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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
We have to find the perimeter of the triangle KLM.
We have been given that the length of the side LM=12, , and
Refer the attached image.
In a triangle sum of three angles should be .
So,
Plugging the values of angle K and angle M, we get:
So,
Now, that we have the measure of angle L, we will apply sine rule to find the length of the sides KL and KM.
Using the sine law for the triangle KLM, we get:
Refer the image. Plugging the value of the sides of the triangle KLM and the angles of the triangle KLM, we get:
Now using,
We get the value of 'y'
So the length of the side KM is 13.38 units.
Now using,
We get the value of 'x'
So the length of the side KL is 9.79 units.
Now, to find the perimeter of the triangle KLM we need to sum up the length of the sides of the triangle KLM.
The perimeter of the triangle KLM units
