Answer:
The maximum potential difference is 186.02 x 10¹⁵ V
Explanation:
formula for calculating maximum potential difference

where;
Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²
k is the dielectric constant = 2.3
b is the outer radius of the conductor = 3 mm
a is the inner radius of the conductor = 0.8 mm
λ is the linear charge density = 18 x 10⁶ V/m
Substitute in these values in the above equation;

Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V
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<span>A+B-C
</span><span>A = 6x - 2y
B = -4x - 8y
C = -3x + 9y
(</span>6x - 2y) + (-4x - 8y) - (-3x + 9y)
(6x - 2y) + (-4x - 8y) + (3x - 9y)
2x -10y + (3x - 9y)
5x - 19y