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zhuklara [117]
2 years ago
9

HELLLLLLP RNNN PLSSSSSSSSSSSS

Mathematics
2 answers:
Rashid [163]2 years ago
8 0

Answer:

Your answer would be number 4, "A and D"

Step-by-step explanation:

I already know that if you multiply a number by less than 1 it will give a product less than the original number so that makes B and C not the correct answer leaving a and d

I hope this helps :D

svetoff [14.1K]2 years ago
5 0

Answer:

D

Step-by-step explanation:

5/6 times 13 is 10 5/6. No

4 times 13 is 52. Yes

3/4 times 13 is 9 3/4. No

10/10 times 13 is 13. Yes

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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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Rhombus has that equiangular and is equilateral but I'm not entirely sure about this...

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