12. \( \frac{e}{f^{2}-2 f+1} \), for \( e=-6 \) and \( f=5 \) 13. \( \frac{17-u^{2}}{v^{2}-4 v} \), for \( u=-3 \) and \( v=2 \) 14. \( \frac{-50}{2 x^{2}-3 x+5} \), for \( x=-1 \) 15. \( \frac{y^{3}-4 y+6}{3} \), for \( y=-2 \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Beyond the Answer
To calculate the first expression, we substitute \( e = -6 \) and \( f = 5 \) into the equation. The denominator simplifies as follows: \( f^2 - 2f + 1 = 5^2 - 2(5) + 1 = 25 - 10 + 1 = 16 \). Therefore, \( \frac{-6}{16} = -\frac{3}{8} \). For the second expression, we replace \( u = -3 \) and \( v = 2 \). The numerator becomes \( 17 - (-3)^2 = 17 - 9 = 8 \) and the denominator simplifies as \( v^2 - 4v = 2^2 - 4(2) = 4 - 8 = -4 \). Thus, \( \frac{8}{-4} = -2 \). Next, for the third expression, substituting \( x = -1 \) gives us \( 2(-1)^2 - 3(-1) + 5 = 2(1) + 3 + 5 = 10 \). Therefore, \( \frac{-50}{10} = -5 \). Finally, in the last expression, substituting \( y = -2 \) results in \( (-2)^3 - 4(-2) + 6 = -8 + 8 + 6 = 6 \). Hence, \( \frac{6}{3} = 2 \).