# How would you use a factor in the real world?

Depending on what you really mean..................

The factor 3 as in $3 + $5 =$8 could be that your friend owes you 3 dollars from

you paying him for nachos at the pool, and 5 more dollars for you paying him to get

a muse song on itunes. Or maybe you have the problem $47+$92 for groceries and water bills. That would equal your entire bill for groceries and water. The answer is

$141. Or maybe subtraction factors..... You have groceries and water and electricity

and rent to pay this month. groceries are 92$, water is 47$,electricity is 400$, and rent is 700$ .You have already payed water and groceries. That's 141$ payed subtracted by the total amount of money which is left to pay. That's 141 dollars subtracted by 1239 =1098 dollars left to pay.

All of the factors I have used in this answer is 3,5,47,92,and 700. the rest of the factors are also answers because I subtracted 141 from 1239 which was the answer of the factors for the bills: 92,47,400, and 700.

### Is contextual intelligence the ability to use and apply your knowledge in a real world setting?

No. The context need not be a real world setting. If I am writing a Science Fiction book, my contextual intelligence would be the ability to use and apply the knowledge that is appropriate to my fictional world. If I were any good as a writer, that would not be the real world setting but a convincing but unreal extension of it.

### Why power factor is used?

Power factor is the ratio of real power over apparent power. In a purely resistive load, real and apparent power are the same, so the power factor is one. In a reactive load, such as an inductive or capacitative load, however, current lags (for inductive) or leads (for capacitative) the voltage. This phase angle means that, at certain portions or phases of the line cycle, the load is feeding power back into the source. A…

### How do you determine the number of real roots in a polynomial?

For a general polynominal, the cubic, quartic, and greater formulĂ¦ are too hellishly hard to work with, so you would need to plot the function or use Newton's/somesuch method to count the real roots by hand. If the polynomial has integral roots, you can use synthetic division to peel off the degrees to see if they factor wholely into binominals; then all roots will be real and explicit. Good luck: