Given the statement "r varies jointly as s and t". if r =16 when s = 4 and t =6, what is the constant of the variation?
1 answer:
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Two ways to solve the problem:
A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)
B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1 [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation
is given by
= 
- h() / h'()....(2)
Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000
Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal
Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748
So the solutions are S={-3.99989997696619,0.6692468328777476} both accurate to 14 places after the decimal
A. By graphing f(x)=10^x and g(x)=x+4.
The intersection(s) will be the solution.
See graph below, approximate solutions: (-4,0),(0.66, 4.67)
B. Refine approximate solutions using Newton's method
Let h(x)=f(x)-g(x) = 10^x-x-4 ...............(1)
we calculate the derivative, h'(x) = log(10)*10^x-1 [ note:log(x) means ln(x) ]
and use Newton's iterative formula to find successive approximations to the root, basically refining the approximate solutions.
The iterative formula for nth approximation
Using initial approximation (-4,0), we have x0=-4
x1
=x0-h(x0)/h'(x0)
=-4 - h(-4)/h'(-4)
=-4 - (10^(-4)-(-4)-4)/(log(10*10^(-4)-1)
=-4 - (1/10000)/(log(10)/10000-1)
=-3.9999000
Repeating the same for second approximation, x2, we get
x2=-3.99989997696619 which is accurate to 14 places after decimal
Now we can refine the other approximate solution, x0=0.66 to find
x1=0.669356
x2=0.6692468481102326
x3=0.669246832877748
x4=0.6692468328777476
So we will accept x=0.669246832877748
So the solutions are S={-3.99989997696619,0.6692468328777476} both accurate to 14 places after the decimal