Answer:
x1 = t, x2 = -t and x3 = 0
Step-by-step explanation:
Given the system of equation
x1 + x2 + x3 = 0 .... 1
x1 + x2 + 9x3 = 0 .... 2
Subtract both equation
x3 - 9x3 = 0
-8x3 = 0
x3 = 0
Substitute x3 = 0 into equation 1
x1 + x2 + 0 = 0
x1+x2 = 0
x1 = -x2
Let t = x1
t = -x2
x2 = -t
Hence x1 = t, x2 = -t and x3 = 0
Answer:
Step-by-step explanation:
the appropriate null and alternative hypotheses's are something
3rd degree polynomial with 4 terms
The given plane, 
, has normal vector 
. Any plane parallel to this one has the same normal vector.
Let 
be any point in the plane we want. The plane contains the point (1, 1, -1), so an arbitrary vector in this plane is
and this is perpendicular to 
.
So the equation of the plane is
or equivalently,
We need Pythagoras theorem here
a^2+b^2 = c^2
a, b = legs of a right-triangle
c = length of hypotenuse
Let S=shorter leg, in cm, then longer leg=S+2 cm
use Pythagoras theorem
S^2+(S+2)^2 = (10 cm)^2
expand (S+2)^2
S^2 + S^2+4S+4 = 100 cm^2 [collect terms and isolate]
2S^2+4S = 100-4 = 96 cm^2
simplify and form standard form of quadratic
S^2+2S-48=0
Solve by factoring
(S+8)(S-6) = 0 means (S+8)=0, S=-8
or (S-6)=0, S=6
Reject nengative root, so
Shorter leg = 6 cm
Longer leg = 6+2 cm = 8 cm
Hypotenuse (given) = 10 cm
