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nikdorinn [45]
2 years ago
11

The normal water level in the river is 0. Now it is 2 feet below normal.

Mathematics
2 answers:
UNO [17]2 years ago
7 0
You need to know how much rain per hour there will be.

Mademuasel [1]2 years ago
6 0

The question can't be answered with the information given.

First of all, we don't know how fast the rain is falling ... how many inches
per hour.

But even of we knew that, that would only tell us how much water would
pile up in our yard or on our street.  You're talking about a river, and
that's a whole different ball game. 

The rain that falls right here is going to flow down the river to somewhere
else.  But the rain that falls up-river from us is going to flow down to where
we are, and the river is going to rise.  The amount it rises depends on . . .

-- how fast the rain is falling,
-- how much area it's falling on,
-- how much forest there is in that area and how many houses and
         parking lots where the ground can't soak up the water and hold it,
-- and also on how wide the river is and how fast the water can flow through it. 


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Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

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3 years ago
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e-lub [12.9K]
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zhenek [66]
207,048 is the answer :)
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2 years ago
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bonufazy [111]

Answer:

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Vcylinder=hpir^2 Vsphere=(4/3)pir^3 Vcone=(1/3)hpir^2 Vcylinder=15*pi*5^2=375pi in^3 Vsphere=(4/3)*pi*6^3=288pi in^3 Vcone=(1/3)*15*pi*8^2=320pi in^3 greatest is Vcylinder at 375pi in^3 answer is A (cylinder)
8 0
2 years ago
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