14M + 2n - 6M + 11n = 8M + 13n
Let the two consecutive integers be n and n+1. Then, n + (n + 1) < 55 2n + 1 < 55 2n < 55 - 1 : 2n < 54 n < 27 The Inequality Statement can therefore be modified to show that for two consecutive integers to be less than 55 then the smaller integer must be less than 27.
10-14m = -4
43km 14m to kilometers
(Term)n = 59 - 2n
The easiest way to do this is to use 'n' as one integer. Now, the next one needs to be consecutive, so we can make the second integer 'n+1' n + (n+1) = 55 Now all we do is start simplifying 2n + 1 = 55 2n = 55 -1 2n = 54 n = 27 Since n = 27, the first integer will be 27. And we stated that the second is 'n+1', so the second would be 28. Double check, 27 + 28 = 55 So, our integers are 27 and 28.
The LCM of 7m-21 and 14m-42 is 14m-42. Since 14m-42 is a multiple of 7m-21, it is also the least common multiple (LCM) of the two expressions.
3p
6p divide by 3p = 2
m² - 14m + 48 = (m - 6)(m - 8)
3p - 6 > 21 3p - 6+6 > 21+6 3p > 27 3p/3 > 27/3 p>9 this WILL help
2n+2n equals 4n