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boyakko [2]
3 years ago
6

Which expression is equivalent to

Mathematics
2 answers:
Damm [24]3 years ago
8 0

Answer:

pq^9

Step-by-step explanation:

Alexus [3.1K]3 years ago
4 0
Pq^9 is your answer hope that helps
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How do you solve this equation for y 5x-10y=-40
Tanya [424]
5x - 10y = -40
-10y = -40 - 5x
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8 0
3 years ago
Question in Picture<br> vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Crank

Answer:

Angle BOC = 28; Angle AOB = 21

Step-by-step explanation:

Because angle AOC = AOB + BOC you can set the equation:

2x+10 + 4x-15 = 49

<solve for x>

6x - 5 = 49

6x = 54

x= 9

<plug in x for the angles>

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6 0
3 years ago
Given cos theta = - 2/5 and sin theta &lt; 0, find the six trigonometric values. I need help with this
Lena [83]

Answer:

\sin\,\theta =-\frac{\sqrt{21} }{5}

\tan\,\theta =\frac{\sqrt{21} }{2}

\sec\,\theta = \frac{-5}{2}

cosec\,\theta =\frac{-5}{\sqrt{21} }

\cot\,\theta =\frac{2}{\sqrt{21} }

Step-by-step explanation:

\cos\theta =\frac{-2}{5}

As both sin\,\theta,

\theta lies in the third quadrant.

In the third quadrant,

\sin\theta

\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}

\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}

\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}

\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }

\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }

3 0
3 years ago
Are these three side lengths a right triangle?<br> 5, 5, 5√s<br> yes or no?
11111nata11111 [884]

Answer:

no

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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