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yan [13]
2 years ago
5

Solve the simultaneous equation

Mathematics
2 answers:
Reika [66]2 years ago
8 0

Answer:

x=\frac{9}{2} =4.5, y=-\frac{5}{2} =-2.5

Step-by-step explanation:

2x-4y=19, multi ply- by 5\\10x-20y=95 , Eq I\\3x+5y=1, multiply - by 4\\12x+20y=4, Eq II\\Add -EQI -and- EqII\\\\10x-20y+12x+20y=95+4\\22x=99\\x=\frac{99}{22}=\frac{9}{2}  =4.5\\3(\frac{9}{2})+5y= 1\\5y=1-\frac{27}{2} =-\frac{25}{2}\\y=-\frac{25}{2}*\frac{1}{5}   =-\frac{5}{2} =-2.5\\

MrMuchimi2 years ago
5 0

Answer:

x=4.5

y=-2.5

Step-by-step explanation:

Multiply 2x-4y=19 by 3

Multiply 3x+5y=1 by 2

Solve simultaneously

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Answer:

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What is each power of i with its multiplicative inverse.
pentagon [3]

Answer:

1) multiplicative inverse of i = -i

2) Multiplicative inverse of i^2 = -1

3) Multiplicative inverse of i^3 = i

4) Multiplicative inverse of i^4 = 1

Step-by-step explanation:

We have to find multiplicative inverse of each of the following.

1) i

The multiplicative inverse is 1/i

if i is in the denominator we find their conjugate

=1/i * i/i\\=i/i^2\\=We\,\, know\,\, that\,\, i^2 = -1\\=i/(-1)\\= -i

So, multiplicative inverse of i = -i

2) i^2

The multiplicative inverse is 1/i^2

We know that i^2 = -1

1/-1 = -1

so, Multiplicative inverse of i^2 = -1

3) i^3

The multiplicative inverse is 1/i^3

We know that i^2 = -1

and i^3 = i.i^2

1/i^3\\=1/i.i^2 \\=1/i(-1)\\=-1/i * i/i\\=-i/i^2\\= -i/-1\\= i

so, Multiplicative inverse of i^3 = i

4) i^4

The multiplicative inverse is 1/i^4

We know that i^2 = -1

and i^4 = i^2.i^2

=1/i^2.i^2\\=1/(-1)(-1)\\=1/1\\=1

so, Multiplicative inverse of i^4 = 1

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