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kkurt [141]
3 years ago
10

Use row operations to solve the system.

Mathematics
1 answer:
Aleks04 [339]3 years ago
3 0

Answer:

x = 2, y = 7, z = 12

Step-by-step explanation:

x + y - z = -3 (1)

3x - y + z = 11 (2)

x - 4y + z = -14 (3)

add (1) and (2)

4x = 8 x = 2

add (1) and (3)

2x - 3y = -17

2(2) - 3y = -17

-3y = -17 - 4

-3y = -21

y = 7

Find z using any equation

x + y - z = -3

2 + 7 - z = -3

9 - z = -3

z = 12

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No habla engles

Step-by-step explanation:

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Help me!!!<br> Is for today!!!<br> And what is its congruency?
solniwko [45]

Answer:

Side-Side-Side (SSS) Congruence Property

Step-by-step explanation:

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7 0
3 years ago
1011+111 in binary form
Ann [662]

Answer:

10010

Step-by-step explanation:

1011=1(2)^3+0(2)^2+1(2)^1+1(2)^0

111=1(2)^2+1(2)^1+1(2)^0

So 1011+111 gives us:

1(2)^3+0(2)^2+1(2)^1+1(2)^0

+

1(2)^2+1(2)^1+1(2)^0

-----------------------------------------------------

Combine like terms:

1(2)^3+(0+1)(2)^2+(1+1)(2)^1+(1+1)(2)^0

1(2)^3+1(2)^2+(2)(2)^1+(2)(2)^0

We aren't allowed to have a coefficient bigger than 1.

I'm going to replace 2^0 with 1 and 2 with (2)^1:

1(2)^3+1(2)^2+(2)^2+(2)^1(1)

I want a 2^0 number:

1(2)^3+1(2)^2+1(2)^2+1(2)^1+0(2)^0

Combine like terms:

1(2)^3+2(2)^2+1(2)^1+0(2)^0

2(2)^2=2^3:

1(2)^3+2^3+1(2)^1+0(2)^0

Combine like terms:

2(2)^3+1(2)^1+0(2)^0

We can rewrite the first term by law of exponents:

2^4+1(2)^1+0(2)^0

1(2)^4+1(2)^1+0(2)^0

So the binary form is:

10010

Maybe you like this way more:

Keep in mind 1+1=10 and that 1+1+1=11:

Setup:

      1     0     1      1

+            1      1      1

------------------------------

     (1)    (1)    (1)

      1     0     1      1

+            1      1      1

------------------------------

     1 0    0     1       0

I had to do some carry over with my 1+1=10 and 1+1+1=11.

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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