8.1/-.3 if you rationalize the denominator you get 81/-3 you can do this because the one number you can multiply anything by and not change it is 1 which can take on many forms. the one we want to use for this is 10/10 because when we multiply 8.1/-.3 with 10/10 we get 81/-3 since a +/- = - and 81/3=27 then our answer is -27 to prove this to yourself take a calculator and plug in both 8.1/-.3 and 81/-3 you will see that you get the same answer in both cases
decimals and fractions are spoken the same way if you look at the last place of the decimal that tells you what the denominator will be. for example .5645 has a last place of ten thousandths so we would say five thousand six hundred forty-five ten thousandths so our fraction would be 5645/10000 which is said the same exact way. the only problem is this fraction is not reduced and in any math class you will be expected to reduce your fraction. your original fraction was 465/1000 but both of these are at least divisible by five so it becomes 93/200 this is as far as this fraction can be reduced keep in mind that the highest you need to go when trying to find a divisor is the numbers square root. sqrt of 93 is 9.64...... so if you can not find a divisor that will go into both 93 and 200 before the number 9 then you can stop and know that you are done.
a fraction is just another way of writing a division problem. so when you see a fraction that looks like this 24/5 (which is improper) we can rewrite it as 24 divided by five. five goes into 24 four times with a remainder of 4. we are still dividing 4 by 5 so our mixed number then becomes 4 4/5 which is said four and four fifths because we are adding 4 with 4/5 and when we add fractions with whole numbers we need a common denominator. 4/1 = 20/5 and when we add 20/5+4/5 we get 24/5 which is what we started with
there are approx. 2.2lbs per 1kg and there are 1000g per 1kg so that means that there are approx. 2.2lbs per 1000g. what you will want to do is multiply .15lbs/1 with 1000g/2.2lbs the lbs increment cancels out and you are left with150/2.2 g = 68.18g approx.
The first thing you want to do is find out how many there are. you can determine this by looking at the sign changes of both f(x) and of f(-x). in your example there is only going to be one real positive zero because it only has one sign change in f(x). Now when you look at f(-x) there are five sign changes so there is a possibly five negative real zeros or three or one. this means that there are possibly no complex zeros or two or three. anytime that you one or none for pos or neg reals then that is what you have and it will not change that is why the pos real zero doesn't have any other possibilities. once you have decided the number of possible zeros then it is time to find them. to do this you will need to know how to do long division of polynomials or synthetic division. it is a bit of trial and error but you will need to start out with the first coefficient of the equation and the constant at the end these are referred to as P and Q respectively. 6 and -4 you will use all combination of these two numbers and their prime components in the fraction P/Q. P= 1,2,3,6 and Q=+-1,+-2,+-4 thus you will need to divide the equation or expression by +-1/1 or +-1, +-1/2, +-1/4, +-2/1 or +-2 , +-2/2 or +-1, +-2/4 or +-1/2, +-3/1 or +-3, +-3/2, +-3/4, +-6/1 or +-6, +-6/2 or +-3, +-6/4 or +-3/2. notice that some of them repeat this means that one of these choices could have a multiplicity higher than one so you will need to try to divide them in more than once. for instance
x^3+3x^2+3x +1 if you divide this once using (x+1) you will get x^2+2x+1 which can then be divided by (x+1) again giving you (x+1) thus your final answer is (x+1)^3 which means you have one real neg zero however it has a multiplicity of 3 and that is why the beginning of this process are only possible quantities of real and complex zeros. just remember to try all of the P/Q in your division so that you are sure to break the expression done completely so you can get all of your zeros. Hope this wasn't to confusing sorry i forgot about the fact that it was a -4 so that doubles the possibilities
The first thing you want to do is find out how many there are. you can determine this by looking at the sign changes of both f(x) and of f(-x). in your example there is only going to be one real positive zero because it only has one sign change in f(x). Now when you look at f(-x) there are five sign changes so there is a possibly five negative real zeros or three or one. this means that there are possibly no complex zeros or two or three. anytime that you one or none for pos or neg reals then that is what you have and it will not change that is why the pos real zero doesn't have any other possibilities. once you have decided the number of possible zeros then it is time to find them. to do this you will need to know how to do long division of polynomials or synthetic division. it is a bit of trial and error but you will need to start out with the first coefficient of the equation and the constant at the end these are referred to as P and Q respectively. 6 and 4 you will use all combination of these two numbers and their prime components in the fraction P/Q. P= 1,2,3,6 and Q=1,2,4 thus you will need to divide the equation or expression by 1/1 or 1, 1/2, 1/4, 2/1 or 2 , 2/2 or 1, 2/4 or 1/2, 3/1 or 3, 3/2, 3/4, 6/1 or 6, 6/2 or 3, 6/4 or 3/2. notice that some of them repeat this means that one of these choices could have a multiplicity higher than one so you will need to try to divide them in more than once. for instance
x^3+3x^2+3x +1 if you divide this once using (x+1) you will get x^2+2x+1 which can then be divided by (x+1) again giving you (x+1) thus your final answer is (x+1)^3 which means you have one real neg zero however it has a multiplicity of 3 and that is why the beginning of this process are only possible quantities of real and complex zeros. just remember to try all of the P/Q in your division so that you are sure to break the expression done completely so you can get all of your zeros. Hope this wasn't to confusing