Answer:
The answer is that the temperature needs to fall 14 degrees to return to the original temperature of 68 degrees BUT ONLY IF the temperature at the time of the calculation is in fact 82 degrees. The question does not state that. The question merely states that "Later that day the temperature was 82 degrees." Beware of trick questions.
Step-by-step explanation:
Answer:
y is 8/5 when x = -4.
Step-by-step explanation:
If the value of "Y" varies directly with "X", we can write y = kx, where k is the constant of proportionality.
If "y=-8,when x=20," then we can find k: -8 = 20k, or k = -2/5.
Then y = (-2/5)x.
If x = -4, then y = (-2/5)(-4) = 8/5
y is 8/5 when x = -4.
Answer:
If you have a quantity X of a substance, with a decay constant r, then the equation that tells you the amount of substance that you have, at a time t, is:
C(t) = X*e^(-r*t)
Now, we know that:
We have 2000g of substance A, and it has a decay constant of 0.03 (i assume that is in 1/year because the question asks in years)
And we have 3000 grams of substance B, with a decay constant of 0.05.
Then the equations for both of them will be:
Ca = 2000g*e^(-0.03*t)
Cb = 3000g*e^(-0.05*t)
Where t is in years.
We want to find the value of t such that Ca = Cb.
So we need to solve:
2000g*e^(-0.03*t) = 3000g*e^(-0.05*t)
e^(-0.03*t) = (3/2)e^(-0.05*t)
e^(-0.03*t)/e^(-0.05*t) = 3/2
e^(t*(0.05 - 0.03)) = 3/2
e^(t*0.02) = 3/2
Now we can apply Ln(x) to both sides, and get:
Ln(e^(t*0.02)) = Ln(3/2)
t*0.02 = Ln(3/2)
t = Ln(3/2)/0.02 = 20.3
Then after 20.3 years, both substances will have the same mass.
Step-by-step explanation:
x² - 14x + 58
= (x - 7)² + 9.
Hence the vertex is (7, 9).
