DB/dt = k*B
dB/(kB) = dt
Integrating both sides:
1/k*ln|B| = t + C
B = C*e^(kt)
At time t = 0, B = 4000 and at time t = 3 (3 hours past 9 AM), B = 4500
4000 = C
4500 = 4000*e^(3k) => k = 1/3*ln(9/8)
At time t = 15 (Midnight), B = 4000*e^(1/3*ln(9/8)*15) = 4000*(9/8)^5 = 7208 bacteria
Answer:
2) -141 hits/da
Step-by-step explanation:
Calculate the values of h(2) and h(4)
h(2) = 2170(0.92)² = 2170 × 0.8464 = 1837
h(4) = 2170(0.92)⁴ = 2170 × 0.7164 = 1555
Calculate the average rate of change
Rate of change = (y₂ - y₁)/(x₂-x₁) = (1555 -1837)/(4 - 2) = -282/2 = -141 hits/da
The average rate of change is -141 hits/da.
In the figure below, the red curve represents the function h(d), while the green dashed line represents the average rate of change over the interval 2 ≤ d ≤4.
Assuming you know the quadratic formula:

You can fill in a, b and c in this and get:

Now, let's create a shorthand
to make the writing easier:

But since one root is three times the other root (but we don't know which one), we also know that:
or 
If you solve this, you get:
or 
However, since q is a square root, it must be postive.
So we know that:

From this you find that

So p = 3/2
The roots are x=1/2 and x=3/2.
Answer:
ok
Step-by-step explanation:
Answer:
Expanded Notation Form:
3
+ 0.6
+ 0.01
+ 0.001
Expanded Factors Form:
3 × 1
+ 6 × 0.1
+ 1 × 0.01
+ 1 × 0.001
Expanded Exponential Form:
3 × 100
+ 6 × 10-1
+ 1 × 10-2
+ 1 × 10-3
Word Form:
three and six hundred eleven thousandths