it depends on the shelf of course
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While massaging the question into some form that can be answered,I'm assuming:-- exactly 4 of the 10 books are math books-- the other 6 are not math books-- you're asking how many different ways all 10 can be arrangedon the shelf while keeping the math books all together.If that's your question, then the answer is as follows. If that's not your question,then the following is the answer to my question but it won't help you at all :The 10 books are really only 7 units to be arranged ... 6 random books, andone unit comprised of math books that might as well be glued together becausethey can't separated. So the shelf arrangement involves lining up 7 things.The first can be any one of the seven. For each of those . . .The second one can be any one of the remaining six. For each of those . . .The third can be any one of the remaining five. For each of those . . ...etc.Total number of possible arrangements is (7 x 6 x 5 x 4 x 3 x 2) = 5,040 ways.
3x - 7 ≤ 9 "Seven less than" means that you subtract seven from the answer. "Three times a number" means that you multiply it. "At most" means that this is the greatest number that it can be, which is, in this case, 9. To solve this, you apply the principle rules of algebra to determine x. 3x ≤ 9 + 7 3x ≤ 16 x ≤ 16/3
It is time for the students to get into the books and quit relying on the internet because they can tell you fake things here on this internet and you can get it wrong because of it.
seventy seven quintillion, seven hundred seventy seven quadrillion, seven hundred seventy seven trillion, seven hundred seventy seven billion, sevenhundred seventy seven million, seven hundred seventy seven thousand, seven hundred seventy seven. :-)
The sum of six and seven is expressed as six plus seven. The full equation would be six plus seven equals thirteen.