For this case we have the following difference equation:
Applying separable variables we have:
Integrating both sides we have:
applying exponential to both sides:
For y (1) = 1 we have:
Thus, the particular solution is:
Whose domain is all real.
Answer: y = 0.2 * exp ((3/2) x ^ 2) Domain: all real numbers
Answer:
Step-by-step explanation:
A) 3x−12≥15
3x ≥ 15 + 12
3x ≥ 27
or x≥ 9
B) 2x−5≤9
2x ≤ 9 +5
2x ≤ 14
or x ≤ 7
C) 5x+8≤53
5x ≤ 53 -8
5x ≤ 45
or x ≤ 9
D) 3x−5≥16
3x ≥ 16 + 5
3x ≥ 21
or x ≥ 7
Have a good day!
Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
No its not because its a repeating decimal for a number to be a irrational number it has to be non terminating or non repeating
It would be e, because it went down 10 instead of 5
