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drek231 [11]
3 years ago
5

How many two digit whole numbers are increased by 18 when their digits are reversed?

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0
To reverse by adding 18, the first digit must be smaller,and the second digit must be two greater.
Start with
13 -> 31
24 -> 42
35 -> 53
46 -> 64
57 -> 75
68 ->86
79 -> 97
That should be all together 7 two digit whole numbers.
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Step-by-step explanation:

3452.2

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There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of
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Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta (1)

Where:

AC - The distance from A to C, measured in miles.

AB - The distance from A to B, measured in miles.

BC - The distance from B to C, measured in miles.

\theta - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta

\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}

\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right) (2)

If we know that BC = 7\,mi, AB = 8\,mi, AC = 5\,mi, then the bearing from island B to island C:

\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]

\theta \approx 38.213^{\circ}

The bearing needed to navigate from island B to island C is approximately 38.213º.

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