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
The answer to this question is answer choice C.
For the SSS postulate, we need 3 pairs of congruent sides. However, in the image, there are only 2 pairs.
The missing pair in the image is AC and DF. If those are congruent, the triangle can be proven congruent by the SSS postulate.
Hope this helps! :)
~AgentCozmo4, Junior Moderator
Answer:
10ibs=1000% 6ibs=600%
Step-by-step explanation:
Answer:
36 and 54
Step-by-step explanation:
54-36=18
54+36= 90
Answer:
Step-by-step explanation:
The graph of the two functions are shown on the same diagram in the attachment above;
The solution to the equation 
is where the two graphs intersected.
These points are
