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3 years ago

Can someone help me with this

Mathematics

2 answers:

Dafna1 [17]3 years ago

5 0

Answer:

18.18 /// 17.9982

Step-by-step explanation:

Multiplying by 100 means moving up the other number by two digits,

soo 100x (or 100(0.1818) is now 18.18

For the next one:

0. 1818

x 99

---------------

1.6362

16.3620

-------------

17.9982

I have no idea about that fraction part

anastassius [24]3 years ago

3 0

It would probably be 17.9982

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Select the action you would use to solve 3x = 12. Then select the property

podryga [215]

Answer:

C: Divide both sides by 3

F: Division property of equality

Step-by-step explanation:

$3x=12\\x=\frac{12}{3}\\x=4$

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3 years ago

A store is having a sale where winter clothes are 60% of the original price. A sweater is on sale for $30. What is the original

Thepotemich [5.8K]

Answer:$50

Step-by-step explanation:

So we have the sweater that’s $30 with a sale of 60% of the original price.

That means $30=60% of the original

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3 years ago

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3 years ago

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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,

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

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

5 0

3 years ago

Can someone help me with this ​

Can someone help me with this