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Novay_Z [31]
2 years ago
14

Point S is on line segment RT

Mathematics
1 answer:
storchak [24]2 years ago
6 0

The numerical length of ST is 8

<h3>How to determine the numerical length of ST?</h3>

The given parameters are:

ST = 5x - 7

RT = 4x + 9

RS = 2x + 7

To calculate the length ST, we use the following equation

RT = RS + ST

Substitute the known values in the above equation

4x + 9 = 5x - 7 + 2x + 7

This gives

4x + 9 = 7x

Evaluate the  like terms

x = 3

Substitute x = 3 in ST = 5x - 7

ST = 5 * 3 - 7

This gives

ST = 8

Hence, the numerical length of ST is 8

Read more about line segments at:

brainly.com/question/2437195

#SPJ1

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If A(−1, 1), B(−4, 5), C(−9, 5), and D(−6, 1) are the vertices of a quadrilateral, do the points form a rhombus? Justify your an
Flura [38]

Answer:

Yes, it is a rhombus because the slope of diagonal segment BD equals 2, the slope of diagonal segment AC equals -1/2, and the diagonals bisect each other.

Step-by-step explanation:

The slope of diagonal BD is ...

  slope BD = (change in y)/(change in x) = (-4/-2) = 2

The slope of diagonal AC is ...

  slope AC = (change in y)/(change in x) = (4/-8) = -1/2

The midpoint of BD is ...

  ((-4, 5) +(-6, 1))/2 = (-10, 6)/2 = (-5, 3)

The midpoint of AC is ...

  ((-1, 1) +(-9, 5))/2 = (-10, 6)/2 = (-5, 3)

so the diagonals bisect each other.

While the third statement is true (both midpoints are the same), that fact alone is not sufficient to allow the figure to be declared a rhombus. The diagonals must also be perpendicular (have slopes whose product is -1). The appropriate answer choice is the <em>second one</em>:

  • Yes, it is a rhombus because the slope of diagonal segment BD equals 2, the slope of diagonal segment AC equals -1/2, and the diagonals bisect each other.
3 0
3 years ago
Use synthetic division to determine whether the number k is an upper or lower bound (as specified) for the real zeros of the fun
anastassius [24]

Answer:

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Step-by-step explanation:

Lower Bound Theorem:

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Given;

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synthetic division;

-1 ]    4       -2        2        4

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---------------------------------------------

        4         -6       8          -4  

(the result has alternating signs, hence k is lower bound)

Thus, k is lower bound for the real zeros of the function.

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