Answer:
The temperature of the part won’t reach or exceed 141℉
Step-by-step explanation:
Substituting the value of T= 141 the given equation becomes
[tex]\begin{array}{l}{141=-0.005 x^{2}+0.45 x+125} \\ {-0.005 x^{2}+0.45 x+125-141=0} \\ {-0.005 x^{2}+0.45 x-16=0}\end{array}[/tex]
we know that discriminant D = [tex]b^{2}-4 a c[/tex]
Three conditions: - D < 0: imaginary roots, D = 0: real and equal roots,
D > 0: two real and unequal roots Substituting a = -0.005, b = 0.045 and c = -16, [tex]\mathrm{D}=0.045^{2}-4(-0.005)(-16) \mathrm{D}=-0.317975 \mathrm{D}=-0.32(\text { approx. })[/tex] Therefore, it won't reach the temperature.
