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:
35
Step-by-step explanation:
A quadratic sequence is a form. of mathematical progression of numbers in which the subsequent differences between each consecutive term differ by the exact amount. This is known as the common second difference.
Hence in this case,
we have
S= 3+8+15+24+…..+Tn - 1.
S=. 3+8+15+……………..Tn-1 + Tn
Deducting 2nd series from the first
We have 0=3+5+7+…….Tn
Hence, we have Tn= n (n+2)
= 24 (9+2)
= 24 + 11
= 35
Answer:
<h2>The function f(x) = (x - 6)(x - 6) has only one x-intercept. But at (6, 0) not at (-6, 0).</h2>
Step-by-step explanation:
The intercept form of a quadratic equation (parabola):

p, q - x-intercepts
Therefore
The function f(x) = x(x - 6) = (x - 0)(x - 6) has two x-intercepts at (0, 0) and (6, 0)
The function f(x) = (x - 6)(x - 6) has only one x-intercept at (6, 0)
The function f(x) = (x + 6)(x - 6) = (x - (-6))(x - 6)
has two x-intercept at (-6, 0) and (6, 0)
The function f(x) = (x + 1)(x + 6) = (x - (-1))(x - (-6))
has two x-intercepts at (-1, 0) and (-6, 0).
Answer:
See below
Step-by-step explanation:
A 30° angle is measured 30° counterclockwise while a -30° angle is measured 30° clockwise. Therefore, you can say -30°=360-30°=330° because 30° is a reference angle located in Quadrant IV.
not completely sure but i think b