Estimate the quotient: 1628 ÷ 8
2 answers:
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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.
Answer:
Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.
Step-by-step explanation:
Given that, the volume of cylindrical can with out top is 25 cm³.
Consider the height of the can be h and radius be r.
The volume of the can is V=
According to the problem,
The surface area of the base of the can is =
The metal for the bottom will cost $2.00 per cm²
The metal cost for the base is =$(2.00× )
The lateral surface area of the can is =
The metal for the side will cost $1.25 per cm²
The metal cost for the base is =$(1.25× )
Total cost of metal is C= 2.00 +
Putting
Differentiating with respect to r
Again differentiating with respect to r
To find the minimize cost, we set C'=0
⇒r=1.71
Now,
When r=1.71 cm, the metal cost will be minimum.
Therefore,
⇒h=2.72 cm
Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.