Question 238135
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1.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ -(3x\ +\ 5)]


Change it to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (-1)(3x\ +\ 5)]


And apply the distributive property:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a(b\ +\ c)\ =\ ab\ +\ bc]


2.  You need to write back for this one.  Not sure whether you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}\ +\ 4m\ =\ 3m\ -\frac{5}{2}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2\ +\ 4m}\ =\ \frac{3m\ -\ 5}{2}]


...and there are other interpretations.


3.  *[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{4}x^3\ +\ 4x^2\ -\ x^3\ +\ 7]


The only like terms are the *[tex \Large x^3] terms.  One has a coefficient of *[tex \Large \frac{3}{4}] and the other has a coefficient of *[tex \Large -1].  Just add: *[tex \Large \frac{3}{4}\ +\ (-1)] and the sum is then the coefficient on your *[tex \Large x^3] term.  When you are done, the exponents should get smaller from left to right.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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