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djverab [1.8K]
2 years ago
5

What is y=x^4/x^3 when values of x= {-2, 0, 3, 5, 7}

Mathematics
1 answer:
VMariaS [17]2 years ago
3 0

For this case we have a function of the formy = f (x) where:

f (x) = \frac {x ^ 4} {x ^ 3}

We simplify eliminating common terms of the numerator and denominator:

f (x) = x

We must find the values of the function when:

x = {-2, 0, 3, 5, 7}

So, we have:

For x = -2:

f (-2) = - 2

For x = 0

f (0) = 0

For x = 3

f (3) = 3

For x = 5

f (5) = 5

For x = 7:

f (7) = 7

Thus, the values of the function are:

y = {-2, 0, 3, 5, 7}

Answer:

y = {-2, 0, 3, 5, 7}

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Step-by-step explanation:

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Solve: 2cos(x)-square root 3=0 for 0 less than x less than 2 pi
Leona [35]

Answer:

The general solution of   cos x = cos(\frac{\pi }{6})   is  

                                                x = 2nπ±\frac{\pi }{6}

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Step-by-step explanation:

Explanation:-

Given equation is  

                              2cosx-\sqrt{3} =0  for 0

                              2cosx =\sqrt{3}

Dividing '2' on both sides, we get

                             cos x =\frac{\sqrt{3} }{2}

                             cos x = cos(\frac{\pi }{6})

<em>General solution of cos θ = cos ∝ is θ = 2nπ±∝</em>

<em>Now The general solution of   </em>cos x = cos(\frac{\pi }{6})<em>   is  </em>

<em>                                                 x = 2nπ±</em>\frac{\pi }{6}<em></em>

put n=0

x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

Put n=1  

x = 2\pi +\frac{\pi }{6} = \frac{13\pi }{6}

x = 2\pi -\frac{\pi }{6} = \frac{11\pi }{6}

put n=2

x = 4\pi +\frac{\pi }{6} = \frac{25\pi }{6}

x = 4\pi -\frac{\pi }{6} = \frac{23\pi }{6}

And so on

But given 0 < x< 2π

The general solution values  

                                 x = - \frac{\pi }{6}  and x = \frac{\pi }{6}

                               

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