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maria [59]
3 years ago
9

Mike had 280 dollars to spend on 8 books.After buying them he had 16 dollars. How much did each book cost?

Mathematics
2 answers:
adell [148]3 years ago
6 0
Well, you can't really determine the price of 1 book, but you can find out the average cost.

280 - 16 = 8x (x = cost of 1 book)
264 = 8x
264/8 = 8x/8
33 = x

So the average cost of one book is 33$( A little expensive if I do say so myself)
Hope this helped!
musickatia [10]3 years ago
3 0
280 subtract 16
264 divided by 8

Each book cost $33 
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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}.

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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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

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

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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