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4 years ago

Factor x^2-10x+25,please show work

1 answer:

5 0

This is relatively easy because that 25 is a perfect square, whose (square) roots are 5 and -5. x^2-10x+25 = (x - 5)(x - 5). Note how (-5)(-5) = +25, and how -5x - 5x = -10x.

The roots are { (x-5), (x-5) }.

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3 years ago

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3 years ago

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NeX [460]

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3 years ago

What is 6x-36=216<br> What is x

stiks02 [169]

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4 years ago

Read 2 more answers

Write a coordinate proof to prove that the segment that joins the vertex angle of an isosceles triangle to the midpoint of its b

Natalija [7]

Answer:

Slope of the base segment line is zero, hence base segment is horizontal

Slope of the segment that joins the vertex angle to the midpoint of its base is undefined hence the line is vertical

Therefore, angle between base segment line and the segment line from the vertex angle to the midpoint is perpendicular

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base

Step-by-step explanation:

Here we prove the required relation as follows;

Let the isosceles be ABC

The coordinates of the points are

C = (0, 0) (Vertex)

A = (-a, b)

C = (a, b)

P = (0, b) (Midpoint of base)

Therefore, the gradient or slope of the base AC is presented as follows;

$Slope, m \, of \. base, \, AC= \frac{dy}{dx} = \frac{b - b}{a - (-a)} = \frac{0}{2\cdot a} = 0$

Hence, segment AC is vertical

$Slope, m \, of \, segment \ CP \, joining \, vertex, \, to \, midpoint, P, on \,AC = \frac{0 - b}{0 - 0} = \frac{-b}{0} = Undifined.$

Hence, segment CP is vertical

Therefore, the segment that joins the vertex angle to an isosceles triangle to the midpoint of its base is perpendicular to the base.

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4 years ago