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
I don’t know how to solve this but I think u have to use the fractions with the other numbers and multiply them then divide
Remark
The key step is just to subtract 5 from both sides. The pointed of the inequality still points away from the variable and towards the number. As long as that remains true, the correct answer can be found.
Solution
2.7 ≤ b + 5 Subtract 5 from both sides.
2.7 - 5 ≤ b
- 2.3 ≤ b Write with the variable on the left.
b ≥ - 2.3 <<<< answer
Answer:
1/9th.
Step-by-step explanation:
Because the area is measured in units squared, then you have to square 1/3 which is 19th.
