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horrorfan [7]
4 years ago
13

does the following system of equations have a solution if so find it. 2x+y+z= 4 x-y+3z= -2 -x+y+z= -2

Mathematics
1 answer:
saul85 [17]4 years ago
5 0
I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).

Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system

 - y + 3z = -2
   y  + z  =  -2
-----------------
         4z = -4, so z = -1.

Next, multiply the 3rd equation by 2:  You'll get -2x + 2y + 2z = -2.

Add this result to the first equation.  The 2x terms will cancel, leaving you with the system

2y + 2z = -2
  y  +  z  = 4

This would be a good time to subst. -1 for z.  We then get:

-2y - 2 = -2.  Then y must be 0.  y = 0.

Now subst. -1 for z and 0 for y in any of the original equations.

For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.

Then a tentative solution is     (-3, -1, 0).

It's very important that you ensure that this satisfies all 3 of the originale quations.
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x=73/Tan56=49.2391=49.2

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There are 12 face cards in a standard deck of 52 cards. How many ways can you arrange a standard deck of 52 cards such that the
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Let $M$, $N$, and $P$ be the midpoints of sides $\overline{TU}$, $\overline{US}$, and $\overline{ST}$ of triangle $STU$, respect
AysviL [449]

Answer:

<NPM + <NZM = 146°

Step-by-step explanation:

Given:

In triangle STU: M, N and P are the midpoints of the line TU, US and ST respectively.

Line UZ  is the altitude of the triangle STU

<TSU =71°, <TSU = 36°, <TUS = 73°

From the diagram:

N is the midpoint of line SU and M is the mid point of line UT.

∴ Line NM is parallel to line ST

P is the mid point of  line ST and M is the mid point of line UT

∴Line PM is parallel to line SU

N is the mid point of  line SU and P is the mid point of line ST

∴Line NP is parallel to line UT

Δ SPN = Δ STU = 36°

<SPN + <NPM + <MPT (Sum of angles in a triangle = 180°)

<UST = <MPT = 71°

36° + <NPM + 71° = 180°

<NPM + 107° = 180°

<NPM= 180°-107°

<NPM= 73°

In ΔSNZ, line SN = line NZ

∴ side SN = side NZ

< NSZ = <NZS = 71°

<MTZ = <MZT = 36°

<NZS + <NZM <MZT = 180°  (Sum of angles in a triangle = 180°)

71° + <NZM + 36° = 180°

107° + <NZM = 180°

<NZM = 180° - 107°

<NZM = 73°

<NPM + <NZM =73° + 73°

<NPM + <NZM = 146°

5 0
3 years ago
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