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densk [106]
3 years ago
8

F(x)=2x – x(2) - 2 Can you also solve it for me

Mathematics
1 answer:
evablogger [386]3 years ago
5 0

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em><em>:</em><em>)</em><em>✌</em>

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This is hard, but I really need help!
sukhopar [10]
Please see pic, I'd solved in it.

4 0
4 years ago
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
For the following geometric sequence find the recursive formula and the 5th term in the sequence. In your final answer, include
bulgar [2K]
We have a geometric sequence----------- > <span>{-4, 12, -36, ...}
</span><span>
the formula is a(r)^(n-1)

a------------- >a is the first term------------ > -4
r--------------- > </span><span>is the common ratio------- > 12/(-4)=(-36/12)=-3
n--------------- > is the number of terms

</span>The fifth term is -4[(-3)^(5-1)]=-4[(81]=-324

the answer is -324
3 0
3 years ago
Read 2 more answers
Question 6
hichkok12 [17]
Türkiye cumhuriyeti ccc
3 0
4 years ago
Solve the absolute value equation: |x|=8
vampirchik [111]

Answer:

x = 8, -8

When taking an absolute value the sign of a number is ignored. Therefore, both 8 and -8 would come out equalling 8.

3 0
3 years ago
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