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DaniilM [7]
3 years ago
15

write in scientific  notation 1. 0.0000364 2. 0.00751 3. 0.10005 4. 1,094 5. 0.00000099 6. 0.04101 7. 10,500 8. 8,9000

Mathematics
1 answer:
solmaris [256]3 years ago
3 0
1.\\0\underbrace{.00003}_{5\to}54=3.54\times10^{-5}\\2.\\0\underbrace{.007}_{3\to}51=7.51\times10^{-3}\\3.\\0\underbrace{.1}_{1\to}0005=1.0005\times10^{-1}\\4.\\1\underbrace{,094}_{\leftarrow3}=1.094\times10^3\\5.\\0\underbrace{.0000009}_{7\to}9=9.9\times10^{-7}
6.\\0\underbrace{.04}_{2\to}101=4.101\times10^{-2}\\7.\\1\underbrace{0,500}_{\leftarrow4}=1.05\times10^4\\8.\\8\underbrace{9000}_{\leftarrow4}=8.9\times10^4
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Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, d_1 = 1\times610=610 miles.

The distance covered at 310 degrees, d_2 = 3\times610=1830 miles.

Total distance covered, d, is the magnitude of the sum of vectors \vec{d_1} and \vec{d_2} as shown in the figure.

The angle between the vectors \vec{d_1} and \vec{d_2}, \theta=310-270=40 degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}

\Rightarrow d=2330.51 miles

Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.

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