Answer:
12 % reduction
Step-by-step explanation:
11800 - 10384 = 1416
1416 / 11800 = 0.12
0.12 × 100 = 12
Answer:
Step-by-step explanation:
Given the angle ∠AOB
It is stated that CO is the angle bisector of ∠AOB.
Given that ∠AOB = 30°
As we know that the angle bisector bisects the angle into two equal angles.
Thus, the angle bisector CO bisects the angle ∠AOB into two equal angles, which are:
as
∠AOB = 30°
Thus, the two formed angles i.e m∠AOC and m∠BOC by the angle bisector would be half of the angle bisector as the angle bisector bisects the angle ∠AOB into two equal angles.
Therefore,
9514 1404 393
Answer:
dy/dx = y/(2x)
Step-by-step explanation:
The product formula can be used, along with the power rule.
d(uv) = du·v +u·dv
__
d(y^2/x) = d(18)
2y·dy/x -y^2/x^2·dx = 0
2x·dy -y·dx = 0 . . . . . . . . multiply by x^2/y
dy/dx = y/(2x) . . . . . . . . add y·dx, divide by 2x·dx
Answer:
2nd one
Step-by-step explanation:
The question is incomplete, here is the complete question:
The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.
When will there be less than 1 g remaining?
<u>Answer:</u> The time required for a radioactive substance to remain less than 1 gram is 168.27 days.
<u>Step-by-step explanation:</u>
All radioactive decay processes follow first order reaction.
To calculate the rate constant by given half life of the reaction, we use the equation:
where,
= half life period of the reaction = 46 days
k = rate constant = ?
Putting values in above equation, we get:
The formula used to calculate the time period for a first order reaction follows:
where,
k = rate constant =
t = time period = ? days
a = initial concentration of the reactant = 12.6 g
a - x = concentration of reactant left after time 't' = 1 g
Putting values in above equation, we get:
Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.