Answer:
1) 2/7
2) 8/11
3) 3/2 or 1 1/2
4) 32/35
Step-by-step explanation:
1.5/7•14/35
= 5/7 × 14/35
Using Cancellation method
= 1/1 × 2/7
= 2/7
2.[4/5][10/11]
= 4/5 × 10/11
Using Cancellation method
= 4/1 × 2/11
= 8/11
3.8 3/4 multiplied by 2/9
= 8 3/4 × 2/9
= 27/4 × 2/9
= 3/2 × 1/1
= 3/2
= 1 1/2
4. [4/5]10/11][7/8]
40/50 ÷ 77/88
= 40/50 × 88/77
= 40/50 × 8/7
= 40/25 × 4/7
= 8/5 × 4/7
= 32/35
Step-by-step explanation:
1/2 -1/4
2*4-1*1/4
=1/4.
1/4is answer
I took the liberty of finding for the complete question.
And here I believe that the problem asks for the half life of Curium. Assuming
that the radioactive decay of Curium is of 1st order, therefore the
rate equation is in the form of:
A = Ao e^(-kt)
where,
A = amount after t years = 2755
Ao = initial amount = 3312
k = rate constant
t = number of years passed = 6
Therefore the rate constant is:
2755/3312 = e^(-6k)
-6k = ln (2755/3312)
k = 0.0307/yr
The half life, t’, can be calculated using the formula:
t’ = ln 2 / k
Substituting the value of k:
t’ = ln 2 / 0.0307
t’ = 22.586 years
or
t’ = 22.6 years
Those are going to change to positive values due to the property being used
the order would be
18 , 18 , 19 , 20 , 22 , 30
Answer:
Step-by-step explanation:
