2010 Exam 1 CAS Methods [Copy]

[3Xamz]

I'll put my solutions up soon, but if anyone wants it.

http://www.mediafire.com/?vu5911nt3461u1t

[3Xamz]

My Solutions, [Not necessarily right]

1) a) (x^2)(e^2x)(3+2x)
b) 4/5

2) a) (1/2)sin(2x+1)
b) p = 0.5

3) a) x^4
b) ginv(16) = 2

4) a) Amplitude = 4, Period = 2pi/3 EDIT: Period = 6pi
b) x = pi/6 and x = -5pi/6

5)a) 0.5
b) b = -2/3

6) a=4/9, b=8/9, c=58/9

7)a) a =6/125
b) Pr(X<3) = (6/125) Definite integral of (5x-x^2) from x=0 to x=3

p=0.5

9)a) 2xloge(x) + x
b) (9/2)loge(3) -2 sq units

10) a=1/6, d=-5/2, c=1/2

11)a) h = 5 - 2.5r
b) S = 10(pi)r - 3(pi)r^2
c) r = 5/3 cm.

[8039]

My Solutions, [Not necessarily right]

1) a) (x^2)(e^2x)(3+2x)
b) 4/5

2) a) (1/2)sin(2x+1)
b) p = 0.5

3) a) x^4
b) ginv(16) = 2

4) a) Amplitude = 4, Period = 2pi/3
b)x = pi/6 and x = 5pi/6

5)a) 0.5
b) b = -2/3

6) a=4/9, b=8/9, c=58/9

7)a) a =6/125
b) Pr(X<3) = (6/125) Definite integral of (5x-x^2) from x=0 to x=3

p=0.5

9)a) 2xloge(x) + x
b) (9/2)loge(x) -2 sq units

10) a=1/6, d=-5/2, c=1/2

11)a) h = 5 - 2.5r
b) S = 10(pi)r - 3(pi)r^2
c) r = 5/3 cm.

Isn't period pi6? and for 10) c= 13/2?

[98.40_for_sure]

Yo dude, i think you made a typo on question 9b) i think that x is supposed to be a 3

[qshyrn]

My Solutions, [Not necessarily right]

1) a) (x^2)(e^2x)(3+2x)
b) 4/5

2) a) (1/2)sin(2x+1)
b) p = 0.5

3) a) x^4
b) ginv(16) = 2

4) a) Amplitude = 4, Period = 2pi/3
b)x = pi/6 and x = 5pi/6

5)a) 0.5
b) b = -2/3

6) a=4/9, b=8/9, c=58/9

7)a) a =6/125
b) Pr(X<3) = (6/125) Definite integral of (5x-x^2) from x=0 to x=3

p=0.5

9)a) 2xloge(x) + x
b) (9/2)loge(x) -2 sq units

10) a=1/6, d=-5/2, c=1/2

11)a) h = 5 - 2.5r
b) S = 10(pi)r - 3(pi)r^2
c) r = 5/3 cm.

Isn't period pi6? and for 10) c= 13/2?

yeah period is 6pi  but c is 1/2

[3Xamz]

Yo dude, i think you made a typo on question 9b) i think that x is supposed to be a 3

Yup, soz meant 3. Thanks for that.

[8039]

My Solutions, [Not necessarily right]

1) a) (x^2)(e^2x)(3+2x)
b) 4/5

2) a) (1/2)sin(2x+1)
b) p = 0.5

3) a) x^4
b) ginv(16) = 2

4) a) Amplitude = 4, Period = 2pi/3
b)x = pi/6 and x = 5pi/6

5)a) 0.5
b) b = -2/3

6) a=4/9, b=8/9, c=58/9

7)a) a =6/125
b) Pr(X<3) = (6/125) Definite integral of (5x-x^2) from x=0 to x=3

p=0.5

9)a) 2xloge(x) + x
b) (9/2)loge(x) -2 sq units

10) a=1/6, d=-5/2, c=1/2

11)a) h = 5 - 2.5r
b) S = 10(pi)r - 3(pi)r^2
c) r = 5/3 cm.

Isn't period pi6? and for 10) c= 13/2?

yeah period is 6pi  but c is 1/2

Right it is... looks like I fucked up even more than I thought

[zhangq]

hey isn't 4 b) pie/6 and -ve 5pie/6

[3Xamz]

hey isn't 4 b) pie/6 and -ve 5pie/6

Thanks, didn't pick up on that

[ninbam1k]

heyy, i dont get how you got that from question 11. I got h = 5r/2 from using simlar triangles :/ care to explain?

[costa94]

thats the ratio of h to r, so you need to minus big h from small h where big h is 5 so u get 5 - 5/2 r

[slee]

cant open rar files.. =(

[8039]

What was A cutoff last year?

[Souljette_93]

for q's 5 a) are we supposed to write 0.5 or 1/2, seeming as an exact answer is required?

[3Xamz]

cant open rar files.. =(

Here's a .zip version
http://www.mediafire.com/?dud7cw64kfu0bny