12x - 16 = 68 Solve for x
2 answers:
You might be interested in
Answer:
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,
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,
Then we first solve for the homogeneous part given by,
we obtain,
This is only the solution to the homogeneous part, The final solution would be given by,
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 , so the overall equation becomes,
where if we just plug in the values given in the question we obtain the answer given below,
Answer:
174 units²
Step-by-step explanation:
There are a few different ways you can find the altitude of the trapezoid. Consider the attached figure with some points and lines added. SQ' is parallel to TQ, so Q'Q = 4 and PQ' = 21. Based on the side length of PS = 13, you can <u>guess</u> that the height is 12. (5-12-13 is a commonly-used Pythagorean triple.) This would make PP' = 5, P'Q' = 16, and triangle SP'Q' have side lengths 12, 16, and 20, corresponding to a 3-4-5 right triangle multiplied by 4.
Another way to find the height is to use Heron's formula for the area of triangle PSQ'. The side lengths are 13, 20, 21, so the half-perimeter is 27 and the area is √(27(27-13)(27-20)(27-21)) = √(9²·14²) = 126. The base of the triangle, PQ', is 21, so the height is ...
... h = 2A/b = 2·126/21 = 12
The area of parallelogram Q'STQ is then ...
... A = bh = 4·12 = 48
and the total area is the triangle area plus the parallelogram area:
trapezoid area = 126 + 48 = 174 . . . . units²
_____
Of course, with the height known, the usual formula for the area of a trapezoid can be used:
A = (1/2)(b1 +b2)h
A = (1/2)(25 +4)·12 = 29·6 = 174 . . . . units²
