Math 17 First Long Exam Solutions V7, W3, Y3

I.

1. FALSE

2. TRUE

3. FALSE

4. TRUE

5. FALSE

II.

1. (2i2009 i2011)1,000,001

= (2i (i))1,000,001 = (i)1,000,001= (1)1,000,001(i)1,000,001=

1(i)1,000,001

=1i ii

= i

2. We are given the following: A and B are subsets of theuniversal set U ; n(U) = 25, n(A) = 22, n(B) = 21, andn(A B) = 23.

n(A B) = n(A) + n(B) n(A B) n(AB) = n(A)+n(B)n(AB) = 22+2123 = 20 n((A B)) = n(U) n(A B) = 25 20 = 5

(((A B))) = 2n((AB)) = 25 = 32

III. (2x2 48)4

= (2x4)4 4(2x4)3( 48) + 6(2x4)2( 48)2 4(2x4)( 48)3 + ( 48)4= 16x16 4(8x12)( 48) + 6(4x8)(22) 4(2x4)(4 42) + 8= 16x16 32x12 48 + 48x82 32x4 42 + 8

IV. Quotient: 14x2 + 3x 1 Remainder: 9x+ 9

V.

1. 27x7y3 + 93x5y5 75x3y7

= 3x3y3(9x4 31x2y2 + 25y4)= 3x3y3[(9x4 30x2y2 + 25y4) x2y2]= 3x3y3[(3x2 5y2)2 x2y2]= 3x3y3[(3x2 5y2 xy)(3x2 5y2 + xy)]= 3x3y3(3x2 xy 5y2)(3x2 + xy 5y2)

2. (p6 1) + (3p5 3p2) + (3p4 3p)= (p3 + 1)(p3 1) + 3p2(p3 1) + 3p(p3 1)= (p3 1)[(p3 + 1) + 3p2 + 3p]= (p3 1)(p3 + 3p2 + 3p+ 1)= (p 1)(p2 + p+ 1)(p+ 1)3

VI.

1.

a+a

a 1ab

a3b a+ a2ba3b2 1

a

=

a+a

a2b 1ab

a3b a+ a2ba4b2 1

a

=a+

a2b

a2b 1a(a3b a+ a2b)

a4b2 1=

a3b a+ a2ba2b 1

a(a3b a+ a2b)(a2b 1)(a2b+ 1)

=a2b+ 1

a

2.

( 6m9m2 1

2m 3m 3m2

) 27m

3 114m2 15m 9

18m3 6m2 2m16m 24

=

( 6m(3m 1)(3m+ 1)

2m 3m(3m 1)

) (3m 1)(9m

2 + 6m+ 1)

(7m+ 3)(2m 3) 8(2m 3)

2m(9m2 + 6m+ 1)=6m2 + 6m2 7m 3m(3m 1)(3m+ 1)

3m 1(7m+ 3)

4m

=(7m+ 3)m(3m+ 1)

(4)m(7m+ 3)

=4

m2(7m+ 3)

3.

624

7 +13

713

33 + 39

=(3 23) 16

72 (13)2

33 39 =

316 2 12

36

33 39

39 + 3 + 3 3

3

332 + 3

3 9 + 392 =

316 2 12 36 143 9 (

39 + 3 + 3 3

3)

=316 2 12 2 12 3 12

6 (39 + 3 + 3 3

3) =

2 396 (

39 + 3 + 3 3

3) =

3 33 + 3 3

9 + 9

3= 33 39 3

1

4.

50(

4249

2

)(3i 2)(2i+ 1) (7 6i)

=

5i2

(42 7i

2

)(2 3i)(2i+ 1) (7 6i) =

20i+ 35

(4i 2 6i2 3i) (7 6i) =5(7 + 4i)

3 + i 3 i3 i

=5(21 + 7i 12i+ 4i2)

10=25 5i)

2=252

+5i2

/ajdporlante 08.09.09/

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