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astra-53 [7]
3 years ago
14

If the system of inequalities y>= 2x + 1 and y > 1/2x -1 is graphed in the xy-plane above, which quadrant contains no solu

tions to the system?

Mathematics
1 answer:
kvv77 [185]3 years ago
7 0

Answer:

View graph

Step-by-step explanation:

We have as a solution of the inequalities the shaded region in quadrant I, that is, there is no solution in quadrants II, III and IV

For the equation y ≥ 2x + 1 the region generated by the values ​​on the line and to the left is taken, and for the inequality y> 1 / 2x-1 the region generated by the values ​​on the curve and to the right is taken

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Determine the standard form of the equation of the line that passes through (6, -9) and (-8, 6)
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9514 1404 393

Answer:

  A.  15x +14y = -36

Step-by-step explanation:

Since we are given two points, we can start with the 2-point form of the equation for a line.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  y = (6 -(-9))/(-8 -6)(x -6) +(-9)

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Multiplying by -14, we have ...

  -14y = 15x -90 +126

Adding 14y-36 to both sides gives ...

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The standard-form equation is ...

  15x +14y = -36

_____

<em>Additional comments</em>

It can be easier to start with the form ...

  (Δy)x -(Δx)y = (Δy)x1 -(Δx)y1 . . . . . where Δx = x2-x1 and Δy = y2-y1

This gives ...

  (6+9)x -(-8-6)y = 15(6) +14(-9)

  15x +14y = -36 . . . simplified

__

You can also start with the slope-intercept form or the point-slope form, if you're more familiar with those. The result will be the same. I find it handy to be familiar with a number of different forms of the equation for a line.

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