Find the value of “K” given that one solution of the equation 5x?- kx + 40 = 0 is
1 answer:
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Expanded, the function is ...
... p(x) = x² -x +1
Compared to
... ax² +bx +1
- The value of a is 1
- The vaule of b is -1
- The value of c is 1
- The value of the discriminant is b²-4ac = 1²-4·1·1 = -3
- The quadratic function will intersect the x-axis 0 times.
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The negative discriminant means there are no real roots, hence no x-intercepts.
Answer:
<em>not</em> a rectangle
Step-by-step explanation:
There are several ways to determine whether the quadrilateral is a rectangle. Computing slope is one of the more time-consuming. We can already learn that the figure is not a rectangle by seeing if the midpoint of AC is the same as that of BD. (It is not.) A+C = (-5+4, 5+2) = (-1, 7). B+D = (1-2, 8-2) = (-1, 6). (A+C)/2 ≠ (B+D)/2, so the midpoints of the diagonals are different points.
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The slope of AB is ∆y/∆x, where the ∆y is the change in y-coordinates, and ∆x is the change in x-coordinates.
... AB slope = (8-5)/(1-(-5)) = 3/6 = 1/2
The slope of AD is computed in similar fashion.
... AD slope = (-2-5)/(-2-(-5)) = -7/3
The product of these slopes is (1/2)(-7/3) = -7/6 ≠ -1. Since the product is not -1, the segments AB and AD are not perpendicular to each other. Adjacent sides of a rectangle are perpendicular, so this figure is not a rectangle.
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Our preliminary work with the diagonals showed us the figure was not a parallelogram (hence not a rectangle). For our slope calculation, we "magically" chose two sides that were not perpendicular. In fact, this choice was by "trial and error". Side BC <em>is perpendicular</em> to AB, so we needed to choose a different side to find one that wasn't. A graph of the points is informative, but we didn't start with that.
