Question 1054893
<pre><b>
Given:  n(x) = x<sup>2</sup>-9 
        m(x) = x+3 
        p(x) = &#8730;<span style="text-decoration: overline">x+1</span>
 
Use the functions to find:
 
a. (n-m)(-6)

First we find (n-m)(x)

(n-m)(x) = 

(n-m)(x) = n(x) - m(x) = 

(n-m)(x) = (x<sup>2</sup>-9) - (x+3) = 

(n-m)(x) = x<sup>2</sup>-9-x-3 = 

(n-m)(x) = x<sup>2</sup>-x-12 

Now substitute (-6) for x

(n-m)(-6) = (-6)<sup>2</sup>-(-6)-12

(n-m)(-6) = 36+6-12

(n-m)(-6) = 30

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b.  (n•p)(0) 

First we find (n•p)(x)

(n•p)(x) = n(x)•p(x)

(n•p)(x) = (x<sup>2</sup>-9)•(&#8730;<span style="text-decoration: overline">x+1</span>)

Now substitute 0 for x

(n•p)(0) = (0<sup>2</sup>-9)•&#8730;<span style="text-decoration: overline">0+1</span>

(n•p)(0) = (0-9)•&#8730;<span style="text-decoration: overline">1</span>

(n•p)(0) = (-9)•1

(n•p)(0) = -9

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c. {{{(p/m)}}}(0) 

First we find {{{(p/m)}}}(x) 

{{{(p/m)}}}(x) = {{{"p(x)"/"m(x)"}}}

{{{(p/m)}}}(x) = {{{sqrt(x+1)/(x+3)}}}

Now substitute 0 for x

{{{(p/m)}}}(0) = {{{sqrt(0+1)/(0+3)}}}

{{{(p/m)}}}(0) = {{{sqrt(1)/3}}}

{{{(p/m)}}}(0) = {{{1/3}}}

Edwin</pre></b>