3 years ago

For any real number x,x2>x3

1 answer:

8 0

No. That's only true if x is less than ' 1 ', including anything negative. If x is more than +1, then the square is LESS than the cube.

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F(x)=|3x+5|+6<br> g(x)=7<br> find (f+g) (x).

horrorfan [7]

Answer:

$(f+g) (x)=\mid 3x+5\mid+13$

Step-by-step explanation:

<u>Operation of Functions</u>

Given:

$f(x)=\mid 3x+5\mid+6$

$g(x)=7$

The sum of (f+g) (x) is:

$(f+g) (x)=\mid 3x+5\mid+6+7$

We cannot operate with the expression inside the absolute bars, thus:

$\boxed{(f+g) (x)=\mid 3x+5\mid+13}$

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Step-by-step explanation:

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Step-by-step explanation:

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