If we are going to talk about growth rate the only thing you need to do is:
1.42%/100 = .0142
I hope this can help a lot

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Posted in Mathematics

What is the maximum value that the graph of y = cos x assumes?

maximum value that the graph y = cos x assumes is 1

Step-by-step explanation:

y =cos x is the function given.

By definition of trignometric ratios, cos x = adjacent side/hypotenuse of a right triangle.

Obviously hypotenuse can never be less than a side and hence cos x can take maximum value as 1 only

cosx is a periodic function with period 2pi and maximum value 1

maximum value that the graph of y = cos x assumes=1

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Posted in Mathematics

What is the best description of an isolated system?

There are many definitions and descriptions of isolated systems, and here are some of them:
1. it can be a physical system which is located very very far from all other systems, so there is absolutely no interaction between them, thus making it isolated
2. it can be a thermodynamic system with rigid walls, which prevents mass and energy to pass through

Posted in Mathematics

Integrate (1/(x^2*sqrt(x^2-9))dx

The first thing you need to do is say that x = 3sec(u) ==> dx = 3sec(u)tan(u) du
If we say this we can proceed in this manner
∫ 1/[x^2√(x^2 – 9)] dx
= ∫ 3sec(u)tan(u)/{9sec^2(u)√[9sec^2(u) – 9]} du, by applying substitutions
= ∫ 3sec(u)tan(u)/{27sec^2(u)tan(u)] du, since tan^2(u) = sec^2(u) – 1
= 1/9 ∫ 1/sec(u) du, by canceling sec(u)tan(u) du
= 1/9 ∫ cos(u) du, since 1/sec(u) = cos(u)
= (1/9)sin(u) + C.

With x = 3sec(u) ==> sec(u) = x/3 and cos(u) = 3/x, we have:
sin(u) = √[1 – cos^2(u)] = √[1 – (3/x)^2] = √(x^2 – 9)/x.

Therefore, back-substituting yields:
∫ 1/[x^2√(x^2 – 9)] dx = (1/9)sin(u) + C
= (1/9)[√(x^2 – 9)/x] + C
= √(x^2 – 9)/(9x) + C.
I hope this helps a lot

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