y = 6 - 4 = 2, thus 3y = 3 x 2 = 6
6(4+y)
Algebraic expressions are useful for translating problems into the language of mathematics. An algebraic expression for the problem "6 times the sum of 4 and y" would be: 6(4+y) = 24 + 6y.
Simply add the two force vectors together and their sum is the resultant force. (4, 8, 4).
2/3 +4/6 =
6
-6 + 7 - 4 = -3
6 4 3 does not equal 2. The sum of 6, 4, and 3 is 13, not 2.
2/3 + 4/6 = 4/3 or 11/3
6 and 4 6, 3 and 1 4, 3, 2 and 1
There are 216 permutations of three standard dice. Of these, there are exactly 10 that sum to 15, namely 3-3-6, 4-5-6, 4-6-5, 5-4-6, 5-5-5, 5-6-4, 6-3-6, 6-4-5, 6-5-4, and 6-6-3. The probability, then, of rolling a sum of 15 with three standard dice is 10 in 216, or 5 in 108, or about 0.04630.
6
Sum = 36 Count = 6 Average = Sum/Count = 36/6 = 6
To solve the expression, first calculate the product of 3 and the sum of 4 and -2. The sum of 4 and -2 is 2, and multiplying that by 3 gives 6. Now, divide 8 by this product: (8 \div 6 = \frac{4}{3}) or approximately 1.33.
It is Sum/Count = 18/3 = 6
24
3(2+4) 3(6) 18 ■
3+6+9+12 = 30