Let KLMN be a trapezoid (see added picture). From the point M put down the trapezoid height MP, then quadrilateral KLMP is square and KP=MP=10.
A triangle MPN is right and <span>isosceles, because
</span>m∠N=45^{0}, m∠P=90^{0}, so m∠M=180^{0}-45^{0}-90^{0}=45^{0}.Then PN=MP=10.
The ttapezoid side KN consists of two parts KP and PN, each of them is equal to 10, then KN=20 units.
Area of KLMN is egual to
sq. units.
First we will convert those radian angles to degrees, since my mind works better with degrees. Let's work one at a time. First,
. If we start at the positive x-axis and measure out 315 we end up in the 4th quadrant with a reference angle of 45 with the positive x-axis. The side across from the reference angle is -1, the side adjacent to the angle is 1, and the hypotenuse is sqrt2. The cotangent of this angle, then is 1/-1 which is -1. As for the second one, converting radians to degrees gives us that
. Sweeping out that angle has us going around the origin more than once and ending up in the first quadrant with a reference angle of 30° with the positive x-axis. The side across from the angle is 1, the side adjacent to the angle is √3, and the hypotenuse is 2. Therefore, the secant of that angle is 2/√3.
For a parallelogram, the area is calculated by the equation,
A = bh
where A is area, b is base, and h is height. From this equation, we can solve for the base of the banner by dividing the area by the height.
base of the banner = 127.5 cm² / 4.25 cm
base of the banner = 30 cm
Thus, the measure of the base of the banner is equal to 30 cm.
Answer:
C. (0,b)
Step-by-step explanation:
If X is a random variable representing the number of children that end up having light-colored eyes, then X follows a binomial distribution with success probability p = 0.25 and number of trials n = 5.
The variance of a binomial distribution with these parameters is n p (1 - p), so the variance in this case would be
5 • 0.25 • (1 - 0.25) = 0.9375 ≈ 0.9
