Find the area of the triangle with vertices (0,0,0),(−4,1,−2), and (−4,2,−3).

0.5*sqrt33

Step-by-step explanation:

A(0,0,0) B(-4,1,-2), c(-4,2,-3)

Vector AB is (-4-0,-1-0, -2-0)= (-4,-1,-2)  The modul of AB is sqrt (4squared+

+(-1) squared+ (-2) squared)= sqrt (16+1+4)=sqrt21

Vector AC is (-4,2,-3) The modul of vector AC is equal to sqrt ((-4)squared+ 2squared+(-3)squared)= sqrt(16+4+9)= sqrt29

Vector BC is equal to (-4-(-4), 2-1, -3-(-2))= (0,1,-1)

The modul of BC is sqrt (1^2+(-1)^2)=sqrt2

Find the angle B

Ac^2= BC^2+AB^2-2*BC*AB*cosB

29= 2+21-2*sqrt2*sqrt21*cosB

29= 2+21-2*sqrt42*cosB

cosB= -3/ sqrt42

sinB= sqrt( 1-(-3/sqrt42)^2)=sqrt33/42= sqrt11/14

s=1/2* (sqrt2*sqrt21*sqrt11/14)=1/2*sqrt(42*11/14)= 0.5*sqrt33

u bout ugly as   a mf

step-by-step explanation: lol

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: