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Maru [420]
4 years ago
15

Sam has 1000 Reputation. He Gets 500 More Reputation. How Much Reputation Does He Have And How? PLEASE HELP I WILL PICK BRAINLIE

ST TO FIRST ANSWER
Mathematics
2 answers:
kow [346]4 years ago
8 0

Answer:1500

Step-by-step explanation:

1000+500=1500

Evgesh-ka [11]4 years ago
8 0
1500
Have a great day today
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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}

3 0
1 year ago
Which graph represents a function?
Firdavs [7]

Answer:

c or d

Step-by-step explanation:

7 0
3 years ago
Find a quadratic function with the given x-values as the only two REAL zeros. x=-2 and x=8. Please show work! Will mark Brainlie
posledela

 f(x) = (x- zero) * (x -zero)

f(x) = (x--2) (x-8)

f(x) = (x+2) (x-8)

if you need it multiplied out

f(x) = x^2 +2x -8x-16

f(x) = x^2 -6x-16

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3 years ago
Tom makes 145.80 in 3 days. How much does he make in 2 days
ArbitrLikvidat [17]
Multiply 145.80 by 2/3, get $97.20 in two days.
6 0
3 years ago
Look at the figure below:
Dmitrij [34]
The answer has to be 10
3 0
3 years ago
Read 2 more answers
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