14M + 2n - 6M + 11n = 8M + 13n
6p divide by 3p = 2
To find out how much 7p is more than 3p, subtract 3p from 7p. This gives you 7p - 3p = 4p. Therefore, 7p is 4p more than 3p.
It is: 3p+5p = 8p
17
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 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.
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.
3p
m² - 14m + 48 = (m - 6)(m - 8)
6p divide by 3p = 2
3p - 6 > 21 3p - 6+6 > 21+6 3p > 27 3p/3 > 27/3 p>9 this WILL help
2n+2n equals 4n