Answer:
A
Step-by-step explanation:
given 
= 
using the method of cross- multiplication then
8P = 42 ( divide both sides by 8 )
P = 
= 5.25
B, because every prime number greater than 2 is odd, and the product of two odd numbers is odd.
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
Answer:
Step-by-step explanation:
+6x-5=0
we divide the coefficient of the X by half :
in this case: 6/2 = 3 , then we do the following
The result obtained is raised to square power: 3^2=9
we sum and subtract by 9 to maintain the balance of the equation:
+6x+9-9-5=0
we have:
-9-5=0
= 14
lets apply square root on both sides of the equation:
we know:
so we have:
abs(x+3)=
from where two solutions are obtained
finally we have:
