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Aneli [31]
3 years ago
14

How to solve the system of equations algebraically for all values of x, y, and z

Mathematics
1 answer:
stich3 [128]3 years ago
4 0
4x -3y + z = -10...............(1)
2x +y + 3z =  0...............(2)
-x +2y - 5z = 17...............(3)

First we multiply 3*(2) and add it to (1) 
6x +3y +9z =0..................+(1)..................> 10x + 10z = -10......(4)

Then we multiply -2*(2) and add it to (3) 
-4x -2y -6z =0 ...................+(3)................> -5x -11z = 17...........(5)

Multiply  2*(5) and add it to (4)
-10x -22z = 34...................+(4).................> -12 z = 24 ..............>>> z = -2

Substitute z in (4)............> 10x +10(-2) = -10.............................>>> x = 1

Substitute x and z in (2).....> 2(1) +y + 3(-2) =  0..................>>> y = 4

Solution (x,y,z) = (1,4,-2)

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Answer:

The sum of an even number plus an odd number is odd.

Step-by-step explanation:

The sum of an even number and an odd number is odd.

For example:

2+3=5

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The end digit of an even number is always 0, 2, 4, 6, or 8

The end digit of an odd number is 1, 3, 5, 7, or 9.

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Hence, the sum of an even number plus an odd number is odd.

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3 years ago
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and t
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Answer:

Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.  

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Step-by-step explanation:

Use the initial condition when dp/dt = 1, p = 10 to get k;

\frac{dp}{dt} =kp^2\\\\1=k(10)^2\\\\k=\frac{1}{100}

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You have 100 rodents when:

100=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{100} \\\\2t=98\\\\t=49\ months

You have 1000 rodents when:

1000=-\frac{200}{2t-100} \\\\2t-100=-\frac{200}{1000} \\\\2t=99.8\\\\t=49.9\ months

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Answer:

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P.S. (5.2 is more than 5.03, btw)

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