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Llana [10]
3 years ago
7

Order the values from least to greatest.

Mathematics
2 answers:
77julia77 [94]3 years ago
7 0

Answer:4/4, 3/4, 1/4 2/4

Step-by-step explanation:

SOVA2 [1]3 years ago
4 0

Answer:

0.03, (3 x 1/100) + (3 x 1/1000), 3 tenths

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8 0
2 years ago
Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!
astra-53 [7]
(a) By DeMoivre's theorem, we have

(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta

On the LHS, expanding yields

\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta

Matching up real and imaginary parts, we have for (i) and (ii),


\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta
\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta

(b) By the definition of the tangent function,

\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}
=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}

=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}
=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}


(c) Setting \theta=\dfrac\pi5, we have t=\tan\dfrac\pi5 and \tan5\left(\dfrac\pi5\right)=\tan\pi=0. So

0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}

At the given value of t, the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.


0=t^5-10t^3+5t\implies0=t^4-10t^2+5

Remember, this is saying that

0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5

If we replace \tan^2\dfrac\pi5 with a variable x, then the above means \tan^2\dfrac\pi5 is a root to the quadratic equation,

x^2-10x+5=0

Also, if \theta=\dfrac{2\pi}5, then t=\tan\dfrac{2\pi}5 and \tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0. So by a similar argument as above, we deduce that \tan^2\dfrac{2\pi}5 is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5

But \tanx>0 for all 0, as is the case for x=\dfrac\pi5 and x=\dfrac{2\pi}5, so we choose the positive root.
3 0
3 years ago
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