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Lyrx [107]
3 years ago
6

the computer lab is removing the three broken computers and adding 13 wireless devices about what percent of the lab are wireles

s devices
Mathematics
2 answers:
andreyandreev [35.5K]3 years ago
5 0
So the fraction would be 13/100 so now divide it and then u should get 7.6923076923



faltersainse [42]3 years ago
3 0

Answer:

13/100

Step-by-step explanation:

The computer lab is removing the three broken computers and adding 13 wireless devices.

So, about 13/100 percent of the lab are wireless devices.

Though it i not given that how many computers were there in total, but you can write like this as a fraction, to solve such questions.

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Each circle (A, B, and C) contain shapes that all share at least one characteristic. Some shapes are contained in more than one circle because they share more than one characteristic. For example, shape 3 fits the rule for circles A and B, but not circle C. It lies within circles A and B, but not circle C.

Step-by-step explanation:

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Amelia is training for a triathlon. She swims 3/5 mile. Then she runs about 6 times farther than she swims. About how far does A
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Step-by-step explanation:


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How many times do you need to run to complete 5/6 of a mile
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Break the mile in to 6 pieces and run 5 parts of the mile
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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

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

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:

e^{z}=e^{x+iy} = e^x(\cos y +i\sin y).

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).

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:

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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