rd equation, which will eliminate c completely and leave you with 2 equations in 2 unknowns.
Continuing this procedure, I determined that a=9, b=6 and c=10. Thus, the volume of the box is V = 9*6*10 = 540 cubic inches (answer)
2(7b + 5) will equal to 14b+10
Explanation:
2 times 7b= 14b
2 times 5= 10
Answer:
6.83 units
Step-by-step explanation:
Let the height of the original pyramid be represented by h. Then the cut off top has a height of (h -2). The scale factor for the area is the square of the scale factor for height, so we have ...
(height ratio)^2 = 1/2
((h -2)/h)^2 = 1/2
(h -2)√2 = h . . . . . . square root; multiply by h√2
h(√2 -1) = 2√2 . . . . add 2√2 -h
h = (2√2)/(√2 -1) ≈ 6.8284 . . . units
The altitude of the original pyramid is about 6.83 units.
