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andrey2020 [161]
3 years ago
8

Determine algebraically whether the function g(x)=x^2 is even odd or neither

Mathematics
2 answers:
irina1246 [14]3 years ago
8 0

g(x) = x^2

g(-x) = (-x)^2 ... replace every x with -x

g(-x) = x^2 ... squaring a negative makes it positive

So because g(x) = g(-x) for every x, this means g(x) is an even function. For polynomials, if every exponent is even, then the entire polynomial is even as well.

Note: if a function is even, then it won't be odd or vice versa unless the function is the zero function. In other words, g(x) = 0 is both even and odd.

julsineya [31]3 years ago
5 0

I think its 2 because its 2 so you will put 2

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For the function F(x) = 1/x+2, which of these could be a value of F(x) when xis
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B 2

Step-by-step explanation:

The computation of the value of f(x) in the case when x = 2 is shown below

As per the question, following function is given

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Based on this, the x = 2

Now put the x value in the above equation

So,

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Given that sin theta = 1/4, 0
GaryK [48]
Answer: cos(Θ) = (√15) / 4

Explanation:

The question states:

1) sin(Θ) = 1/4

2) 0 < Θ < π / 2

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This is how you solve it.

1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).

(cos \alpha )^2 + (sin \alpha )^2 =1

2) From which you can find:

(cos \alpha )^2 = 1 - (sin \alpha )^2

3) Replace sin(α) with 1/4

=> (cos \alpha )^2 = 1 - (1/4)^2 = 1 - 1/16 = 15/16

=> cos \alpha =+/- \sqrt{15/16} = +/- (\sqrt{15} )/4

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cos(Θ) = \sqrt{15} /4.

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