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3 years ago

15

The width of this rectangle is measured as 19.4mm correct to 1 decimal place.

1 answer:

marishachu [46]3 years ago

7 0

We need to find the shortest possible width and length to get the smallest possible area.
To get the boundaries for 19.4, we go on to the next significant figure (the hundredths) and <span>± 5 of them.
The boundaries are, therefore: 19.35 - 19.45
As for the length, we can see they've added 5 units as the measurement is correct to 2 sig' figures, which is the tens.
And so, if we do as we did before, we go to the next sig' figure (the units) and </span>± 5 of them, we get the boundaries to be 365 - 375.

Now, we just multiply the lower bounds of the length and width to get the minimal/lower-bound area:
365 * 19.35 = 7062.75 mm²

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The water level is rising at a rate of approximately 0.1415 meters per minute.

Step-by-step explanation:

Water is flowing into a right cylindrical-shaped swimming pool at a rate of 4 cubic meters per minute. The radius of the base is 3 meters.

And we want to determine the rate at which the water level of the pool is rising.

Recall that the volume of a cylinder is given by:

$\displaystyle V = \pi r^ 2h$

Since the radius is a constant 3 meters:

$\displaystyle V = 9\pi h$

Water is flowing at a rate of 4 cubic meters per minute. In other words, dV/dt = 4 m³ / min.

Take the derivative of both sides with respect to <em>t: </em>

<em /> $\displaystyle \frac{d}{dt}\left[ V\right] = \frac{d}{dt}\left[ 9\pi h\right]$ <em />

Implicitly differentiate:

$\displaystyle \frac{dV}{dt} = 9\pi \frac{dh}{dt}$

The rate at which the water level is rising is represented by dh/dt. Substitute and solve:

$\displaystyle \left(4 \right) = 9\pi \frac{dh}{dt}$

Therefore:

$\displaystyle \frac{dh}{dt} = \frac{4}{9\pi} \approx 0.1415\text{ m/min}$

In conclusion, the water level is rising at a rate of approximately 0.1415 meters per minute.

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