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Sergeu [11.5K]
3 years ago
7

What relation does not have an initial value of 50

Mathematics
1 answer:
Mamont248 [21]3 years ago
4 0
I guess itz B because with the rest, a number can be added to get exactly 50. Example C. y= 50x. x could be 1 which is still 50 and D. y=50-x and x could be 0 which is still 50 whiles A. y=50 remains 50

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2x+4y+2z= -8<br> 5y-2z= 0<br> 4z= -20
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Answer:

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Step-by-step explanation:

Here, the given set of equations are:

2 x + 4 y +2 z = -8

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Now, simplifying each question one by one ,we get:

Simplifying third equation for the value of z:

4 z = -20 ⇒ z  = -20 / 4 = - 5

or, z =  -5

Simplifying second equation for the value of y:

Substitute the value of z  = -5 in (2) , we get

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or, 5 y + 10 = 0

or, 5 y = -10

or, y  =  -10/5 = -2

⇒ y = -2

Simplifying first equation for the value of x:

Substitute the value of z  = -5 and y  = -2 in (1) , we get

2 x + 4 y + 2 z = -8  ⇒ 2 x + 4 (-2) + 2 (-5) = -8

or,  2 x - 8 - 10 = -8

or, 2 x = 10

x = 10 / 2  = 5, or x  = 5

Hence, the given solution for the system is x = 5, y = -2 and z = -5

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Answer:

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Step-by-step explanation:

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