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aev [14]
2 years ago
5

WILL GIVE BRAINLIEST: Which function has a removable discontinuity?

Mathematics
1 answer:
elena55 [62]2 years ago
8 0

Answer:

First choice

Step-by-step explanation:

The discontinuity is removable if by reducing the fraction that discontinuity doesn't continue to exist as a discontinuity.

Example of, (x-1)/(x-1) has a a discontinuity at x=1 and it's removable because the fraction reduces to 1 which doesn't have a discontinuity at x=1.

Example not of, (x-1)/(x-2) has a discontinuity at x=2 and it is not removable because we can't get rid of the x-2 factor in the denominator.

The first choice has a discontinuity at x=-1 and it is removable because x^2-x-2=(x-2)(x+1) and the x+1's will cancel on top and bottom making the point at x=-1 a removable discontinuity.

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(2y-at)(y+2at)


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Osama starts with a population of 1,000 amoebas that increases 30% in size every hour for a number of hours, h. The expression 1
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Answer: The correct option is A, itis the product of the initial population and the growth factor after h hours.

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Initial population = 1000

Increasing rate or growth rate = 30% every hour.

No of population increase in every hour is,

1000\times \frac{30}{100} =1000\times 0.3

Total population after h hours is,

1000(1+0.3)^h

It is in the form of,

P(t)=P_0(t)(1+r)^t

Where P_0(t) is the initial population, r is increasing rate, t is time and [tex(1+r)^t[/tex]  is the growth factor after time t.

In the above equation 1000 is the initial population and (1+0.3)^h is the growth factor after h hours. So the equation is product of of the initial population and the growth factor after h hours.

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3 years ago
Find e^cos(2+3i) as a complex number expressed in Cartesian form.
ozzi

Answer:

The complex number e^{\cos(2+31)} = \exp(\cos(2+3i)) has Cartesian form

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Step-by-step explanation:

First, we need to recall the definition of \cos z when z is a complex number:

\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}.

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\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}. (I)

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Thus,

e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)

Now we group conveniently in the above expression:

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Thus,

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)

\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right].

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