Answer:
There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400
That's a huge number of points.
First: Calculate the value of 62 two pointers
62 * 2 = 124
Second: find the number of points that were due to 3 pointers.
151 - 124 = 27
Third: Divide by 3 to find the number of 3 pointers.
27÷3 = 9
Answer: There were 9 three pointers.
It can help because 9×4 =36, and 9×8=72. if you divide 72 by 2 you get 36 which is the answer to 9 ×4. basically because nine is the same in both problems you are multiplying by 4 (and you know 4 is half of 8) you can assume that multiplying the sum of 4 and 9 that you will get the sum of 8 and 9.
Answer:
44
Step-by-step explanation:
6 times 3 is 18 so.. 18 minus 69 would be -36 plus 12 is a positive 1 times 14 would be 14 minus 7 would be 7 7 times 27 would 38 plus 7 would be 45 minus one would be 44
The answer would be 150 as each ten bar is si worth 10 and 10x15=150.
