x5=4+8(5-1)+5(5-1)(5-2)/2=4+8(4)+5(4)(3)/2=

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Guide :

# x5=4+8(5-1)+5(5-1)(5-2)/2=4+8(4)+5(4)(3)/2=

x5=4+8(5-1)+5(5-1)(5-2)/2=4+8(4)+5(4)(3)/2=

## Research, Knowledge and Information :

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### 2.5 Multiplying Fractions - Innovative Learning Solutions ...

2.5 Multiplying Fractions 2.5 OBJECTIVES 1. Multiply two fractions 2. ... 3the same nonzero number. 5 8 2 3 1 4 5 5 8 2 3 9 5 5 8 23 11 13 1 114 Write any mixed or ...

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### Amazon.com : Avery Mini Durable Binder for 5.5 x 8.5 Inch ...

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... and read about the BMW X5 4.4i at Car and Driver. New Cars. Acura; ... 29.6 sec Street start, 5-60 mph: 7.3 sec Standing 1/4-mile: 15.4 sec @ 92 mph Top speed ...

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Multiplying Fractions and Mixed Numbers e.g. 4 x 3 6/7; Multiplying Mixed Numbers e.g. 4 2/3 x 5 3/8; Similar to the above listing, ...

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## Suggested Questions And Answer :

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### what are the possible combination of making 30 with 1, 3, 5, 7, 9

I started off by figuring how many ways to use nines: 30   3 nines, 27 total   2 nines, 18 total   1 nine, 9 total   0 nines, 0 total I then figured how many ways to use sevens: 30   3 nines, 27 total     0 sevens, 27 total &nb

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### what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices

what is the answer to 2x-y-2z=1, 4x+y-z=5, x+y+4z=-5 using matrices how to solve this math problem using matrices?   2  -1  -2  │  1   4   1  -1  │  5   1   1   4  │ -5 Multiply line 2 by 2. &nbs

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### how many ways can you have coins that total 18p using 1p , 2p, 5p, and 10p coins

in how many ways can you have coins that total exactly 18pence using 1p, 2p, 5p and 10p coins but you may wish to use as many of each sort as you wish. Start with the biggest coin, then the next biggest, etc. This is so that you always end up adding on just single coins of 1p. 10   &

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### how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0

how do you solve this using a gausian reduction using the three equations? 4x+z=3,2x-y=2,and 3y+2z=0 I think gaussian reduction is similar to back substitution, but with all the equation shaving different variables... just not sure how to do it. You create a matrix using the constants in the equati

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### is there a solution to (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format (Gauss method): ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1:

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### Solve by Gauss-Jordan elimination (a) x1 + 2x2 + 3x3 = 1 2x1 + 3x2 + 4x3 = 3 x1 + 2x2 + x3 = 3

Matrix format: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 1 2 1 | 3 ) R3-R1: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 -2 | 2 ) R3/-2: ( 1 2 3 | 1 ) ( 2 3 4 | 3 ) ( 0 0 1 | -1 ) R2-R1: ( 1 2 3 | 1 ) ( 1 1 1 | 2 ) ( 0 0 1 | -1 ) R1⇔R2: ( 1 1 1 | 2 ) ( 1 2 3 | 1 ) ( 0 0 1 | -1 ) R2-R1: ( 1 1 1 | 2 )

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### show that Z15 is isomorphic to Z5 external Z3

Z15 can be represented as a pair of quantities a and b so that a has a range of 0 to 4 (Z5) and b has a range of 0 to 2 (Z3). So we have a scheme of representation: (0,0)=0 (1,1)=11 (2,2)=7 (0,3)=3 (1,4)=14 (2,0)=10 (0,1)=6 (1,2)=2 (2,3)=13 (0,4)=9 (1,0)=5 (2,1)=1

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### solve these equations using matrices without a calculator. x=y=z=6, 2x=y-4z=-15, 5x-3y+z=-10

x-y-z=6 2x-y-4z=-15 5x-3y+z=-10 Your equations, as written, had typos in them. I've assumed that some of the equal-signs should have been minus-signs and altered then appropriately. In matrix form the equations would be AX = b Where A is the matrix 1 -1 -1 2 -1 -4 5 -3 1

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### x+y+z=9,2x-3y+4z=13,2x-3y+4z=13,3x+4y+5z=40. solve

There are 4 equations but only 3 unknowns. This means one equation is unnecessary or there is inconsistency. Call the equations in order A, B, C and D. B and C are the same so we can remove one. Let's remove C. 3A+B: 3x+3y+3z+2x-3y+4z=27+13; 5x+7z=40. So 5x=40-7z. We have x in terms of z.

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### how to solve gauss jordan reduction using three equations?

Augmented matrix format: ( 2 -3 1 | 2 ) ( 1 1 1 | 8 ) ( 3 -1 -1 | 0 ) R1⇔R2: ( 1 1 1 | 8 ) ( 2 -3 1 | 2 ) ( 3 -1 -1 | 0 ) R2-2R1 then R2*-1: ( 1 1 1 | 8 ) ( 0 5 1 | 14 ) ( 3 -1 -1 | 0 ) R3-3R1 then R3*(-1/4): ( 1 1 1 | 8 ) ( 0 5 1 | 14 ) ( 0 1 1 | 6 ) R2-R3 then R2/4: ( 1 1 1 | 8

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