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Allushta [10]
1 year ago
15

Type the correct answer in each box. If necessary, use / for the fraction bar. Find the solution to this system of equations. It

y = 1 21-y+z = 1 1+2y+z IL y = = Reset Next

Mathematics
1 answer:
sertanlavr [38]1 year ago
5 0

ANSWER

• x = 1/3

,

• y = 2/3

,

• z = 1

EXPLANATION

There are many methods to solve a linear system of equations, but in this case we have to use the substitution method -which consists in clearing one variable as a function of the other/s and replace in another equation. For a system of three variables such as this one, we have to do this twice:

1° clear x from the first equation:

\begin{gathered} x+y=1 \\ x=1-y \end{gathered}

Replace x by this expression in the second equation:

\begin{gathered} 2x-y+z=1 \\ 2(1-y)-y+z=1 \end{gathered}

Note that now we have two variables, y and z. Before the next step we have to rewrite the equation above so that we only have one y:

\begin{gathered} 2\cdot1-2y-y+z=1 \\ 2-3y+z=1 \end{gathered}

2° clear y from the equation above:

\begin{gathered} -3y+z=1-2 \\ -3y=-1-z \\ y=\frac{-(1+z)}{3} \\ y=\frac{1+z}{3} \end{gathered}

And replace y by this expression in the last equation. Note that the third equation also contains x, so we have to replace first x as a function of y like in the first step:

\begin{gathered} x+2y+z=\frac{8}{3} \\ (1-y)+2y+z=\frac{8}{3} \end{gathered}

Rewrite it so we only se one y:

\begin{gathered} 1-y+2y+z=\frac{8}{3} \\ 1+y+z=\frac{8}{3} \end{gathered}

And now we replace y by the expression we found in the second step:

1+\frac{1+z}{3}+z=\frac{8}{3}

So now we have one equation with one variable. Let's solve for z:

\begin{gathered} 1+\frac{1}{3}+\frac{z}{3}+z=\frac{8}{3} \\ \frac{4}{3}+\frac{4z}{3}=\frac{8}{3} \\ \frac{4z}{3}=\frac{8}{3}-\frac{4}{3} \\ \frac{4z}{3}=\frac{4}{3} \\ z=1 \end{gathered}

We have that z = 1.

The next steps are to back replace and find the other variables. Remember that in the second step we had y as a function of z:

y=\frac{1+z}{3}

Replace z = 1 and solve:

y=\frac{1+1}{3}=\frac{2}{3}

y = 2/3

And finally, replace y = 2/3 in the expression of the first step, where we found x as a function of y:

x=1-y=1-\frac{2}{3}=\frac{1}{3}

and we got x = 1/3

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We know that<span>
<span>Figures can be proven similar if one, or more, similarity transformations (reflections, translations, rotations, dilations) can be found that map one figure onto another. 
In this problem to prove circle 1 and circle 2 are similar, a translation and a scale factor (from a dilation) will be found to map one circle onto another.

we have that</span>
<span> Circle 1 is centered at (5,8) and has a radius of 8 centimeters
 Circle 2 is centered at (1,-2) and has a radius of 4 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the center of the circle 2
the transformation has the following rule</span>
(x,y)--------> (x-4,y-10)
so
(5,8)------> (5-4,8-10)-----> (1,-2)
so
center circle 1 is now equal to center circle 2 
<span>The circles are now concentric (they have the same center)
</span>
step 2
<span>A dilation is needed to decrease the size of circle 1 to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle 1-----> 4/8----> 0.5

radius circle 1 will be=8*scale factor-----> 8*0.5-----> 4 cm
radius circle 1 is now equal to radius circle 2 

<span>A translation, followed by a dilation will map one circle onto the other, thus proving that the circles are similar

the answer is
</span>
</span>The circles are similar because you can translate Circle 1 using the transformation rule (x-4,y-10) and then dilate it using a scale factor of (0.5)
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