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hoa [83]
3 years ago
8

Evaluate 5 - t/3 when t = 12.

Mathematics
2 answers:
Nady [450]3 years ago
4 0

5 - t/3

5 - 12 /3

-7/3

Iteru [2.4K]3 years ago
3 0

I"m guessing you want us to evaluate 5-\frac[t}{3} and not \frac{5-t}{3}


if t=12, then 5-(t/3)=5-(12/3)=5-4=1

if it is the other one then

if t=12, then (5-t)/3=(5-12)/3=7/3=2.33333333


so answer is 1

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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:

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Then,

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

Now, recall the definition of the complex exponential:

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So,

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

e^{-2i+3} = e^{3}(\cos 2-i\sin 2) (we use that \sin(-y)=-\sin y).

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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:

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

Now, substituting this equality in (I) we get

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\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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