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
10≤n<100
So the number of two digit numbers is 99-10+1=90
The only combinations that sum to 5 are 14, 23, 32, 41, 50 so
P(sum=5)=5/90
P(sum=5)=1/18
Answer: -1/2
Step-by-step explanation: to find slope, use the equation (y2-y1) / (x2-x1)
(-6 - -3) / (7 - 1)
-3 / 6
= -1/2
Answer: 22 hours 11 minutes
<u>Step-by-step explanation:</u>

![22\ hours + \bigg[\dfrac{19}{100}=\dfrac{x}{60}\bigg]\\\\22\ hours + 11\ minutes](https://tex.z-dn.net/?f=22%5C%20hours%20%2B%20%5Cbigg%5B%5Cdfrac%7B19%7D%7B100%7D%3D%5Cdfrac%7Bx%7D%7B60%7D%5Cbigg%5D%5C%5C%5C%5C22%5C%20hours%20%2B%2011%5C%20minutes)