Question

PLEASE HELP. WILL FAN AND MEDAL.
During a manufacturing process, a metal part in a machine is exposed to varying temperature conditions. The manufacturer of the machine recommends that the temperature of the machine part remain below 141°F. The temperature T in degrees Fahrenheit x minutes after the machine is put into operation is modeled by T = – 0.005x2 + 0.45x + 125. Will the temperature of the part ever reach or exceed 141°F? Use the discriminant of a quadratic equation to decide.
A. yes
B. no

Answer 1

@1DEA could you please help me if you have time? if not, its fine. thanks :)

Answer 2

sure I can help but it's already been an hr, are you still there?

Answer 3

yes please

Answer 4

T = – 0.005x^2 + 0.45x + 125
T should be less than 141 degrees. Therefore,
– 0.005x^2 + 0.45x + 125 < 141
– 0.005x^2 + 0.45x + 125 - 141 < 0
– 0.005x^2 + 0.45x - 16 < 0
Let us see what x will male the left hand side zero. That is,
– 0.005x^2 + 0.45x - 16 = 0
Find the discriminant. D = b^2 - 4ac
D = (0.45)^2 - 4 * (-0.005) * (-16) = 0.2025 - 0.32 = -0.1175
Since the discriminant is negative, the roots for this quadratic are complex.
So there is no real value of x that will make – 0.005x^2 + 0.45x - 16 equal zero.
This is a parabola that opens downward due to negative leading coefficient of -0.005 and it never crosses the x-axis. Therefore, there is NO value of x for which T = – 0.005x^2 + 0.45x + 125 will reach or exceed 141 degrees.

Answer 5

looks like aum beat me to the punch lol.

Answer 6

Thank you