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11Alexandr11 [23.1K]

3 years ago

10

So, I have a math problem, the problem is,16x^0+2x^2*y^-1,How do I solve if x = 2 and y = 4?

1 answer:

Mekhanik [1.2K]3 years ago

7 0

Answer:

Given x = 2 and y =4

so 16 $x^{0}$ +2 $x^{2}$ $y^{-1}$

= 16* $2^{0}$ + 2* $2^{2}$ * $4^{-1}$

= 16 + (2*4)/4

=16 + 8/2

16 + 2

=18

Step-by-step explanation:

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I don't know what to do. This is about odd and even functions.

Alexxandr [17]

Answer:

C.

Step-by-step explanation:

We are given $f,g$ are odd which means:

$f(-x)=-f(x)$

$g(-x)=-g(x)$

We can tell if a function, $h$ , is even if $h(-x)=h(x)$ .

We can tell if a function, $h$ , is odd if $h(-x)=-h(x)$ .

So let's test your I,II,III.

We will be replacing x with -x to find out.

I.

$p(x)=f(g(x))$

$p(-x)=f(g(-x))$

$p(-x)=f(-g(x))$

$p(-x)=-f(g(x))$

$p(-x)=-p(x)$

So $p$ is odd.

II

$r(x)=f(x)+g(x)$

$r(-x)=f(-x)+g(-x)$

$r(-x)=-f(x)+-g(x)$

$r(-x)=-(f(x)+g(x))$

$r(-x)=-(r(x))$

$r(-x)=-r(x)$

So $r$ is odd.

III

$s(x)=f(x)\cdot g(x)$

$s(-x)=f(-x) \cdot g(-x)$

$s(-x)=-f(x) \cdot -g(x)$

$s(-x)=f(x) \cdot g(x)$

$s(-x)=s(x)$

So $s$ is even.

So I and II are odd and III is even.

C. is the answer.

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Answer:

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Answer:

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Step-by-step explanation:

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Plz help me thanks so much :( :)

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Step-by-step explanation:

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