SOLUTION: How would this sequence of numbers be solved? I don't know how to type in in.Open parenthesis, two, b to the fourth power, c to the third, close parenthesis, Open parenthesis, thre

Algebra ->  Algebra  -> Sequences-and-series -> SOLUTION: How would this sequence of numbers be solved? I don't know how to type in in.Open parenthesis, two, b to the fourth power, c to the third, close parenthesis, Open parenthesis, thre      Log On

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Question 30728: How would this sequence of numbers be solved? I don't know how to type in in.Open parenthesis, two, b to the fourth power, c to the third, close parenthesis, Open parenthesis, three a to the second power, b to the negative second power. (2b(to the fourth power)c(to the third)(3a(to the second power)b(to the -2))
Answer by Cintchr(475) About Me  (Show Source):
You can put this solution on YOUR website!
+%282b%5E4c%5E3%29%283a%5E2b%5E-2%29%282b%5E4%29%28c%5E3%29%283a%5E2%29%28b%5E-2%29+ Assuming this is what you were trying to convey ....
multiply out all the numbers first... then a, then b, then c. remember to add the exponents.
+%282b%5E4c%5E3%29%283a%5E2b%5E-2%29%282b%5E4%29%28c%5E3%29%283a%5E2%29%28b%5E-2%29+
+2%2A3%2A2%2A3+=+36+
+a%5E2a%5E2+=+a%5E4+
+b%5E4b%5E-2b%5E4b%5E-2+=+b%5E4+
+c%5E3+
bring these pieces back together.
+36a%5E4b%5E4c%5E3+