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3 years ago

Help me please I’m confused

Mathematics

1 answer:

sergiy2304 [10]3 years ago

5 0

Answer:

Its c

Step-by-step explanation:

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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

Are These shapes similar or congruent

Klio2033 [76]

Answer:

is it congruent !!!!!!!

4 0

3 years ago

Why do we say that a number is squared when raised to the power of 2

yawa3891 [41]

Answer:

We say that it is squared because we are multiplying that number twice. A square has two sides, therefore we are squaring the number.

I hope this helps. Please mark me Branliest if It did! Thank you and have a nice day!

4 0

3 years ago

Read 2 more answers

I NEED HELP ASAP PLS

xeze [42]

Answer:mot sire

Step-by-step explanation:

4 0

3 years ago

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 14 of them

Mariulka [41]

Answer:

<em>Two possible answers below</em>

Step-by-step explanation:

<u>Probability and Sets</u>

We are given two sets: Students that play basketball and students that play baseball.

It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.

This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

$\displaystyle P=\frac{19}{29}$

P = 0.66

Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:

We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.

Thus, there 19-2=17 students who play only one of the sports. The probability is:

$\displaystyle P=\frac{17}{29}$

P = 0.59

3 0

3 years ago