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Oksanka [162]
3 years ago
8

Help with 8 and 9??? Please help me out! Correct answers please

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
4 0

Answer:

  8.  (-∞, 0) ∪ (0, 1/3) ∪ (1/3, ∞)

  9.  f(g(x)) = 4x/(1+6x)

Step-by-step explanation:

8. For f(g(x)) to be defined, g(x) must be defined and f(g(x)) must be defined.

g(x) will be defined for all x≠0. f(g(x)) will be defined for g(x)≠4. Solving g(x)=4, we find the value of x is 1/3. (The answer choices give a clue.)

So, the domain of f(g(x)) is all x that is not 0 or 1/3. Only one answer choice makes those particular exclusions:

  (-∞, 0) ∪ (0, 1/3) ∪ (1/3, ∞)

__

9. Substituting the definition of g(x) into the expression for f(x), we get ...

\displaystyle (f\circ g)(x)=f(g(x))=\frac{2}{g(x)+3}=\frac{2}{\frac{1}{2x}+3}\\\\=\frac{2}{\left(\frac{1+6x}{2x}\right)}\\\\(f\circ g)(x)=\frac{4x}{1+6x}

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The amount of weeks is your independent variable, or X variable, and the height is you dependents variable.

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What is the solution to the proportion 3y-8/12=y/5? can someone help me do this?
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\displaystyle\\
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Expressed as a product of its prime factors in index form, a number N is
Elena-2011 [213]

<u>ANSWER</u>



5N^2=3^{2} \times 5^{5} \times x^{6}



<u>EXPLANATION</u>


N=3\times5^2 \times x^3.


5N^2=5(3\times5^2 \times x^3)^2


Recall this property of exponents;


(a^m)^2=a^{m} \times a^m



So our product becomes;


5N^2=5(3\times5^2 \times x^3) \times (3\times5^2 \times x^3)



5N^2=5\times 3\times 3 \times 5^2 \times 5^2 \times x^3 \times x^3


5N^2=3\times 3\times 5 \times 5^2 \times 5^2 \times x^3 \times x^3



Recall this law of exponents:


a^m \times a^n =a ^{m+n}


5N^2=3^{1+1} \times 5^{1+2+2} \times x^{3+3}


5N^2=3^{2} \times 5^{5} \times x^{6}




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