1.

Multiply the reciprocal of 9/11 by the additive inverse of -18/55

Answer»

Answer:

Step-by-step explanation:

To solve this we first NEED to know what is reciprocal and additive inverse.

  • Every NONZERO rational number a/b has its reciprocal b/a. So that,

\sf{\circ\;\dfrac{a}{b}\times\dfrac{b}{a}=\dfrac{b}{a}\times\dfrac{a}{b}=1}

  • Every rational number a/b (or -a/b), there is a additive inverse -a/b (or a/b). So that,

\sf{\circ\;\dfrac{a}{b}+\dfrac{-a}{b}=0}

So,

\sf{\circ\;Reciprocal\;of\;\dfrac{9}{11}=\dfrac{11}{9}}

\sf{\circ\;Additive \;inverse \;of \;\dfrac{-18}{55}=\dfrac{18}{55}}

Now,

According to the question we have to MULTIPLY 11/9 by 18/55.

\sf{\longmapsto\dfrac{11}{9}\times\dfrac{18}{55}}

Reducing the numbers,

\sf{\longmapsto\dfrac{1}{1}\times\dfrac{2}{5}}

Multiplying the numbers,

\sf{\longmapsto\dfrac{2}{5}}

Hence,

  • The product 11/9 and 18/55 is 2/5.


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