Solve 3(x + 2) > x.

1 answer:

5 0

3(x + 2) > x
3x + 6 > x
3x - 3x + 6 > x - 3x
6 > -2x
6/-2 < -2x/-2
-3< x
(When dividing by a negative, reverse the direction of the inequality side, reverse it EVERY time you divide by a negative)

ANSWER: -3 < x

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A 20-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 30 o

Dafna1 [17]

Answer:

$i(t)=(2/3)(1-e^{-300t})$

Step-by-step explanation:

Before we even begin it would be very helpful to draw out a simple layout of the circuit. Then we go ahead and apply kirchoffs second law(sum of voltages around a loop must be zero) on the circuit and we obtain the following differential equation,

$-V +Ldi/dt+Ri=0$

where V is the electromotive force applied to the LR series circuit, Ldi/dt is the voltage drop across the inductor and Ri is the voltage drop across the resistor. we can re write the equation as,

$di/dt+Ri/L=V/L$

Then we first solve for the homogeneous part given by,

$di/dt+Ri/L=0$

we obtain,

$i(t)_{h} =I_{max}e^{-Rt/L}$

This is only the solution to the homogeneous part, The final solution would be given by,

$i(t)=i(t)_{h} +c$

where c is some constant, we added this because the right side of the primary differential equation has a constant term given by V/R. We put this in the main differential equation and obtain the value of c as c=V/R by comparing the constants on both sides.if we put in our initial condition of i(0)=0, we obtain $I_{max} =V/R$ , so the overall equation becomes,