Answer:
(n-5)^2 - (6n-35)=(n-10)(n-6)
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- n²-10n+25-6n+35 =
- n²-16+60 = n²- 10n - 6n + 60 =
- n(n-10) - 6(n-10) =
- (n-10)(n-6)
so what I do is caunt or just add 7 +8 it will equal 15
Answer: a - 4.512 hours
b - 1.94 hours
Step-by-step explanation:
Given,
a) A(t) = 10 (0.7)^t
To determine when 2mg is left in the body
We would have,
A(t) = 2, therefore
2 = 10(0.7)^t
0.7^t =2÷10
0.7^t = 0.2
Take the log of both sides,
Log (0.7)^t = log 0.2
t log 0.7 = log 0.2
t = log 0.2/ 0.7
t = 4.512 hours
Thus it will take 4.512 hours for 2mg to be left in the body.
b) Half life
Let A(t) = 1/2 A(0)
Thus,
1/2 A(0) = A(0)0.7^t
Divide both sides by A(0)
1/2 = 0.7^t
0.7^t = 0.5
Take log of both sides
Log 0.7^t = log 0.5
t log 0.7 = log 0.5
t = log 0.5/log 0.7
t = 1.94 hours
Therefore, the half life of the drug is 1.94 hours
85/5
(50+35)/5
50/5 + 35/5
10+7=17
The answer is 3 because Pythagorean triplets consist of 3,4 and 5.
