Answer:
-35.626 < temperature < 733.928 . . . . degrees F
(-35.626, 733.928)
Step-by-step explanation:
It is convenient to let a calculator evaluate the expression for converting °C to °F.
-35.626 < temperature < 733.928 . . . . degrees F
(-35.626, 733.928) . . . . . . . . . . . . . . . . . in interval notation
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The given relation is C = 5/9(F -32). The inverse relation is F = 9/5C +32. That is the relation we need to use to answer this question.
The answer is the 4th option,
x+y=6 and x-y= -10
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
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F(x) = 5x - 2
f(-3/5) = 5(-3/5) - 2
f(-3/5) = -15/5 - 2
f(-3/5) = -3 - 2
f(-3/5) = -5
