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Which of these formulas is correctly rewritten from a equals 2bc plus d to solve for b?
(a-d)/2c=b
Factor this expression 4ac plus 2ad plus 2bc plus bd?
4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)
What is the original formula to solve for b in this problem a equals 2bc plus b?
Not sure about the original formula, but one that works is b = a/(2c + 1)
What is the answer to b plus b plus c plus c in algebra?
2b + 2c or 2(b + c)
Factor ab-2ac plus b2-2bc?
(a + b)(b - 2c)
Which if the following is a factor b squared plus bc munus 2c squared equals?
(b + 2c)(b - c)
Which of these formulas is correctly rewritten from a equals 2bc plus d to solve for b?
(a-d)/2c=b
Factor this expression 4ac plus 2ad plus 2bc plus bd?
4ac + 2ad + 2bc +bd = 2a*(2c + d) + b*(2c + d) = (2c + d)*(2a + b)
What is the original formula to solve for b in this problem a equals 2bc plus b?
Not sure about the original formula, but one that works is b = a/(2c + 1)
If a plus a plus b plus b equals 26 and b plus b plus c equals 22 and c plus c plus c plus c plus b plus b equals 46 then a equals?
2a + 2b = 26 2b + c = 22 ===> b = 11 - c/2 4c + 2b = 46 ===> b = 23 - 2c From the 2nd and 3rd equations: 11 - c/2 = 23 - 2c 22 - c = 46 - 4c 22 - 46 = -4c + c ===> -24 = -3c ===> c = 8 Substitute c=8 into the 3rd equation: b = 23 - 16 = 7 Substitute b=7 into the 1st equation: 2a + 14 = 26 2a = 12 ===> a = 6
What does a plus b plus c plus f equals?
That's going to depend on the values of 'a', 'b', 'c', and 'f'.
What is the answer to b plus b plus c plus c in algebra?
2b + 2c or 2(b + c)
What is 2c plus 4-3c equals -9 plus c plus 5?
2c + 4 - 3c = -9 + c + 5 -c + 4 = c - 4 -2c = -8 2c = 8 c = 4
Which of these formulas is correctly rewritten from the original formula to solve for b a equals 3b - 2c?
1
4c-7 equals 2c plus 11?
4c-7 = 2c+11 4c-2c = 11+7 2c = 18 c = 9
2a plus 1b equals 2c plus 2d-----2a plus 1b equals 1c plus 2d----- 1a plus 2b equals 3d-----what does 4a equal?
I--2a+b=2c+2d II--2a+b=c+2d III--a+2b=3d * Here if we perform (I)-(II), we get, c=0 * Using the c we can prove b=0 using (I)-(III). * Now replace 'b' & 'c' as 0 in (I), then we get a=d. Email me at yash2008gates@gmail.com for any queries. I will be very thankful to receive your queries.
What is the factor for ab-2ac plus b2-2bc?
(a + b)(b - 2c)
