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Expressed powers are those that aref(x) = 5x + 3; g(x) = 6x - 5 Find f/g. (f/g)(x) = Quantity six x minus five divided by five x plus three. domain {x|x ≠ Five over six. } (f/g)(x) = Quantity five x plus three divided by six x minus five. domain {x|x ≠ Three over five } (f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. } (f/g)(x) = Quantity six x minus five divided by five x plus three. domain {x|x ≠ - Three over five. }

Respuesta :

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Answer: (f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. }

Explanation:

1) (f/g)(x) = f(x) / g(x)

2)

  f(x)         5x + 3
-------- = ------------
  g(x)        6x - 5

3) Since the division by 0 is not defined, the domain is restricted to 6x - 5 ≠ 0.

Therefore you must solve 6x - 5 = 0 to exclude the value of x for which the denominator becomes 0:

6 x - 5 = 0 => 6x = 5 => x = 5 / 6.

So, the domain is all x | x ≠ 5/6.

Answer:

(f/g)(x) = Quantity five x plus three divided by six x minus five.domain {x|x ≠ Five over six. }

Step-by-step explanation:

Given functions,

[tex]f(x)=5x+3[/tex]

[tex]g(x) =6x-5[/tex]

[tex](\frac{f}{g})(x)=\frac{f(x)}{g(x)}[/tex]

[tex]=\frac{5x+3}{6x-5}[/tex]                        ( By substitution )

Let [tex]h(x)=\frac{5x+3}{6x-5}[/tex]    

Which is the rational function,

Since, a rational function is defined for all real number except those for which denominator = 0,

[tex]6x-5=0[/tex]

[tex]6x=5[/tex]

[tex]x=\frac{5}{6}[/tex]

Thus, h(x) is defined on all real numbers except 5/6,

Hence, domain of [tex]\frac{f}{g}(x)[/tex] is {x| x ≠ [tex]\frac{5}{6}[/tex] }

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