Respuesta :
[tex]\frac{x^2-16}{x^2+5x+6} / \frac{x^2+5x+4}{x^2-2x-8}[/tex]
We can begin by rearranging this into multiplication:
[tex]\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}[/tex]
Now we can factor the numerators and denominators:
[tex]\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}[/tex]
The factors (x+4) and (x+2) cancel out, leaving us with:
[tex]\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}[/tex]
Our answer comes out to be:
[tex]\frac{(x-4)^{2} }{(x+3)(x+1)} [/tex] or [tex]\frac{ x^{2} -8x+16}{ x^{2}+4x+3 }[/tex]
Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4
We can begin by rearranging this into multiplication:
[tex]\frac{x^2-16}{x^2+5x+6} * \frac{x^2-2x-8}{x^2+5x+4}[/tex]
Now we can factor the numerators and denominators:
[tex]\frac{(x+4)(x-4)}{(x+3)(x+2)} * \frac{(x-4)(x+2)}{(x+4)(x+1)}[/tex]
The factors (x+4) and (x+2) cancel out, leaving us with:
[tex]\frac{(x-4)}{(x+3)} * \frac{(x-4)}{(x+1)}[/tex]
Our answer comes out to be:
[tex]\frac{(x-4)^{2} }{(x+3)(x+1)} [/tex] or [tex]\frac{ x^{2} -8x+16}{ x^{2}+4x+3 }[/tex]
Based on the numerator of the second fraction (since we used its inverse), the denominators of both, and the factors we canceled out earlier, the restrictions are x ≠ -4, -3, -2, -1, 4
Answer:
Full...Solving Rational Equations Quiz part 1.
1.c. n^2-6/n^2-2 ; n = +/- sqrt5, n= +/- sqrt2
2.B. 4a/7b^2 , a = 0, b = 0
3.C. (x-4)^2/(x+3)(x+1) ; x= -4,-3,-2,-1,4
4.B. (x+1)(x-1)(x^2+1)
5.A. 7a-49/(a-8)(a+8)
6.A. 21a-28/(A-6)(a+8)
7.C. 4x/3x^2+10x+3
8.C. 3x^2(y+4)/7y
9.D. -11/3
10.D. 14
11. D. 9 mi/h downstream, 6 mi/h upstream
Step-by-step explanation:
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