we have

Solve for c--------> that means that clear variable c
so
Divide by
both sides

Adds
both sides

Multiply by
both sides
![a[(R/5)+0.3]=c](https://tex.z-dn.net/?f=a%5B%28R%2F5%29%2B0.3%5D%3Dc)
so
![c=a[(R/5)+0.3]](https://tex.z-dn.net/?f=c%3Da%5B%28R%2F5%29%2B0.3%5D)
therefore
<u>the answer is</u>
![c=a[(R/5)+0.3]](https://tex.z-dn.net/?f=c%3Da%5B%28R%2F5%29%2B0.3%5D)
B = 7
7 x 6 = 42
42 + 6 = 48.
Answer:
7.25
Step-by-step explanation:
2 7/8 + 3 2/8 * 1 1/8 = 7.25
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.
Answer:
A 5.5% = 5400 B 5% = 1800
Step-by-step explanation:
A + B = 7200
A * 0.055 + B * 0.05 = $387
A = 7200 - B
(7200 - B)*0.055 + B*0.05 = 387
396 - 0.055B + 0.05B = 387
396 - 0.005B = 387
0.005B = 9
B = 1800
A = 7200 - B = 7200 - 1800
A = 5400