document.write( "Question 142600: Simplify:\r
\n" ); document.write( "\n" ); document.write( " $\"c%5E4+%2A+d%5E5+%2A+a+%2A+d%5E-1+%2A+a%5E9+%2A+c%5E3\"$ \n" ); document.write( "

Algebra.Com's Answer #104025 by Fruglemiester1234(13) $\"\"$ $\"About$

You can put this solution on YOUR website!
This problem is only hard if you don't know how to combine numbers to powers.
\n" ); document.write( "Lets start with the a's:
\n" ); document.write( "___ $\"a%5E1+%2A+a%5E9\"$ can be written as: a*a*a*a*a*a*a*a*a*a, if you have a problem seeing how they ad together. One a timesed together by nine a's is a total of ten a's multiplied together_______ $\"a%5E10....a%5E10+%2A+c%5E4+%2A+d%5E5+%2A+d%5E-1+%2A+c%5E3\"$
\n" ); document.write( "now the c's:
\n" ); document.write( "___ $\"c%5E4+%2A+c%5E3\"$ = c*c*c*c*c*c*c = $\"c%5E7\"$ ...... $\"a%5E10+%2A+c%5E7+%2A+d%5E5+%2A+d%5E-1\"$
\n" ); document.write( "Now the d's:
\n" ); document.write( "___d^5 * d^-1 = This one is harder to show then the others because of the negative, but another way to look at it is to just add the powers together = 5-1 = 4 = $\"d%5E4\"$
\n" ); document.write( "....... $\"a%5E10+%2A+c%5E7+%2A+d%5E4\"$
\n" ); document.write( "And we're as simple as we can get. \n" ); document.write( "

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