Choose whether it's always, sometimes, never  an integer added to an integer is an integer a polynomial subtracted from a polynomial is a polynomial a polynomial divided by a polynomial is a polynomial a polynomial multiplied by a polynomial is a polynomial

1) an integer added to an integer is an integer - Always

(adding two integers always results in an integer, if the two integers are positive, their sum will be positive, if two integers are negative, they will yield a negative sum)

2) a polynomial subtracted from a polynomial is a polynomial - Always

(if the polynomials are subtracted vertically, then the signs of the subtracted polynomial's need to be flipped to their opposites)

3) a polynomial divided by a polynomial is a polynomial - Sometimes

(this depends on the degree of polynomials in the numerator and the denominator)

4) a polynomial multiplied by a polynomial is a polynomial - Always

(when two polynomials are added, the product function is the addition of degree to both the polynomials)

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

49/6 is going to equal 8 1/6