Divide. Assume that no denominator equals zero.
sorry @Jackienvdm last one
Flip the right side so that it turns into a multiplication problem. \[8d ^{3} \div 6c \times 24 \div c ^{3}d ^{4}\] From here, cancel out what you can. |dw:1400820805429:dw| The restrictions are: \[c \neq 0\] \[d \neq 0\]
A. http://static.k12.com/bank_packages/files/media/mathml_ce00b54d2d0fa6419ebbade7c21b46b96886771a_1.gif B. http://static.k12.com/bank_packages/files/media/mathml_7e4d92fbfa9e65bbd13862195ef509566a50175e_1.gif C. http://static.k12.com/bank_packages/files/media/mathml_4351d40c78df223d3146f94592712c7e8014ada5_1.gif D. http://static.k12.com/bank_packages/files/media/mathml_934f03c958e8c7af59a8350c2f10989bfa425c66_1.gif
So it would be B.
oh,thanks and can i get help on just two more questions? @Jackienvdm
Go ahead. (:
thanks
Divide and simplify completely. Assume that no denominator equals zero.
@Jackienvdm
So flip the right side and factor out the binomials. |dw:1400821425485:dw| Restrictions: \[d \neq 0, -1\]
A. http://static.k12.com/bank_packages/files/media/mathml_84282efc819bd0701e59b129f243034d9887b447_1.gif B.1 C. http://static.k12.com/bank_packages/files/media/mathml_38af02216b8ccb49e6f39bd1b9c83fd168c201f8_1.gif D. http://static.k12.com/bank_packages/files/media/mathml_e46103edc75972a370201d86c117b12591dee51b_1.gif
Yup so it would be A.
oh,thanks
last one
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