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3 years ago

EASY MATH QUESTION HELP! Under the Angle Addition Postulate, which equation would be valid?

Mathematics

2 answers:

Licemer1 [7]3 years ago

6 0

Answer:

B for the first one , and C for the second

Step-by-step explanation:

for the first question DEG which is 60, and GEF which is 120, would equal to 180 degrees, which would be DEF because its a linear pair.

For the second question the are alternate exterior angles, so the are opposite of each other and on the outside.

Dmitriy789 [7]3 years ago

5 0

Answer: Whats True about 2 angles and 7 anglesis that They are not the same number and one of them is less and great

Step-by-step explanation: This is not a easy math question.

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Please help! And 20 points! :3<br> I need help with these 3 questions-<br> -3-||

Katyanochek1 [597]

Answer:

18. Its 2 meters.

19. Its 0.2 grams

20. Its 1,920

Step-by-step explanation:

When you convert 200 centimeters to meters its 2.

When you convert 2,000 milligrams in grams its 0.2

15% of 12, 800 is 1,920

Hope this helps:)

8 0

3 years ago

Directions in picture.

densk [106]

Use PEMDAS, the answer that I got is 19.

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3 years ago

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I need help on this one question. Please help me. Find the area of the shaded region. It says round to the nearest tenth if nece

schepotkina [342]

30.44 but it depends how you do the problem

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3 years ago

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$