The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.
Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1
Answer:
2x²y³
<em>good luck, i hope this helps :)</em>
Answer:
Step-by-step explanation:
4). a). If the diagonals of a parallelogram are congruent, then it must be a RECTANGLE.
b). If the diagonals of a parallelogram are perpendicular, then it must be a SQUARE.
c). If the diagonals of a parallelogram bisect the angles of the parallelogram, then it must be a RHOMBUS.
d). If the diagonals of a parallelogram are perpendicular and congruent, then it must be a SQUARE.
e). If a parallelogram has four congruent sides, then it must be a SQUARE.
5). Given quadrilateral SELF is a rhombus.
a). All sides of a rhombus are equal,
Therefore, ES = EL = 25
b). Diagonals of a rhombus bisects the opposite angles,
Therefore, m∠ELS = m∠FLS
3x - 2 = 2x + 7
3x - 2x = 7 + 2
x = 9
c). Diagonals of the rhombus bisect the opposite angles, and adjacent angles are supplementary.
m∠ELF = 2(m∠ELS) = 2(2y - 9)
m∠LES = 2(m∠LEF) = 2(3y + 9)
And 2(2y - 9) + 2(3y + 9) = 180
(2y - 9) + (3y + 9) = 90
5y = 90
y = 18
Answer:
not sure, sorry : p
Step-by-step explanation:
