Search results
Results from the WOW.Com Content Network
To test for divisibility by D, where D ends in 1, 3, 7, or 9, the following method can be used. [12] Find any multiple of D ending in 9. (If D ends respectively in 1, 3, 7, or 9, then multiply by 9, 3, 7, or 1.) Then add 1 and divide by 10, denoting the result as m. Then a number N = 10t + q is divisible by D if and only if mq + t is divisible ...
This can be seen in Figure 6 by the decrease in depth from y 1,q=30 to y 1,q=10 and the increase in depth between y 2,q=30 and y 2,q=10. From this analysis of the change in depth due to a change in flow rate, we can also imagine that the energy lost in a jump with a value of q = 10 ft 2 /s would be different from that of a jump with q = 30 ft 2 /s.
For instance, if the number π is rounded to 4 decimal places, the result is 3.142 because the following digit is a 5, so 3.142 is closer to π than 3.141. [107] These methods allow computers to efficiently perform approximate calculations on real numbers.
The French scale measures and is proportional to the outer diameter of a catheter, with 1 French (Fr) defined as 1 ⁄ 3 millimeter, making the relationship: 1 mm = 3 Fr. Thus, the outer diameter of a catheter in millimeters can be calculated by dividing the French size by 3. [ 2 ]
For example, the following algorithm is a direct implementation to compute the function A(x) = (x−1) / (exp(x−1) − 1) which is well-conditioned at 1.0, [nb 12] however it can be shown to be numerically unstable and lose up to half the significant digits carried by the arithmetic when computed near 1.0.
If the intervals are readjusted for these frequencies, the entropy of the message would be 4.755 bits and the same NEUTRAL NEGATIVE END-OF-DATA message could be encoded as intervals [0, 1/3); [1/9, 2/9); [5/27, 6/27); and a binary interval of [0.00101111011, 0.00111000111).
For example, 1 and 1.02 are within an order of magnitude. So are 1 and 2, 1 and 9, or 1 and 0.2. However, 1 and 15 are not within an order of magnitude, since their ratio is 15/1 = 15 > 10. The reciprocal ratio, 1/15, is less than 0.1, so the same result is obtained.
1 + 2 = 3, 3 + 3 = 6, 6 + 4 = 10, 10 + 5 = 15. This difficulty results from subtly different uses of the sign in education. In early, arithmetic-focused grades, the equal sign may be operational ; like the equal button on an electronic calculator, it demands the result of a calculation.