An equation where c = 5
That would be 10
Yes! So in conclusion, 1Q, 2C, and 1M are all equivalent. Goodbya, Koda! :P
2c 2(5) 10 The answer is 10. You plug in the value of 5 for c. When the number and variable touch directly it implies multiplication.
12-2c = 2c add 2c to both sides 12 = 4c divide both sides by 4 3 = c
Without any information about a or b or c, all that can be said is: a + b + 2c = a + b + 2c Not particularly illuminating!
That would be 10
Yes! So in conclusion, 1Q, 2C, and 1M are all equivalent. Goodbya, Koda! :P
2c 2(5) 10 The answer is 10. You plug in the value of 5 for c. When the number and variable touch directly it implies multiplication.
12-2c = 2c add 2c to both sides 12 = 4c divide both sides by 4 3 = c
4c-7 = 2c+11 4c-2c = 11+7 2c = 18 c = 9
Without any information about a or b or c, all that can be said is: a + b + 2c = a + b + 2c Not particularly illuminating!
(b + 2c)(b - c)
If 9 - 2c + 3 = -13, then c would be 12.5
2c-10
To evaluate the expression (4c , 1 - (5 , 2c)), we first need to clarify the notation. Assuming (c) represents a variable and (n , k) represents the binomial coefficient "n choose k," we have (4c , 1 = 4) and (5 , 2c = 10). Therefore, the expression simplifies to (4 - 10), which equals (-6).
2c + 3 = 15 Subtract 3 from both sides: 2c = 12 Divide both sides by 2: c = 6
16 oz