-5 · 12 = -5 · 12 = -5 · 12 = -5 · 12 =

Mathematics

2 answers:

Inessa05 [86]4 years ago

6 0

Answer:

-60 if you meant all of them its -180

Step-by-step explanation:

avanturin [10]4 years ago

3 0

You add them all together

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If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's ru

aleksandrvk [35]

Answer:

Step-by-step explanation:

If a sequence c1,c2,c3,...has limit K then the sequence ec1,ec2,ec3,...has limit e^K. Use this fact together with l'Hopital's rule to compute the limit of the sequence given by

bn=(n)^(5.6/n).

$L = \lim_{n \to \infty} b_n \\\\\\L= \lim_{n \to \infty} n^{\frac{5.6}{n} }$

Log on both sides

$In (L) = \lim_{n \to \infty} In (n)^{\frac{5.6}{n} }\\\\= \lim_{n \to \infty} \frac{5.6}{n} In(n)$

$=5.6 \lim_{n \to \infty} \frac{d}{dn} In(n)/\frac{d}{dn} (n)\\\\=5.6 \lim_{n \to \infty} \frac{1}{n} /1 \\\\=5.6 \lim_{n \to \infty} \frac{1}{n} \\\\=5.6 \times 0\\\\In(L) =0\\\\L=e^0\\\\L=1$

$\therefore \lim_{n \to \infty} (n)^{\frac{5.6}{n} =1$

5 0

4 years ago

Solve using demoivres theorem

pashok25 [27]

It's difficult to make out, but I think the task is to expand

$(\sqrt3-i)^4$

Write the number in polar form first:

$\sqrt3-i=2e^{-i\pi/6}$

By DeMoivre's theorem, you have

$(\sqrt3-i)^4=\left(2e^{-i\pi/6}\right)^4=2^4e^{-i4\pi/6}=16e^{-i2\pi/3}$

and converting back to Cartesian form, this number is equivalent to

$16\left(\cos\dfrac{2\pi}3-i\sin\dfrac{2\pi}3\right)=16\left(-\dfrac12+-\dfrac{\sqrt3}2\right)=-8(1+i\sqrt3)$