40 pts ! approximate the solution to the equation above using three iterations of successive approximation. use the graph below as a starting point.

Option B. 7/16

Step-by-step explanation:

Approximation of Roots of Equations

Newton-Raphson's method is widely used to calculate approximate values of the roots of equations which cannot be solved with algebraic methods.

Let's suposse we need to solve the equation

To find the approximate value of x, we use the following iterative formula

where f' is the first derivative of f

Each value will determine the next one which should be closer to the solution f(x)=0. We need an initial value to start the iterations

We want to solve this equation:

We construct a function f(x) which can be equated to zero and find its roots. Thus we define

To solve the original equation, it's the same as finding the roots of f(x)=0

The starting point will be obtained from the graph provided in the figure

Lets find the derivative

Evaluating in the point

We find the next value:

Let's perform another iteration

Evaluating in the point

We find the next value

This is a very good approximation since

We need to pick one of the options and find none of them is close enough to 0.409. So we'll just choose the closest

Option B. 7/16=0.4375

29.5977692753

step-by-step explanation:

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18