MATHEMATICS COLLEGE

Disk requests come in to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. A seek takes 6 msec per cylinder. How much seek time is needed for (a) First-come, first served. (b) Closest cylinder next. (c) Elevator algorithm (initially moving upward). In all cases, the arm is initially at cylinder 20.

Answers

Answer 1
Answer:

Answer:

Step-by-step explanation:

The order for FCFS is: 20->10->22->20->2->40->6->38.

Distance is

10+12+2+18+38+34+32 = 146

cylinders, so time is

146* 6 = 876 sec.

The order for elevator is:

20->20->22->38->40->10->6->2.

Distance is

0+2+16+2+30+4+4 = 58

cylinders, so time is

58 * 6 =348 msec.

The order for CCN is:

20->20->22->10->6->2->38->40.

Distance is

0+2+12+4+4+36+2 = 60

cylinders, so time is 60 * 6 =360 msec.

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Answers

Answer:

79.8%

Step-by-step explanation:

I believe this is correct, if not feel free to let me know and I will fix it. I'm sorry in advance if it's incorrect.

What is the total surface area please help

Answers

Answer:

167

Step-by-step explanation:

Write the number 0.0049 in scientific notation.

Answers

Answer:

= 4.9*10^-3

Step-by-step explanation:

Hope it helpzzz

Area of Composite Figures.

Answers

The area is 21

The area of a triangle is base*hight/2
Take the area of each triangle and divide them. Hope this helped!

Answer:

21

Step-by-step explanation:

Pleeease open the image and hellllp me

Answers

1. Rewrite the expression in terms of logarithms:

y=x^x=e^(\ln x^x)=e^(x\ln x)

Then differentiate with the chain rule (I'll use prime notation to save space; that is, the derivative of y is denoted y' )

y'=e^(x\ln x)(x\ln x)'=x^x(x\ln x)'

y'=x^x(x'\ln x+x(\ln x)')

y'=x^x\left(\ln x+\frac xx\right)

y'=x^x(\ln x+1)

2. Chain rule:

y=\ln(\csc(3x))

y'=\frac1{\csc(3x)}(\csc(3x))'

y'=\sin(3x)\left(-\cot^2(3x)(3x)'\right)

y'=-3\sin(3x)\cot^2(3x)

Since \cot x=(\cos x)/(\sin x), we can cancel one factor of sine:

y'=-3(\cos^2(3x))/(\sin(3x))=-3\cos(3x)\cot(3x)

3. Chain rule:

y=e^{e^(\sin x)}

y'=e^{e^(\sin x)}\left(e^(\sin x)\right)'

y'=e^{e^(\sin x)}e^(\sin x)(\sin x)'

y'=e^{e^(\sin x)+\sin x}\cos x

4. If you're like me and don't remember the rule for differentiating logarithms of bases not equal to e, you can use the change-of-base formula first:

\log_2x=(\ln x)/(\ln2)

Then

(\log_2x)'=\left((\ln x)/(\ln 2)\right)'=\frac1{\ln 2}

So we have

y=\cos^2(\log_2x)

y'=2\cos(\log_2x)\left(\cos(\log_2x)\right)'

y'=2\cos(\log_2x)(-\sin(\log_2x))(\log_2x)'

y'=-\frac2{\ln2}\cos(\log_2x)\sin(\log_2x)

and we can use the double angle identity and logarithm properties to condense this result:

y'=-\frac1{\ln2}\sin(2\log_2x)=-\frac1{\ln2}\sin(\log_2x^2)

5. Differentiate both sides:

\left(x^2-y^2+\sin x\,e^y+\ln y\,x\right)'=0'

2x-2yy'+\cos x\,e^y+\sin x\,e^yy'+\frac{xy'}y+\ln y=0

-\left(2y-\sin x\,e^y-\frac xy\right)y'=-\left(2x+\cos x\,e^y+\ln y\right)

y'=(2x+\cos x\,e^y\ln y)/(2y-\sin x\,e^y-\frac xy)

y'=(2xy+\cos x\,ye^y\ln y)/(2y^2-\sin x\,ye^y-x)

6. Same as with (5):

\left(\sin(x^2+\tan y)+e^(x^3\sec y)+2x-y+2\right)'=0'

\cos(x^2+\tan y)(x^2+\tan y)'+e^(x^3\sec y)(x^3\sec y)'+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\cos(x^2+\tan y)(2x+\sec^2y y')+e^(x^3\sec y)(3x^2\sec y+x^3\sec y\tan y\,y')+2-y'=0

\left(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1\right)y'=-\left(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2\right)

y'=-(2x\cos(x^2+\tan y)+3x^2\sec y\,e^(x^3\sec y)+2)/(\cos(x^2+\tan y)\sec^2y+x^3\sec y\tan y\,e^(x^3\sec y)-1)

7. Looks like

y=x^2-e^(2x)

Compute the second derivative:

y'=2x-2e^(2x)

y''=2-4e^(2x)

Set this equal to 0 and solve for x :

2-4e^(2x)=0

4e^(2x)=2

e^(2x)=\frac12

2x=\ln\frac12=-\ln2

x=-\frac{\ln2}2

I need help please with the work?

Answers

Answer with Step-by-step explanation:

m<A=m<J=20

In JOE,

80+20+m<O=180[Angles of a trinagle]

m<O=80=m<M

m<E=m<Y=80
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