-2x + 2y + 3z = 0 → 2x - 2y - 3z = 0 → 2x - 2y - 3z = 0
-2x - 1y + 1z = -3 → 2x + 1y - 1z = 3 → 2x + 1y - 1z = 3
2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 -3y - 2z = -3
-2x + 2y + 3z = 0 → 2x - 2y - 3z = 0
-2x - 1y + 1z = -3 → 2x + 1y - 1z = 3 → 2x + 1y - 1z = 3
2x + 3y + 3z = 5 → 2x + 3y + 3z = 5 → 2x + 3y + 3z = 5
-2y - 4z = -2
-3y - 2z = -3 → -6y - 4z = -6
-2y - 4z = -2 → -2y - 4z = -2
-4y = -4
-4 -4
y = 1
-3y - 2z = -3
-3(1) - 2z = -3
-3 - 2z = -3
+ 3 + 3
-2z = 0
-2 -2
z = 0
-2x + 2y + 3z = 0
-2x + 2(1) + 3(0) = 0
-2x + 2 + 0 = 0
-2x + 2 = 0
- 2 - 2
-2x = -2
-2 -2
x = 1
(x, y, z) = (1, 1, 0)
-x + 7 = x - 5
add x to both sides of the equation
7 = 2x - 5
add 5 to both sides of the equation
12 = 2x
divide 2 from both sides of the equation
6 = x
Answer:
LHS.= Sin 2x /( 1 + cos2x )
We have , sin 2x = 2 sinx•cosx
And. cos2x = 2cos^2 x - 1
i.e . 1+ cosx 2x = 2cos^2x
Putting the above results in the LHSwe get,
Sin2x/ ( 1+ cos2x ) =2 sinx•cosx/2cos^2x
=sinx / cosx
= Tanx
.•. sin2x/(1 + cos2x)= tanx
Step-by-step explanation:
Answer:
s=p−(p⋅d) . 0.75p
Step-by-step explanation: