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slamgirl [31]
3 years ago
8

You have a 45-gallon jug and a 124-gallon jug. neither of the jugs has any markings (although you do know their capacities). des

cribe a way to measure exactly one gallon of water.
Mathematics
2 answers:
katrin [286]3 years ago
8 0
4*124 = 496
11*45 = 495

Fill the 124-gallon jug 4 times, emptying it each time into the 45-gallon jug. After the 45-gallon jug has been filled 11 times, there will be 1 gallon remaining in the 124-gallon jug.

_____
Mathematically, this works fine. In practice, you're looking for a 1-gallon difference after handling 495 gallons of water. That's about 0.2% of the total amount of water transferred. Any error along the way will substantially affect the outcome.
Alja [10]3 years ago
5 0
Pour 4  times of 124-gallon into a big jug.
Amount of water = 124 x 4 = 496 gallons

Then use the 45-gallon jug to scoop out the water 11 times.
Amount of water scooped out = 11 x 45 = 455 gallons

Water remaining = 496 - 450 = 1 gallon.

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3 years ago
The width of a container is 5 feet less than its height. Its length is 1 foot longer than its height. The volume of the containe
juin [17]

The height of container is 8 feet

<em><u>Solution:</u></em>

Let "w" be the width of container

Let "l" be the length of container

Let "h" be the height of container

The width of a container is 5 feet less than its height

Therefore,

width = height - 5

w = h - 5 ------ eqn 1

Its length is 1 foot longer than its height

length = 1 + height

l = 1 + h ---------- eqn 2

<em><u>The volume of container is given as:</u></em>

v = length \times width \times height

Given that volume of the container is 216 cubic feet

216 = l \times w \times h

Substitute eqn 1 and eqn 2 in above formula

216 = (1 + h) \times (h-5) \times h\\\\216 = (h+h^2)(h-5)\\\\216 = h^2-5h+h^3-5h^2\\\\216 = h^3-4h^2-5h\\\\h^3-4h^2-5h-216 = 0

Solve by factoring

(h-8)(h^2+4h+27) = 0

Use the zero factor principle

If ab = 0 then a = 0 or b = 0 ( or both a = 0 and b = 0)

Therefore,

h - 8 = 0\\\\h = 8

Also,

h^2+4h+27 = 0

Solve by quadratic equation formula

\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

\mathrm{For\:} a=1,\:b=4,\:c=27:\quad h=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:27}}{2\cdot \:1}

h = \frac{-4+\sqrt{4^2-4\cdot \:1\cdot \:27}}{2}=\frac{-4+\sqrt{92}i}{2}

Therefore, on solving we get,

h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i

<em><u>Thus solutions of "h" are:</u></em>

h = 8

h=-2+\sqrt{23}i,\:h=-2-\sqrt{23}i

"h" cannot be a imaginary value

Thus the solution is h = 8

Thus the height of container is 8 feet

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Mice21 [21]

Answer:

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Step-by-step explanation:

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