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lawyer [7]
3 years ago
15

How do I establish this identity??

Mathematics
1 answer:
Aloiza [94]3 years ago
8 0
\frac{cosec\theta+cotan\theta} {sec\theta+tan\theta}=\frac{cosec\theta+cotan\theta} {sec\theta+tan\theta}\times\frac{cosec\theta-cotan\theta}{cosec\theta-cotan\theta}\times\frac{sec\theta-tan\theta} {sec\theta-tan\theta}

And use that,

\sec^2\theta-\tan^2\theta=1

cosec^2\theta-cotan^2\theta=1
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Rewrite the inequality into slope Intercept form. <br> 6x-2y&gt;-2<br> PLEASE PLEASE PLEASE HELP!!!
BARSIC [14]

Answer:

I think it's y < 3x +1

Step-by-step explanation:

subtract 6x to the right side. that's -2y > -6x - 2. then divide -2 out of y, now you have to flip the sign. Now you have y < 3x + 1

5 0
3 years ago
Read 2 more answers
Mary is an years old. What expression represents her age 7 years from now? Two years ago?
Rudiy27

Answer

What expression represents her age 7 years from now?

y =x+7x

Two years ago?

z=y-2x

Step-by-step explanation:

7 0
3 years ago
Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to
eduard
The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079 
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
ln(0.8) = (-0.0079 * t) * ln(e)
-0.223 = -0.0079 * t
t = -0.223 / -0.0079
t = 28.23
t ≈ 28 years
8 0
3 years ago
Two angles have a common vertex but no common side. How can you tell whether the angles are vertical angles?
pentagon [3]
Vertical angles is an X but that's how you tell if they're vertical angles
8 0
3 years ago
I need help really bad on 1 and 2
Dmitry [639]
For question one:
First class goes from 8:00 to 9:00 second class starts 10 mins later at 9:10 and goes to 10:10. Ten minute break again 3rd class starts at 10:20, goes to 11:20. 10 min break again, fourth class starts 10 mins later at 11:30 and goes to 12:30. The answer for 1 is 12:30
8 0
3 years ago
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