J/4 - 8 < 2
J/4 < 2 + 8
J/4 < 10 ...multiply both sides by 4
J < 10 * 4
J < 40
Using the Pythagorean Theorem:
a^2 + b^2 = c^2
we have:
8^2 + 6^2 = C^2
64 + 36 = C^2
100 = c^2
c = SQRT(100)
C = 10
The hypotenuse is 10. The answer is B.
Answer:
P = 6200 / (1 + 5.2e^(0.0013t))
increases the fastest
Step-by-step explanation:
dP/dt = 0.0013 P (1 − P/6200)
Separate the variables.
dP / [P (1 − P/6200)] = 0.0013 dt
Multiply the left side by 6200 / 6200.
6200 dP / [P (6200 − P)] = 0.0013 dt
Factor P from the denominator.
6200 dP / [P² (6200/P − 1)] = 0.0013 dt
(6200/P²) dP / (6200/P − 1) = 0.0013 dt
Integrate.
ln│6200/P − 1│= 0.0013t + C
Solve for P.
6200/P − 1 = Ce^(0.0013t)
6200/P = 1 + Ce^(0.0013t)
P = 6200 / (1 + Ce^(0.0013t))
At t = 0, P = 1000.
1000 = 6200 / (1 + C)
1 + C = 6.2
C = 5.2
P = 6200 / (1 + 5.2e^(0.0013t))
You need to change the exponent from negative to positive.
The inflection points are where the population increases the fastest.
Answer:
5
Step-by-step explanation:
it can be found in two ways
f(n)-f(n-1) is common difference of the sequence.
knowing f(n) = f(n-1) + 5
f(n)-f(n-1)=5
so 5 is difference of the sequence.
other method is
f(1)=4
f(2)=?
f(n)=f(n-1)+5 n=2
f(2)=f(2-1)+5
f(2)=f(1)+5
f(2)=4+5=9
f(2)-f(1)=9-4=5
so 5 is difference of the sequence.
Answer:
That is the third one :
456.222
because it has the most numbers 4,5,6 and point . 2,2,2