Answer:
f(N) = 2+ N/2 + 6N² units of time.
Step-by-step explanation:
Assigning 0 to the variable sum takes one unit of time.
Each time you increment sum by B, you need to call the value of sum, sum it to B and assign it to sum, which takes three units of time in total. You are repeating this process for each value of B which ranges from 1 to 4n and for each value of A which ranges from 1 to n/2. Opening the FOR takes also another unit of time, so, as a result, we have
f(N) 1 + 1 (open the FOR in A)+ N/2*(1 (open the FOR in B) + 4N*3) = 2+ N/2 + 6N² units of time. It has order complexity O(N²).
Combine like terms
Add to both sides
Divide both sides
I only see three steps, man
For this case, the first thing we must take into account is the following definition:
d = v * t
Where,
d: distance
v: speed
t: time
Substituting values we have:
Trip 1:
50 = v * t1
We cleared t1:
t2 = (50) * (1 / v)
Trip 2:
300 = (3 * v) * t2
We cleared t2:
t2 = (300/3) * (1 / v)
t2 = (100) * (1 / v)
Rewriting:
t2 = 2 (50) * (1 / v)
t2 = 2 * t1
Answer:
His new time compared with the old time was:
Twice the old time.
t2 = 2 * t1.
A - B - C =
-3 - 4 + 5 =
-7 + 5 =
-2
negative three minus four plus five is negative two
welcome :/
Answer:
Domain = (-∞, -2) ∪ (-2, ∞).
Range = (-∞, 0) ∪ (0, ∞).
Step-by-step explanation:
t can't have the value -2 because that would make 1/(t + 2) = 1/0 which is undefined.
So the domain is All real values of t. except t = -2.
As t approaches infinity or negative infinity f(t) approaches zero, and as t approaches -2 from below or above f(t) approaches negative infinity or positive infinity.
