It is probably B x has to be more than 7 but can be equal 7 at the same time.
It is 5! Hope I can help! :)
Answer:
5 english students in each team
Step-by-step explanation:
So there are multiple ways to do this, you could do trial and error or you could factor out common numbers between the two.
Through trial and error, the lowest possible number of groups that could be divided between the number of students whilst still being able to maintain a whole # was 16 groups
- 128 math students ÷16 groups = 8 math students/group
- 80 english students ÷ 16 groups = 5 english students/group
The other way to solve this is to factor out common numbers between the two:
Answer:
a) 8.13
b) 4.10
Step-by-step explanation:
Given the rate of reaction R'(t) = 2/t+1 + 1/√t+1
In order to get the total reaction R(t) to the drugs at this times, we need to first integrate the given function to get R(t)
On integrating R'(t)
∫ (2/t+1 + 1/√t+1)dt
In integration, k∫f'(x)/f(x) dx = 1/k ln(fx)+C where k is any constant.
∫ (2/t+1 + 1/√t+1)dt
= ∫ (2/t+1)dt+ ∫ (1/√t+1)dt
= 2∫ 1/t+1 dt +∫1/+(t+1)^1/2 dt
= 2ln(t+1) + 2(t+1)^1/2 + C
= 2ln(t+1) + 2√(t+1) + C
a) For total reactions from t = 1 to t = 12
When t = 1
R(1) = 2ln2 + 2√2
≈ 4.21
When t = 12
R(12) = 2ln13 + 2√13
≈ 12.34
R(12) - R(1) ≈ 12.34-4.21
≈ 8.13
Total reactions to the drugs over the period from t = 1 to t= 12 is approx 8.13.
b) For total reactions from t = 12 to t = 24
When t = 12
R(12) = 2ln13 + 2√13
≈ 12.34
When t = 24
R(24) = 2ln25 + 2√25
≈ 16.44
R(12) - R(1) ≈ 16.44-12.34
≈ 4.10
Total reactions to the drugs over the period from t = 12 to t= 24 is approx 4.10
Answer:
p=2
Step-by-step explanation:
p=2
