A store manager receives a delivery of 2 boxes of lightbulbs. Each box contained 2 broken light bulbs.
1 answer:
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Answer:
3:2
Step-by-step explanation:
Given that <span>Line WX is congruent to Line XY and Line XZ bisects Angle WXY.
We prove that triangle WXZ is congruent to triangle YXZ as follows:
![\begin{tabular} {|c|c|} Statement&Reason\\[1ex] \overline{WX}\cong\overline{XY},\ \overline{XZ}\ bisects\ \angle WXY&Given\\ \angle WXY\cong\angle YXZ & Deifinition of an angle bisector\\ \overline{XZ}\cong\overline{ZX}&Refrexive Property of \cong\\ \triangle WXZ\cong\triangle YXZ&SAS \end{tabular}](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%0A%7B%7Cc%7Cc%7C%7D%0AStatement%26Reason%5C%5C%5B1ex%5D%0A%5Coverline%7BWX%7D%5Ccong%5Coverline%7BXY%7D%2C%5C%20%5Coverline%7BXZ%7D%5C%20bisects%5C%20%5Cangle%20WXY%26Given%5C%5C%0A%5Cangle%20WXY%5Ccong%5Cangle%20YXZ%20%26%20Deifinition%20of%20an%20angle%20bisector%5C%5C%0A%5Coverline%7BXZ%7D%5Ccong%5Coverline%7BZX%7D%26Refrexive%20Property%20of%20%5Ccong%5C%5C%0A%5Ctriangle%20WXZ%5Ccong%5Ctriangle%20YXZ%26SAS%0A%5Cend%7Btabular%7D)
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We prove that triangle WXZ is congruent to triangle YXZ as follows:
Answer:
(a) -3/4
(b) -0.75
(c) -0.75
Step-by-step explanation:
It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...
- evaluating the function at the midpoint of the bracketing interval
- choosing a smaller bracketing interval
- identifying the x-value known to be closest to the solution
Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.
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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.
For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.
For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.
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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.
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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.
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See the attached table for function values.
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<em>Comment on bisection iteration</em>
Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.
Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.
