Q: What is the greatest common factor of 2c2 2c?

Write your answer...

Submit

Still have questions?

Continue Learning about Basic Math

(3a - 2c)(b - d)

(2a + b)(2c + d)

The GCF is 2c.

2a + 4b + 8c = 2(a + 2b + 4c) You could also continue by factoring the inside of the parentheses a bit: 2a + 4b + 8c = 2(a + 2b + 4c) = 2(a + 2[b + 2c])

12, 10, 8 and 6 (in descending order).Let the highest number equal a.We are told that ab - cd = 72However, as the numbers are consecutive even integers we also know that:b = a - 2c = a - 4d = a - 6thus we can use this to write the above equation in terms of only one variable, a, as:(a * (a - 2)) - ((a-4) * (a-6)) = 72(a2 - 2a) - (a2 - 10a + 24) = 72 (Remember that as we expand the second bracket in the next step we will multiply each term by the minus outside the bracket giving:)a2 - 2a - a2 + 10a - 24 = 72 (Now we can continue to simplify to get:)8a - 24 = 728a = 96a = 12Once we have a value for "a" we have a value for all the variables:a = 12b = 10c = 8d = 6As a proof, if we wish, we can substitute these values back into the original equation to get:ab - cd = 72(12 * 10) - (8 * 6) = 72120 - 48 = 72.

Related questions

48 http://www57.wolframalpha.com/input/?i=GCD%28240+%2C+672%29

To balance the equation 4H2 + 2C, we need to adjust the number of atoms on each side of the equation. Adding a coefficient of 2 in front of C on the left side will balance the carbon atoms. The balanced equation will be 4H2 + 2C2.

You must mean, either,(1) sec2 x = sec x + 2, or(2) sec(2x) = sec x + 2.Let's first assume that you mean:sec2 x = sec x + 2;whence,if we let s = sec x, we have,s2 = s + 2,s2 - s - 2 = (s - 2)(s + 1) = 0, ands = 2 or -1; that is,sec x = 2 or -1.As, by definition,cos x = 1/sec x ,this means thatcos x = ½ or -1.Therefore, providing that the first assumption is correct,x = 60°, 180°, or 300°; or,if you prefer,x = ⅓ π, π, or 1⅔ π.Now, let's assume, instead, that you mean:sec(2x) = sec x + 2;whence,1/(cos(2x) = (1/cos x) + 2.If we let c = cos x,then we have the standard identity,2c2 - 1 = cos (2x); and,thus, it follows that1/(cos(2x) = 1/(2c2 - 1)= (1/c) + 2 = (1 + 2c)/c.This gives,1/(2c2 - 1) = (1 + 2c)/c;(2c2 - 1)(2c + 1) = 4c3 + 2c2 - 2c - 1 = c; and4c3 + 2c2 - 3c - 1 = (c + 1)(4c2 - 2c - 1) = 0.As our concern is only with real roots,c = cos x = -1; and,therefore, providing that the second assumption is correct,x = 180°; or,if you like,x = π.

-((3c - 4)(3c + 4)(2c - 5)(2c + 5))

24c2 + 96c + 90 = 6*(4c2 + 16c + 15) = 6*(2c + 3)*(2c + 5)

(b + 2c)(b - c)

4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)

38

(a + b)(b - 2c)

(2a + b)(2c + d)

(a + b)(b - 2c)

(3a - 2c)(b - d)