The mistake made by George is; D:George should have averaged the two differences instead of the two bounds.
<h3>How to Solve Successive Approximations?</h3>
In Mathematics, successive approximation can be defined as a classical method that is used in Calculus for solving integral equations or initial value problems.
In this question, George started the first iteration of successive approximation by using the lower and upper bounds of the graph. However, we can deduce that George made a mistake instep 5 because he should have used x = 3/2 as the new upper bound.
Read more about Successive Approximations at; brainly.com/question/25219621
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Answer:
![y=-\dfrac{1}{3}x-1](https://tex.z-dn.net/?f=y%3D-%5Cdfrac%7B1%7D%7B3%7Dx-1)
Step-by-step explanation:
<u>Given: </u>
line y=3x-1
point (-3,0)
<u>Write:</u> equation of the line that is perpendicular to the given and passes through the point (-3,0)
<u>Solution:</u>
The slope of the given line is ![m=3](https://tex.z-dn.net/?f=m%3D3)
If
is the slope of perpendicular line, then
![m\cdot m_1=-1\\ \\3m_1=-1\\ \\m_1=-\dfrac{1}{3}](https://tex.z-dn.net/?f=m%5Ccdot%20m_1%3D-1%5C%5C%20%5C%5C3m_1%3D-1%5C%5C%20%5C%5Cm_1%3D-%5Cdfrac%7B1%7D%7B3%7D)
So, the equation of the needed line is ![y=-\dfrac{1}{3}x+b.](https://tex.z-dn.net/?f=y%3D-%5Cdfrac%7B1%7D%7B3%7Dx%2Bb.)
Find b. This line passes through the point (-3,0), so its coordinates satisfy the equation:
![0=-\dfrac{1}{3}\cdot (-3)+b\\ \\0=1+b\\ \\b=-1\\ \\y=-\dfrac{1}{3}x-1](https://tex.z-dn.net/?f=0%3D-%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%28-3%29%2Bb%5C%5C%20%5C%5C0%3D1%2Bb%5C%5C%20%5C%5Cb%3D-1%5C%5C%20%5C%5Cy%3D-%5Cdfrac%7B1%7D%7B3%7Dx-1)
i would say h or j, but I’m more towards j :)))
For the law of sines, you would apply it in this particular problem like so:
Since P is 27, its angle is 33 and Q's length is 40; you would set it up like this
<u />40/SinQ = 27/Sin33, multiply 40 with Sin33, then it would be 40Sin33, then divide it by 27. The result should be 40Sin33/27 = X
The formula for the Pythagorean therom is: a[tex] a^{2} + b^{2} = c^{2}
So, If a^{2} and b^{2} add up to c^{2} than it is a Pythagorean triple.