Which value of x would make line segment t v is parallel to line segment q s? 3 8 10 11

The value of x that would make the line segment TV to be parallel to the line segment QS is;

Option C; 10

The image of the line segment is missing and so i have attached it.

    From the attached image of the triangle, we can use the theorem of similar triangles to get;

RT/RQ = RV/RS

We see that;

RT = x + 4

RQ = x + 4 + x - 3 = 2x + 1

RV = x + 10

RS = x + 10 + x = 2x + 10

Thus;

(x + 4)/(2x + 1) = (x + 10)/(2x + 10)

Cross multiply to get;

(x + 4)(2x + 10) = (x + 10)(2x + 1)

2x² + 18x + 40 = 2x² + 21x + 10

2x² will cancel out to give;

18x + 40 = 21x + 10

21x - 18x = 40 - 10

3x = 30

x = 30/3

x = 10

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if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

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