Write a linear equation that intersects y = x2 at two points. then write a second linear equation that intersects y = x2 at just one point, and a third linear equation that does not intersect y = x2. explain how you found the linear equations.

Step-by-step explanation:

1. Any linear equation that describes a line with non-zero slope through the vertex of the parabola will intersect the parabola at two points (the vertex being one of them). A simple equation for such a line is y=x.

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2. Differentiating the equation, you find that the slope of the curve y = x^2 is 2x, so if we choose a line with a slope of 1, it will go through the point on the curve with x-value equal to 1/2. The y-value at that point is y = (1/2)^2 = 1/4, so the y-intercept of the line must be -1/4.

The line that intersects the curve at one point (1/2, 1/4) is tangent at that point. It has equation y = x -1/4.

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3. Any line with the same slope as the tangent line, but a more negative y-intercept, will not intersect the parabola at all. Such a line is y = x -1/2.

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Truth be told, I found the line y = x -1/2 did not intersect the parabola at all when I thought I was writing the equation for the tangent line. It was an answer to part of your question, just not the part I originally intended.

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -

the answer is:

mean: 6

mad: 2.14