Area= base x altura dividido entre 2
Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
Art [367]
Answer:
Step-by-step explanation:
A'={2,5,7}
B'={1,2,4,7}
A'UB'={1,2,4,5,7}
Slope is rise/run
gains 4m height for every 3secons
rise=4m
run=3sec
slope=4/3
if takes 54seconds t
54 times 4/3=72 meters
max height is 72 meteres assuming that it starts from the ground level
so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.
in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.
now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B7%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B%5Cfrac%7B19%7D%7B2%7D%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D-7%20%5Cright%29%5E2%2B%5Cleft%28%20%5Cfrac%7B7%7D%7B2%7D-4%20%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B5%7D%7B2%7D%5Cright%29%5E2%2B%5Cleft%28%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Cboxed%7BAM%5Capprox%202.549509756796392%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
20kg of $0.89 candy
10kg of $1.10 candy
Step-by-step explanation:
Candy 1 = 0.89 per kg
Candy 2 = 1.10 per kg
Total kilogram, kg = 30
Let candy 1 = x ; candy 2 = (30 - x) ;
0.89x + 1.10(30 - x) = 0.96(30)
0.89x + 33 - 1.10x = 28.8
0.89x - 1.10x = 28.8 - 33
-0.21x = - 4.2
x = 4.2 / 0.21
x = 20
20kg of $0.89 candy
(30 - x) = (30 - 20) = 10kg
10kg of $1.10 candy
