Find the area of the parallelogram with vertices A(−1,2,3), B(0,4,6), C(1,1,2), and D(2,3,5).
cupoosta [38]
Answer:
5*sqrt3
Step-by-step explanation:
The vector AB= (0-(-1), 4-2,6-3) AB= (1,2,3)
The modul of AB is sqrt(1^2+2^2+3^2)= sqrt14
The vector AC is (1-(-1), 1-2, 2-3)= (2,-1,-1)
The modul of B is sqrt (2^2+(-1)^2+(-1)^2)= sqrt6
AB*AC= modul AB*modul AC*cosA
cosA=( 1*2+2*(-1)+3*(-1))/ sqrt14*sqrt6= -3/sqrt84=
sinB= sqrt (1- (-3/sqrt84)^2)= sqrt75/84= sqrt 25/28= 5/sqrt28
s= modul AB*modul AC*sinA= sqrt14*sqrt6* 5/ sqrt28= 5*sqrt3
Step-by-step explanation:
Dunno. wish there were more points than this though. bye
if his wasn't algebra, I would've answered answered. hope you have your answer by now though.
Answer:
or 
Step-by-step explanation:
we have
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
Answer:
no solution
Step-by-step explanation:
x2-x1/y2-y1
8-5/-4+4= 3/0
3/0 = N/0 solution
