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0
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6
avatar+870 

Perform the following calculation without using a calculator :

$${\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{\left({\mathtt{\,-\,}}{{\mathtt{1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}}$$

 May 1, 2015

Best Answer 

 #5
avatar+118677 
+5

Einstein is right :)

 May 2, 2015
 #1
avatar+11912 
+5

lets do it then

 

$${\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{\left({\mathtt{\,-\,}}{{\mathtt{1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}}$$

 

$$=\frac{20}{(1 + 5)}$$

 

$$=\frac{20}{6}$$

 

$$=\frac{5}{3}$$

 

now i hope you can do the rest yourself!

 May 1, 2015
 #2
avatar+118677 
+5

no Rosala

 

$$-1^2=-1$$

 

think about it, what is   $$100-3^2$$

it is 100-9=91

the - sign is NOT squared :)

 

The rest of your answer is correct :)

 May 1, 2015
 #3
avatar+870 
+5

Precisely.

$${\mathtt{\,-\,}}\left({{\mathtt{1}}}^{{\mathtt{2}}}\right) = -{\mathtt{1}}$$ but

$${\left(-{\mathtt{1}}\right)}^{{\mathtt{2}}} = {\mathtt{1}}$$

So $${\frac{\left({\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{5}}\right)}{\left({\mathtt{\,-\,}}{{\mathtt{1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}} = {\frac{{\mathtt{20}}}{\left({\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}} = {\frac{{\mathtt{20}}}{{\mathtt{4}}}} = {\mathtt{5}}$$

 May 1, 2015
 #4
avatar+11912 
0

So melody he gave another suggestion? Is that right?

 May 2, 2015
 #5
avatar+118677 
+5
Best Answer

Einstein is right :)

Melody May 2, 2015
 #6
avatar+870 
0

I'm ALWAYS right ! Just kidding.

 May 2, 2015

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