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A equals 2bc plus d solve for d?

Updated: 9/18/2023

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∙ 14y ago

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d = a - 2bc

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Which of these formulas is correctly rewritten from the original formula to solve for d a equals 2bc plus d?

d=a-2bc

Which of these formulas is correctly rewritten from a equals 2bc plus d to solve for b?

(a-d)/2c=b

Which of these formulas is correctly rewritten from the original formula to solve for c if a equals 2bc - d?

(a+d)/2b=c

What formula is correctly rewritten from the original formula to solve for d?

d = a - 2bc

Factor this expression 4ac plus 2ad plus 2bc plus bd?

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

Related questions

Which of these formulas is correctly rewritten from the original formula to solve for d a equals 2bc plus d?

d=a-2bc

Which of these formulas is correctly rewritten from a equals 2bc plus d to solve for b?

(a-d)/2c=b

Which of these formulas is correctly rewritten from the original formula to solve for b a equals 2bc plus d?

If a=2bc+db= (a - d) / 2c If a=2b+c b=1/2(a-c)

A equals 2bc - d what is the formula to solve for c?

Type your answer here... a = 2bc-d add d to both sides a + d = 2bc divide both sides by 2b (a+d)/2b = c

Which of these formulas is correctly rewritten to solve for d?

a+2bc + d is d= a - 2bc

A equals 2bc - d how do you rewrite the original formula to solve for c?

A = 2bc - dAdd 'd' to each side of the equation:A + d = 2bcDivide each side by 2b :(A + d)/2b = c

Which of these formulas is correctly rewritten from the original formula to solve for c if a equals 2bc - d?

(a+d)/2b=c

What formula is correctly rewritten from the original formula a 2bc d to solve for d?

d = a - 2bc

Which of these formulas is correctly rewritten from the original formula to solve for c a equals 2bc - d?

c= (a+d) / 2b

What formula is correctly rewritten from the original formula to solve for d?

d = a - 2bc

How do you solve a equals b plus c plus d solve for c?

If: a = b+c+d Then: c = a-b-d

Factor this expression 4ac plus 2ad plus 2bc plus bd?

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

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