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miv72 [106K]
3 years ago
8

Use geometry to evaluate the integral from 0 to 6 of the function f of x, dx for f of x equals 3 for x less than or equal to 3 a

nd equals the quantity 6 minus x for x greater than 3.
27
13.5
12
10.5

Mathematics
1 answer:
allochka39001 [22]3 years ago
6 0
Check the picture below.

\bf f(x)=
\begin{cases}
3&x\le 3\\
6-x&x\ \textgreater \ 3
\end{cases}\qquad \displaystyle \int_{0}^{6}~f(x)\cdot dx

so, you simply have a 3x3 square and a triangle with a base of 3 and a height of 3, get their areas, sum them up, it'd come out to 9 + 4.5 or 13.5.

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Emily thinks the perfect tomato sauce has 8 cloves of garlic in every 500{ mL} of sauce. Raphael's tomato sauce has 12 cloves of
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Where should the point P be chosen on line segment AB so as to maximize the angle θ? (Assume a = 4 units, b = 5 units, and c = 9
taurus [48]
From the figure, let the distance of point P from point A on line segment AB be x and let the angle opposite side a be M and the angle opposite side c be N.

Using pythagoras theorem,
\tan M= \frac{a}{b-x} \\ \\ M=\tan^{-1}\left(\frac{a}{b-x}\right)
and
\tan N= \frac{c}{x} \\ \\ N=\tan^{-1}\left(\frac{c}{x}\right)

Angle θ is given by
\theta=180-M-N \\  \\ =180-\tan^{-1}\left(\frac{a}{b-x}\right)-\tan^{-1}\left(\frac{c}{x}\right)

Given that a = 4 units, b = 5 units, and c = 9 units, thus
\theta=180-\tan^{-1}\left(\frac{4}{5-x}\right)-\tan^{-1}\left(\frac{9}{x}\right)

To maximixe angle θ, the differentiation of <span>θ with respect to x must be equal to zero.
i.e.
\frac{d\theta}{dx} = -\frac{4}{x^2-10x+41} + \frac{9}{x^2+81} =0 \\  \\ -4(x^2+81)+9(x^2-10x+41)=0 \\  \\ -4x^2-324+9x^2-90x+369=0 \\  \\ 5x^2-90x+45=0 \\  \\ x^2-18x+9=0 \\  \\ x=9\pm6 \sqrt{2}

Given that x is a point on line segment AB, this means that x is a positive number less than 5.

Thus
x=9-6 \sqrt{2}=0.5147

Therefore, The distance from A of point P, so that </span>angle θ is maximum is 0.51 to two decimal places.
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4 years ago
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