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mr_godi [17]
3 years ago
9

What is the domain of validity for csc theta = 1 / sin theta?

Mathematics
1 answer:
Rashid [163]3 years ago
4 0
De finition\ of\ csc\theta=\frac{1}{sin\theta}\\-----------------------\\therefore\ csc\theta=\frac{1}{sin\theta}\to\frac{1}{sin\theta}=\frac{1}{sin\theta}\\\\the\ denominator\ must\ be\ different\ from\ zero,\ therefore\\\\sin\theta\neq0\iff\theta\neq k\pi\ where\ k\in\mathbb{Z}\\\\Answer:\boxed{D}
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Alex_Xolod [135]
Hope this helps you!

7 0
3 years ago
Find the lowest common denominator of and . A. x3y4 B. x2y3 C. xy4 D. x4y5
Blizzard [7]
Lowest Common Denominator refers to lowes t common multiple. These expressions have two terms 'x' and 'y' and we want to choose the expression that has the highest power such that the other expressions can be multiplied into the common denominator.

For the 'x' term, the highest power is x⁴ and for the 'y' term, the highest power is y⁵

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5 0
3 years ago
14. Find the value of X round the nearest degree 12 15
satela [25.4K]

Answer:

37

Step-by-step explanation:

6 0
2 years ago
Let $f(x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equ
Mademuasel [1]
We have to find the values of F.
In this case. F is unlikely to be a polynomial.
But the problem is, we can’t calculate the values of F directly.
There is no real value of x for which x = x−1 x because F isn’t defined at 0 or 1. so,
substituting x = 2.
F(2) + F(1/2) = 3.

Substitute, x = 1/2
F(1/2) + F(−1) = −1/2.
We still are not getting the required value,
therefore,
Substitute x = −1

As, F(2) +F(−1) = 0.
now we have three equations in three unknowns, which we can solve.
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F(3) = 17/12
F(4) = 47/24
and
F(5) = 99/40

Setting
g(x) = 1 − 1/x
and using
2 → 1/2
to denote
g(2) = 1/2
 we see that :
x → 1 - 1/x → 1/(1-x) →x

so that:
g(g(g(x))) = x.

Therefore, whatever x 6= 0, 1 we start with, we will always get three equations in the three “unknowns” F(x), F(g(x)) and F(g(g(x))).
Now solve these equations to get a formula for F(x)

As,
h(x) = (1+x)/(1−x)
which satisfies h(h(h(h(x)))) = x

Now, mapping x to h(x) corresponds to rotating the circle by ninety degrees.

7 0
2 years ago
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