No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.
Consider polynomial equations x^2 = 3 and x^2 = 1.
Obviously they have real solutions.
Subtract the two polynomial equations:
(x^2 - x^2) = (3 - 1)
0 = 2...
We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.
Obviously 0 = 2 has no real solution.
Ok so to find the ratio just do 12/16 = 3/4. that's your common ratio.
The formula for sum of infinite :
a / (1 - r)
a is the first term and r is the ratio
which is
16 / (1 - (3/4))
which is
16 / (1/4)
16 * 4
which is 64
Explanation:
a. The line joining the midpoints of the parallel bases is perpendicular to both of them. It is the line of symmetry for the trapezoid. This means the angles and sides on one side of that line of symmetry are congruent to the corresponding angles and sides on the other side of the line. The diagonals are the same length.
__
b. We observe that adjacent pairs of points have the same x-coordinate, so are on vertical lines, which have undefined slope. KN is a segment of the line x=1; LM is a segment of the line x=3. If the trapezoid is isosceles, the midpoints of these segments will be on a horizontal line. The midpoint of KN is at y=(3-2)/2 = 1/2. The midpoint of LM is at y=(1+0)/2 = 1/2. These points are on the same horizontal line, so the trapezoid <em>is isosceles</em>.
__
c. We observed in part (b) that the parallel sides are KN and LM. The coordinate difference between K and L is (1, 3) -(3, 1) = (-2, 2). That is, segment KL is the hypotenuse of an isosceles right triangle with side lengths 2, so the lengths of KL and MN are both 2√2.
_____
For part (c), we used the shortcut that the hypotenuse of an isosceles right triangle is √2 times the leg length.
Answer:
Step-by-step explanation: the first time you get 0.75 and the second time you get 0.125 hope this helped