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IceJOKER [234]
4 years ago
12

The graph below shows the change in temperature in Troy, NY.

Mathematics
1 answer:
KengaRu [80]4 years ago
7 0

Answer:

Average 3am - 9am: \frac{5}{6}

Average 9pm - 3am: \frac{1}{2}

Step-by-step explanation:

Average Rate of Change: \frac{f(b)-f(a)}{b-a}

Step 1: Define coordinates

3am - 9am; (3, -3) and (9, 2)

9pm - 3am; (9, 0) and (3, -3)

Step 2: Find rate of change

\frac{2-(-3)}{9-3}=\frac{2+3}{6} =\frac{5}{6}

\frac{-3-0}{3-9}=\frac{-3}{-6} =\frac{1}{2}

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Z^4-5(1+2i)z^2+24-10i=0
mixer [17]

Using the quadratic formula, we solve for z^2.

z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2

Taking square roots on both sides, we end up with

z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}

Compute the square roots of -171 + 140i.

|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221

\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)

By de Moivre's theorem,

\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}

as well as the fact that

0

\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}

(whose signs are positive because of the domain of \frac t2).

This leaves us with

z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}

Compute the square roots of 5 + 12i.

|5 + 12i| = \sqrt{5^2 + 12^2} = 13

\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)

By de Moivre,

\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i

and its negative, -3 - 2i. We use similar reasoning as before:

t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}

1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4

\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}

\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}

Lastly, compute the roots of -2i.

|-2i| = 2

\arg(-2i) = -\dfrac\pi2

\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i

as well as -1 + i.

So our simplified solutions to the quartic are

\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}

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