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RideAnS [48]
3 years ago
12

Given six memory partitions of 300 kb, 600 kb, 350 kb, 200 kb, 750 kb, and 125 kb (in order, how would the first-fit, best-fit,

and worst-fit algorithms place processes of size 115 kb, 500 kb, 358 kb, 200 kb, and 375 kb (in order?
Mathematics
1 answer:
NNADVOKAT [17]3 years ago
6 0

Answer:

Step-by-step explanation:

first fit:

115 -> 300

500-> 600

358 -> 750

200 -> 350

375 -> not able to allocate  

Best fit:

115 -> 125

500 -> 600

358 -> 750

200 -> 200

375 -> not able to allocate

worst fit:

115 -> 750

500 -> 600

358 -> not able to allocate

200 -> 350

375 -> not able to allocate

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From a window 20 feet above the ground, the angle of elevation to the top of a building across
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Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

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1 . What is the distance between the two points. (0,5 1,3), (-0,4, -1,3) ?
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