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kirill115 [55]
3 years ago
11

14. The figure shows a slice of an apple pie in the shape of a sector of a circle with centre O and radius 8cm. A cut is made al

ong AB to remove some of the crust
Given that AB = 7 cm find:
(i) AOB in radians
(ii) the maximum number of slices that can be obtained from one full pie.
(iii) the area of the shaded segment that has been removed​

Mathematics
1 answer:
Strike441 [17]3 years ago
4 0

Step-by-step explanation:

(i) Using the law of cosines, we can find that

cos(O) = \frac{8^{2}+8^{2} - 7^{2}}{8*8*2} ≈ 0.62, so O ≈ 0.9

(ii) Since the whole pie has 2π radians (or 360 degrees), we can divide 0.9 by 2π to get around 6.9 slices from the pie

(iii) The area of the shaded segment can be found by finding sector OAB and subtracting triangle OAB from that. Sector OAB can be found by finding the area of the pie (π*8², since 8 is the radius) and dividing that by 6.9, so we get 64π/6.9 ≈ 29. Then, the area of the triangle is √p(p-a)(p-b)(p-c), with p being the perimeter and a, b, and c being the sides. Plugging 7, 8, and 8 in, we get 25.1781 as our area, so the difference is around 3.8

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nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

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Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

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=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

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