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What else can I help you with?
How do you simplify 14M plus 2n-6M plus 11n?
14M + 2n - 6M + 11n = 8M + 13n
What is 6p divide by 3p?
6p divide by 3p = 2
How much is 7p more than 3p?
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.
What is the value of 3p 5p?
It is: 3p+5p = 8p
What is the range of measurements 20m 10m 22m 12m 14m 14m 5m 27m 45m 32m 24m?
17
How do you simplify 14M plus 2n-6M plus 11n?
14M + 2n - 6M + 11n = 8M + 13n
The sum of two consecutive integers is less than 55?
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.
What is 10-14m?
10-14m = -4
43km 14m to kilometers?
43km 14m to kilometers
What is the nth term for the sequence 57 55 53 51?
(Term)n = 59 - 2n
What is the LCM of 7m-21 and 14m-42?
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.
What are two consecutive integers whose sum is 55?
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.
What is the answer to -3p plus 6p?
3p
What is the factorization of m² - 14m plus 48?
m² - 14m + 48 = (m - 6)(m - 8)
What is 6p divide by 3p?
6p divide by 3p = 2
Solve this inequality 3p - 6 21?
3p - 6 > 21 3p - 6+6 > 21+6 3p > 27 3p/3 > 27/3 p>9 this WILL help
How much is 7p more than 3p?
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.
