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2 years ago

Find the measure of the angle labeled a in the diagram below

Mathematics

1 answer:

gogolik [260]2 years ago

7 0

Where is the diagram.

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HELP PLEASE ASAP! Tell how you found 2k + 10! 20 points!

Ivan

Answer:

2(k+5)

Step-by-step explanation:

Factor 2k+10

2k+10

=2(k+5)

Answer:

2(k+5)

8 0

3 years ago

What is the coefficient of 9B in 5A

Norma-Jean [14]

If your asking what 9B coefficient is it's B

3 0

3 years ago

True or false: equilateral triangles can be classified as acute right or obtuse

Ne4ueva [31]

False, since all sides are equal all angles must be equal and when you have an obtuse or right triangle only on angle is obtuse or right. since one is different it cannot be equal on all sides making this statement false

8 0

2 years ago

Which of the following are examples of inferential statistics?

alexgriva [62]

The answer is B. Testing random samples of wild animals to decide whether they are healthy

5 0

3 years ago

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

$\frac{4(sin(420-140)) }{sin280}$

= $\frac{4sin280}{sin280}\\ = 4 \ Proved!$

3 0

2 years ago