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
Answer:
2.1 , 2.103 , 2.13 , 2.1312
Step-by-step explanation:
Look at the first place, and see if it's the same, if it is, then you move to the next place. If the place is empty, fill it with a 0 and go from there.
Answer:
-33/10z - 11
Step-by-step explanation:
since all of these terms are being added, there are no need for the parenthesis
-7/2z +4 + 1/5z -15
add like terms (to add the fractions, convert them to a common denominator)
-7/2z + 4 + 1/5z - 15
-35/10z + 2/10z - 11
-33/10z - 11