Questions 1, 2, 3, and 4 are exercises to give you practice with
common denominators. For each of these questions, change all
the fractions to common denominators, and the answers jump out at you.
#1). 5/12 = 25/60
2/5 = 24/60
Make um negative, and then you'll have the answer right away.
#2). The one that's negative is obviously the least.
Both positive ones must be bigger than the negative one.
For the positive ones:
2/5 = 6/15
2/3 = 10/15 .
Now it's easy.
#3). This is tough. The least common denominator is 2,520 !
It's probably easier to just do the divisions and get the decimals
for each fraction.
-5/8 = -0.625
-7/9 = -0.777...
-4/5 = -0.8
-3/7 = -0.428...
Now it's easy to line um up.
#4). Sneaky one.
Look closely at each fraction.
B, C, and D are all less than 1, so they're not between 1 and anything more than 1.
8/5 is the only one that's more than 1.
#5). A fraction is just a short way to write a division problem.
When you see a fraction, it means
"the top number divided by the bottom number" .
When you actually do the division, the quotient you get
is the decimal form of the fraction.
To change a decimal into a percent,
move the decimal point two places that way ==> .
The numbers in the boxes at the bottom of #5 are the correct numbers,
but they both should be negative. (because the -3/8 is negative)
HEY mate here is your answer
I think the correct option is equivalent...
I am from India..
hope it helps you.
be brainly
This question is in reverse (in two ways):
<span>1. The definition of an additive inverse of a number is precisely that which, when added to the number, will give a sum of zero. </span>
<span>The real problem, in certain fields, is usually to show that for all numbers in that field, there exists an additive inverse. </span>
<span>Therefore, if you tell me that you have a number, and its additive inverse, and you plan to add them together, then I can tell you in advance that the sum MUST be zero. </span>
<span>2. In your question, you use the word "difference", which does not work (unless the number is zero - 0 is an integer AND a rational number, and its additive inverse is -0 which is the same as 0 - the difference would be 0 - -0 = 0). </span>
<span>For example, given the number 3, and its additive inverse -3, if you add them, you get zero: </span>
<span>3 + (-3) = 0 </span>
<span>However, their "difference" will be 6 (or -6, depending which way you do the difference): </span>
<span>3 - (-3) = 6 </span>
<span>-3 - 3 = -6 </span>
<span>(because -3 is a number in the integers, then it has an additive inverse, also in the integers, of +3). </span>
<span>--- </span>
<span>A rational number is simply a number that can be expressed as the "ratio" of two integers. For example, the number 4/7 is the ratio of "four to seven". </span>
<span>It can be written as an endless decimal expansion </span>
<span>0.571428571428571428....(forever), but that does not change its nature, because it CAN be written as a ratio, it is "rational". </span>
<span>Integers are rational numbers as well (because you can always write 3/1, the ratio of 3 to 1, to express the integer we call "3") </span>
<span>The additive inverse of a rational number, written as a ratio, is found by simply flipping the sign of the numerator (top) </span>
<span>The additive inverse of 4/7 is -4/7 </span>
<span>and if you ADD those two numbers together, you get zero (as per the definition of "additive inverse") </span>
<span>(4/7) + (-4/7) = 0/7 = 0 </span>
<span>If you need to "prove" it, you begin by the existence of additive inverses in the integers. </span>
<span>ALL integers each have an additive inverse. </span>
<span>For example, the additive inverse of 4 is -4 </span>
<span>Next, show that this (in the integers) can be applied to the rationals in this manner: </span>
<span>(4/7) + (-4/7) = ? </span>
<span>common denominator, therefore you can factor out the denominator: </span>
<span>(4 + -4)/7 = ? </span>
<span>Inside the bracket is the sum of an integer with its additive inverse, therefore the sum is zero </span>
<span>(0)/7 = 0/7 = 0 </span>
<span>Since this is true for ALL integers, then it must also be true for ALL rational numbers.</span>
