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Mashcka [7]
3 years ago
9

prove or disprove that if you have an 8-gallon jug of water and two empty jugs with capacities of 5 gallons and 3 gallons, respe

ctively, then you can measure 4 gallon by successively pouring some of or all of the water in a jug into another jug.
Mathematics
1 answer:
Serjik [45]3 years ago
4 0

Answer:

We can measure 4 gallon by successively.

Step-by-step explanation:

We have been given that you have an 8-gallon jug of water and two empty jugs with capacities of 5 gallons and 3 gallons, respectively. We are asked to prove or disprove that you can measure 4 gallon by successively pouring some of or all of the water in a jug into another jug.

First of all, we will pour 5 gallons of water from 8 gallon jug. Then we will pour 5 gallons of water from 5 gallon jug to 3 gallon jug. That will result in 2 gallons of water in 5 gallon jug.

Now, we will pour 3 gallons of water from 3 gallon jug to 8 gallon jug and pour 2 gallons from 5 gallon jug into 3 gallon jug. That will give us 6 gallons in 8 gallon jug and 2 gallons in 3 gallon jug and empty 5 gallon jug.

In our last step, we will fill 5 gallons jug from 8 gallon jug. After filling 5 gallon jug, we will fill 3 gallons jug from 5 gallon jug. Since 3 gallon jug already contains 2 gallons water, so it can only take one more gallon.

After pouring one gallon from 5 gallon jug, we will get 4 gallons water in 5 gallon jug.

Therefore, we can measure 4 gallon by successively pouring some of the water in a jug into another jug.

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f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int
eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

as

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0
3 years ago
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