Write the number with the same value as 28 tens

The right answer is C. y> 1

The solution's average current density is 1 ampere per square decimeter.

We need to convert this value to ampere per square meter.

Since 1 decimeter equals meters,
squaring gives us:

1 square decimeter equals square meters.

Hence, the current density now is .

Therefore, the current density of the electrolyte solution is 100 amperes per square meter.

Answer:

The shaped region accounts for 7/18 of the area of ACIG.

Step-by-step explanation:

Refer to the attached diagram for further clarity on the problem.

Step 1

Determine the length of one side of square ABED.

We know that

AB=BE=ED=AD

The area of a square can be calculated as

where b is the side length.

We have

So we substitute

Thus,

Step 2

Calculate the area of ACIG.

The area of rectangle ACIG is determined by

Substituting the given values yields

Step 3

Determine the area of the shaded rectangle DEHG.

The area of rectangle DEHG is given by

We find

and substitute

Step 4

Calculate the area of shaded rectangle BCFE.

The area of rectangle BCFE equals

We see that

and substitute

Step 5

Add the areas of the shaded regions together.

Step 6

Divide the area of the shaded region by the area of ACIG.

Simplify this fraction by dividing both the numerator and denominator by 5.

Hence, the shaped region represents 7/18 of the area of ACIG.

To determine the rates at which the inlet and outlet pipes fill and empty the reservoir, we remember that work done equals rate multiplied by time. Let’s denote the inlet rate as i and for the outlet pipe as 0. Therefore,
i(24) = 1
o(28) = 1
In this context, the '1' represents the total number of reservoirs, since the problem states the time needed for each pipe to either fill or empty a singular reservoir. Solving for rates yields:
i = 1/24 reservoirs/hour
o = 1/28 reservoirs/hour

Over the first six hours, the inlet pipe fills (1/24)(6) = 1/4 reservoirs and during the same period, the outlet pipe empties (1/28)(6) = 3/14 reservoirs. To calculate the net volume of the reservoir filled, we subtract the emptying total from the filling total:
1/4 - 3/14 = 1/28 reservoirs (note that if emptying exceeds filling, a negative value results. In such cases, treat that negative value as zero, indicating that the outlet rate surpasses the inlet rate, leading to an empty reservoir).
Now we need to find out how long it will take to fill up one reservoir since we’ve already partially filled 1/28 of it, after closing the outlet pipe. In simpler terms, we need to determine the time required for the inlet pipe to finish filling the remaining 27/28 of the reservoir. Fortunately, we have already established the filling rate for the inlet pipe, leading to the equation:
(1/24)t = 27/28
Solving for t gives us 23.14 hours. Remember to add the initial 6 hours to this result since the question seeks the total time. Thus, the final total is 29.14 hours.

Please ask me any questions you may have!

Answer:

Wyjaśnienie krok po kroku:

Załóżmy, że prostokąt ma wierzchołki (x,y,z) w dodatnim oktancie.  Prostokątny pudełko musi być symetryczne względem wszystkich trzech osi.

Wtedy jego boki to

Objętość =

Maksymalizuj objętość przy założeniu

tj.

Użyj mnożników Lagrange'a, mamy

przy maksimum

Dzieląc otrzymujemy

Podobnie

Zatem otrzymujemy

Stąd wymiary to

(2x,2y,2z)

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