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Gnesinka [82]
3 years ago
9

Can someone help me plz

Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0
Use hypotenuse leg theorem.

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If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num
weeeeeb [17]

Answer:

<h2>x = 0, y = 5, z = 3</h2>

Step-by-step explanation:

x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right

\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right

\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}

7 0
3 years ago
Johanna used tiles to build a rectangler array with an array with an areavof 54.list all the possible dimensions of the array
lyudmila [28]
Assuming that all tiles must stay whole it would be:
1×54
2×27
3×18
6×9
4 0
3 years ago
Can someone help me please
NikAS [45]
Justin rides 550 kilometers
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justin rides 17 meter
6 0
4 years ago
Write an equation of the line that passes through a pair of points:
vredina [299]

Answer:

b)

The  equation of the line that passes through a pair of points

   y  = \frac{-1}{8}  ( x ) -\frac{11}{8}

Step-by-step explanation:

 Explanation:-

Given points are ( 5 ,-2) and ( 3,-1)

slope of the line

                        m = \frac{y_{2} -y_{1} }{x_{2}-x_{1}  } = \frac{-1-(-2)}{-3-(5)} = \frac{1}{-8}

The  equation of the line that passes through a pair of points

              y - y_{1} = m ( x - x_{1} )

           y - (-2) = \frac{-1}{8}  ( x - (5) )

          y  = \frac{-1}{8}  ( x ) +\frac{5}{8} -2

        y  = \frac{-1}{8}  ( x ) +\frac{5-16}{8}

      y  = \frac{-1}{8}  ( x ) -\frac{11}{8}

3 0
3 years ago
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