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There is no answer because there is no specific question.

The expression can be factorised, or if given v or q it can be evaluated.

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Q: What is the answer to v2 plus 3v plus 9vq plus 27q?
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What is 3v plus 15v - 11v2 - 14v plus 5v2 plus 8v3 - 4v3?

Assuming that the numbers immediately succeeding each "v" in the expression are intended to be exponents, in this type of problem only the coefficients of those terms with the same exponents can be added. Therefore in the sum, the coefficient of "v" will be (3 + 15 - 14) = 4; the coefficient of "v2" will be (-11 + 5) = -6; and the coefficient of "v3" will be (8 - 4) = 4. The final answer is therefore 4 v3 - 6 v2 + 4 v.


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What resistors are required to convert 12volts Dc to 3 volts dc?

In general terms, the voltage across a resistor can be calculated by ohms law: V = I . R voltage across the resistor = current through the resistor multiplied by the value of the resistor. So, if the resistor has a value of 100 ohms and the current flowing through the resistor is 10mA then the voltage across the resistor will be 100 x 0.01 = 1 volt. If the current flows through two resistors connected in series, the voltage will be split over the two resistors according to their resistance. V = V1+V2 = I . R1 + I . R2 You want V2/(V1+V2)=3V/12V Using Ohm's law for each voltage: V2/(V1+V2) = I . R2 / (I . R1 + I . R2) = R2 / (R1+R2) You see that the answer to your question is: Any combination of resistors will do the job, provided that R2/(R1+R2) = 3V/12V e.g. R2=3Ohm and R1=8Ohm or R2=3000Ohm and R1=8000Ohm Depending on your choice, you will get a different current flowing. In the real world you might need to control the current flowing into your circuit. For example if the device to be powered is a 3V LED and the required current is 15mA then we can calculate as follows: 12V - 3V = 9V. 9V = 0.015A x R Therefore R = 9V / 0.015A Therefore R = 600 ohms. However, in real life this is usually a terrible way to control voltage for several reasons. First, because the current in the most circuits is not constant (it would be constant for an LED but not for most gadgets). Second, this method wastes energy heating the resistor. Third, it is often not possible to find a resistor of exactly the right value. So, in real life this problem is usually solved by using a "voltage regulator" which is a simple integrated circuit (often with just three legs).


How do you solve boyles law equation for V2?

To solve Boyle's Law equation for V2, first write the equation as P1V1 = P2V2. Then rearrange it to isolate V2 on one side, dividing both sides by P2 to solve for V2, which will be V2 = (P1 * V1) / P2.


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#include<iostream> #include<cassert> struct A { int val1; int val2; A (int v1, int v2): val1 (v1), val2 (v2) {} A operator- () const { return A (-val1, -val2); } }; int main() { A a (12, -42); A b = -a; assert (b.val1 42); }