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Since 2 represents 20, all numbers in that scale are multiplied by 10. Thus, any answer in the second set of numbers is multiplied by 100. Since 8.8 in the top scale represents 88, the answer must additionally be multiplied by 10. The answer directly reads 1.76. Multiply by 100 and then by 10 to get the actual answer: 1,760.
In other words, to preserve n digits to the right of the decimal point, it is necessary to multiply the entire number by 10 n. In computers, which perform calculations in binary, the real number is multiplied by 2 m to preserve m digits to the right of the binary point; alternatively, one can bit shift the value m places to the left. For ...
If a positional numeral system is used, a natural way of multiplying numbers is taught in schools as long multiplication, sometimes called grade-school multiplication, sometimes called the Standard Algorithm: multiply the multiplicand by each digit of the multiplier and then add up all the properly shifted results.
In algebraic notation, widely used in mathematics, a multiplication symbol is usually omitted wherever it would not cause confusion: "a multiplied by b" can be written as ab or a b. [1] Other symbols can also be used to denote multiplication, often to reduce confusion between the multiplication sign × and the common variable x.
492 is close to 500, which is easy to multiply by. Add and subtract 8 (the difference between 500 and 492) to get 492 -> 484, 500. Multiply these numbers together to get 242,000 (This can be done efficiently by dividing 484 by 2 = 242 and multiplying by 1000). Finally, add the difference (8) squared (8 2 = 64) to the result: 492 2 = 242,064
Figure 2 is used for the multiples of 2, 4, 6, and 8. These patterns can be used to memorize the multiples of any number from 0 to 10, except 5. As you would start on the number you are multiplying, when you multiply by 0, you stay on 0 (0 is external and so the arrows have no effect on 0, otherwise 0 is used as a link to create a perpetual cycle).
For example, to multiply 7 and 15 modulo 17 in Montgomery form, again with R = 100, compute the product of 3 and 4 to get 12 as above. The extended Euclidean algorithm implies that 8⋅100 − 47⋅17 = 1, so R′ = 8. Multiply 12 by 8 to get 96 and reduce modulo 17 to get 11. This is the Montgomery form of 3, as expected.
An ingredient's mass is obtained by multiplying the formula mass by that ingredient's true percentage; because an ingredient's true percentage is that ingredient's baker's percentage divided by the formula percentage expressed as parts per hundred, an ingredient's mass can also be obtained by multiplying the formula mass by the ingredient's ...