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ANEK [815]
3 years ago
9

Which of the following expressions represents a function?

Mathematics
2 answers:
erica [24]3 years ago
6 0

Answer:

y=x²

Step-by-step explanation:

Function is a map from a non-empty set A to a non-empty set B such that corresponding to each point of A there exist a unique image of it in B.

y=x² corresponding to each x there exist a unique image i.e. value of y

Hence, it is a function.

{(9, 2), (4, –2), (5, 3), (9, –4)}

since, for x=9 there exists two images 2 and -4

Hence, it is not a function.

x=-3

it is not a function

x² = 25 − y²

for x=0 y can be 5 or -5

i.e. there exists two images of a function under x=0

Hence, it is not a function.

igor_vitrenko [27]3 years ago
5 0
<span>A function is a relation where there is only one y-value for each x-value. According to this definition, the only given expression that represents a function is y = x^2</span>
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2^3 = 8
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5^2 = 25
2^5 = 32

(5^2 = 5*5 = 25)
(2^5 = 2*2*2*2*2 = 32)
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(First gets brainleist) The elevation of a coral reef is 12 feet below sea level. The elevation of a snorkeler is 2 feet below s
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-12 < -2

Step-by-step explanation:

below sea level means under water which is negative. the snorkler is only 2 feet below sea level which is -2 but the coral reef is 12 feet below sea water which in this case -2 is greater than -12 since its closer to 0

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Elenna [48]

1. By de Moivre's theorem,

\left(4\left(\cos\left(\dfrac{7\pi}9\right) + i \sin\left(\dfrac{7\pi}9\right)\right)\right)^3 = 4^3 \left(\cos\left(\dfrac{21\pi}9\right) + i \sin\left(\dfrac{21\pi}9\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac{7\pi}3\right) + i \sin\left(\dfrac{7\pi}3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\cos\left(\dfrac\pi3\right) + i \sin\left(\dfrac\pi3\right)\right) \\\\ ~~~~~~~~ = 64 \left(\dfrac12 + i\dfrac{\sqrt3}2\right) \\\\ ~~~~~~~~ = \boxed{32 + 32\sqrt3\,i}

2. First write the given number in exponential/trigonometric form.

z = 5-5\sqrt3\,i

has modulus

|z| = \sqrt{5^2 + \left(-5\sqrt3\right)^2} = \sqrt{100} = 10

and since it lies in the second quadrant of the complex plane, its argument is

\arg(z) = \pi + \tan^{-1}\left(-\dfrac{5\sqrt3}5\right) = \pi + \tan^{-1}\left(-\sqrt3\right) = \pi - \dfrac\pi3 = \dfrac{2\pi}3

So, we have

z = 5 - 5\sqrt3\,i = 10 e^{i2\pi/3} = 10 \left(\cos\left(\dfrac{2\pi}3\right) + i \sin\left(\dfrac{2\pi}3\right)\right)

Now we apply de Moivre's theorem again, and make sure to account for the multivalued-ness of the exponential function. For k\in\{0,1,2,3,4\}, the fifth roots of z are

z^{1/5} = 10^{1/5} e^{i(2\pi/3 + 2\pi k)/5}

k=0 \implies z^{1/5} = 10^{1/5} e^{i2\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{2\pi}{15}\right) + i \sin\left(\dfrac{2\pi}{15}\right)\right)}

k=1 \implies z^{1/5} = 10^{1/5} e^{i8\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{8\pi}{15}\right) + i \sin\left(\dfrac{8\pi}{15}\right)\right)}

k=2 \implies z^{1/5} = 10^{1/5} e^{i14\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{14\pi}{15}\right) + i \sin\left(\dfrac{14\pi}{15}\right)\right)}

k=3 \implies z^{1/5} = 10^{1/5} e^{i20\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{4\pi}3\right) + i \sin\left(\dfrac{4\pi}3\right)\right)} k=4 \implies z^{1/5} = 10^{1/5} e^{i26\pi/15} = \boxed{10^{1/5} \left(\cos\left(\dfrac{26\pi}{15}\right) + i \sin\left(\dfrac{26\pi}{15}\right)\right)}

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