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a sample of a radioactive substance has a mass m at time t, where m=ae^-bt where a and b are positive constants.
so ln m= -bt +ln a
I can't do question v
these are my answers so far:
ii) the graph ln m against t passes (0,3) and (5,1). find values a and b.
ln a = e^3 = 20
a= 20
b: 1-3/3-0 = -0.4 so b =0.4
iii) At what time will m have reduced to 0.01?
t= 19
iv) my answer was dy/dx = -0.3pe^-0.3t-0.2qe^0.2t
I can't do question V
v) this is what I've done...
(0,7)(4.25,0)
m= -4
y-17=-4(x-0)
y= -4x+17
when x=0
so dy/dx = -0.3pe^0 - 0.2qe^0
= -0.3p - 0.2q
comparing this gradient to the gradient -4
0.3p+0.2q =4
but I don't know how to work out the other simultaneous equation?
the answers are meant to be p = 6 and q = 11
so ln m= -bt +ln a
I can't do question v
these are my answers so far:
ii) the graph ln m against t passes (0,3) and (5,1). find values a and b.
ln a = e^3 = 20
a= 20
b: 1-3/3-0 = -0.4 so b =0.4
iii) At what time will m have reduced to 0.01?
t= 19
iv) my answer was dy/dx = -0.3pe^-0.3t-0.2qe^0.2t
I can't do question V
v) this is what I've done...
(0,7)(4.25,0)
m= -4
y-17=-4(x-0)
y= -4x+17
when x=0
so dy/dx = -0.3pe^0 - 0.2qe^0
= -0.3p - 0.2q
comparing this gradient to the gradient -4
0.3p+0.2q =4
but I don't know how to work out the other simultaneous equation?
the answers are meant to be p = 6 and q = 11
(edited 11 years ago)
Original post by Jckc123
For the third part, you use back the equation of In m = In a -bt
Since m=0.01, a = e^3, and b = 0.4
Substituting the value:
In 0.01 = 3 - 0.4t.
Just a matter of rearranging the formula:
t= (3- In 0.01) / 0.4
Hope this helps
Since m=0.01, a = e^3, and b = 0.4
Substituting the value:
In 0.01 = 3 - 0.4t.
Just a matter of rearranging the formula:
t= (3- In 0.01) / 0.4
Hope this helps
Thank you so much! This has really helped me
Hi! I've got stuck again.
I worked out iv, my answer was dy/dx = -0.3pe^-0.3t-0.2qe^0.2t
I can't do question V
v) this is what I've done...
(0,7)(4.25,0)
m= -4
y-17=-4(x-0)
y= -4x+17
when x=0
so dy/dx = -0.3pe^0 - 0.2qe^0
= -0.3p - 0.2q
comparing this gradient to the gradient -4
0.3p+0.2q =4
but I don't know how to work out the other simultaneous equation?
the answers are meant to be p = 6 and q = 11
Original post by 5amera
Hi, can someone please help me on part ii) the graph of lm against t passess through (0,3) and (5,1) find a and b.
Many Thanks
Many Thanks
You just have to sub in both coordinates into the equation to get a and b. The equation has to be rearranged to become a linear equation of y = mx+ c first before you sub in. (As shown in the first post)
hope this helps.
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