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joja [24]
3 years ago
8

I need help solving these.

Mathematics
2 answers:
Lelechka [254]3 years ago
4 0
I did some math on the first question, so <span>x+<span>x<span>(3)</span></span></span>=<span>88. x=22 and x*3=66. 
Mr.Klein is 66 years old and Mr.Meyer is 22. I don't know the answers for the other questions but hope this helps. If it does send me a thanks or something. If I'm wrong lemme know too.</span>
marin [14]3 years ago
3 0
If Mr. K is 3 times older than Mr. M, the expression we use to solve an equation would be "K is 3x Mr. M" which translates to this mathematical expression: K=3M.  Their ages added together is 88.  So K + M = 88.  But we already know that K = 3M, so we will sub it into the second equation to get 3M + M = 88 and 4M=88.  M = 22.  So if M = 22, then K = 3(22) which is 66.  Therefore, Mr. K is 66 and Mr. M is 22.  In number 11a, you have to follow the power rule for exponents which says that if we raise a power to a power, we multiply the powers together to get the new power.  5msquared all squared is 5m^4.  n to the fifth all squared is n^(5*2) which is 10.  So our new expression is 5m ^{4}  n^{10}.  In 11b, the rule is that when we multiply like bases (here our bases are a and b), we add the exponents.  Since that whole thing in 11b represents everything being multiplied together, we will add the exponents on all the a's and put them with a single a, and then we will add the exponents on all the b's and put them with a single b.  The exponents on all the a's add up to 8 (don't forget that an a without an exponent indicates a "1", NOT a 0!), and all the exponents on the b's add up to 11.  So our simplified expression is a^{8} b^{11}.  In 11c, we have a quotient, and the rule for that is we will subtract the exponents on like bases, denominator from numerator.  First of all, 35/7 = 5.  That was the easy part.  The exponent on the top x is 3 and the bottom x is 8, so we have the exponent as 3-8 which is -5.  The exponent on the top y is 7 and the bottom y is 4 so our exponent is 7-4=3.  Rewriting that as a single expression (I'm going to keep the negative exponent in it since it doesn't say to express all the exponents as positive) is \frac{5y ^{3} }{ x^{-5} }.  In the last one we deal with negative exponents.  In order to make a negative exponent positive, we put the base it's on under a 1 (keep in mind that bases can be variable OR numbers).  In general, x^{-1} = \frac{1}{ x^{1} } = \frac{1}{x}.  Our base is a 4, so we can rewrite it as 4^{-3}= \frac{1}{ 4^{3} }  = \frac{1}{64}
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