Let the lengths of the bottom of the box be x and y, and let the length of the squares being cu be z, then
V = xyz . . . (1)
2z + x = 16 => x = 16 - 2z . . . (2)
2z + y = 30 => y = 30 - 2z . . . (3)
Putting (2) and (3) into (1) gives:
V = (16 - 2z)(30 - 2z)z = z(480 - 32z - 60z + 4z^2) = z(480 - 92z + 4z^2) = 480z - 92z^2 + 4z^3
For maximum volume, dV/dz = 0
dV/dz = 480 - 184z + 12z^2 = 0
3z^2 - 46z + 120 = 0
z = 3 1/3 inches
Therefore, for maximum volume, a square of length 3 1/3 (3.33) inches should be cut out from each corner of the cardboard.
The maximum volume is 725 25/27 (725.9) cubic inches.
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Answer:
A) 1.3×10^37 ergs
B) 1.435×10^3 mm
Step-by-step explanation:
A) The amount of energy will be the product of the energy rate and time:
(3.9×10^33 ergs/s)×(3.25×10^3 s) =12.675×10^(33+3) ergs
= 1.2675×10^(1+36) ergs
= 1.2675×10^37 ergs ≈ 1.3×10^37 ergs
The mantissa of the result is the product 3.9×3.25, adjusted to have one digit left of the decimal point. The exponent of the result is the sum of the exponents of the factors, adjusted by 1 to match the adjustment in the mantissa.
The final value should be rounded to 2 significant figures, reflecting the precision of the sun's energy production.
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B) A millimeter is a small fraction of an inch. 10^-3 mm is a small fraction of the width of a human hair, so 1.435×10^-3 mm is not a reasonable estimate of the distance between railroad tracks.
On the other hand, 1.435×10^3 mm is 1.435 m, almost 56.5 inches. This is a much more reasonable measurement of the distance between railroad rails.
1.435×10^3 mm is more reasonable
30% of the chairs are red. 75/250 x x/100
Look in the images it is solved.
Answer:
D. Neither table represents a function
Step-by-step explanation:
Functions require a key to go by. In this case, none of the two tables have a key from the start going down.
