Three tanks are capable of holding 361,841, and 901 of the greatest vessel winch can be used to fill each one of them on exact n
1 answer:
Answer:
Tank which is capable of holding 1 can fill each one of the tanks an exact number of time.
Step-by-step explanation:
Three tanks are capable of holding
To find which tank can be used to fill each one of them an exact number of time, find
Therefore,
So, tank which is capable of holding 1 can fill each one of the tanks an exact number of time.
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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,
(* 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