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d2/(d - c) + c2/(c - d) = d2/(d - c) - c2/(d - c) = (d2 - c2)/(d - c) = (d + c)(d - c)/(d - c) = d + c

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How do you factor C² - D²?

C^ (2) - D^(2) Factors to (C -D )(C + D) If we apply FOIL to these bracketed terms. (C -D )(C + D), then we have F ; C^(2) O = CD I = -DC L = -D^(2) 'Stringing out' C^(2) + CD - DC - D^(2) NB Remember CD= DC ; just like 2 x 3 = 6 & 3 x 2 =6 Hence C^(2) + CD - CD _ D^(2) Adding terms we have C^(2) - D^(2) NB THe (+)CD - CD = 0 This is the inverse function, done to show how C^(2) - D^(2) factors. NB Remember two squared terms with a negative(-) between WILL Factor. However, two squared with a positive(+) between them does NOT factor. As a n example, take the Pythagorean Eq'n. h^(2) = x^(2) + y^(2) This does NOT factor . However, h^(2) - y^(2) = x^(2) Does factors to (h - y)(h + y) = x^(2) Hope that helps!!!!! d


How many permutations of the letters C and D are possible?

Only 2: CD and DC.


What is the derivative of 2x-squared plus 4x plus 8?

y = 2x^(2) + 4x + 8 dy.dx = 2(2)x^(2-1) + 4x^(1-1) +8x^(0) dy/dc = 2(2)x^(1) + (1)4x^(0) + (0)8x^(0-1) dy/dx = 4x + 4 + 0 dy/dx = 4x + 4 *The derivative). NB #1 The original index number becomes a coefficient ( in brackets). Then subtract '1' from the original index value, for the new index value. #2 Any term raised to the power of '0' = 1. #3 Any term multiplied by '0' = 0.


What is the equation for wattage?

Wattage = voltage times amperage. That's for DC. For AC there is a power factor PF = cos phi you have do multiply with.AnswerThe above answer suggests that power ('wattage') is an electrical unit, which it is not. In fact, power is defined as the rate of doing work, so the basic equation is work divided by time.


What is the diameter of a circle if BC equals 13 and DC equals 17?

21.4