All Answers 358523 Answers

$${\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\mathtt{0.2}}$$

Simply divide 1 by 5, then multiply by 100

$${\mathtt{0.2}}{\mathtt{\,\times\,}}{\mathtt{100}} = {\mathtt{20}}\%$$

I'm not totally confident about this answer, but I'll give it a shot.....!!!!

The number of posters he can print per hour equals the number of impressions he can make per hour x the number of plates used in each impression = 1000x

And the number of hours it will take him is given by 100,000/(1000x) = 100/x

And the cost per hour equals 202.80

I'm assuming that each plate costs $30 to prepare....so the total cost of the plates is just $30x

So the total cost is given by:

C =[100/x)]202.80 + 30x     simplify

C = 20280/x  + 30x

So.....take the derivative and set it to 0

C' = -20280/x² + 30

Set to 0

-20280/x² + 30 = 0     simplify

20280 = 30x²

x² = 2028/3

x² = 676

x = 26   .....so 26 plates should be used

Makes more sense now...I wasn't sure what "16 from" meant

8.4948

From means to take out, or subtract from, so you would find the square root of 600 first [which is about 24.495 or 10*sqrt(6)] and then subtract 16 from that number. so 24.495 - 16

$${\sqrt{{\mathtt{600}}}} = {\mathtt{24.494\: \!897\: \!427\: \!8}}$$

Other than that I'm not sure what you mean...

y=(x-2)*(x-2) = x^2-4x+4

to solve this you simply FOIL. multiply the First parts in each parentheses (x and x), multiply the Outer of each parentheses (x and -2), multiply in Insides of each parentheses (-2 and x), mulitply the Last in each parentheses (-2 and -2). This yields (in order) x^2, -2x, -2x, and +4. Then you add each part of your FOIL. x^2+(-2x)+(-2x)+4

I was wrong...

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{9}}}{{\mathtt{7}}}}$$

Very simple. Multiply 3 by -3 in the numerator to get -9. this works for any multiplication with fractions

$${\frac{{\mathtt{1}}}{{\mathtt{5}}}}{\mathtt{\,\times\,}}{\mathtt{200}} = {\frac{{\mathtt{40}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{\mathtt{2}}}} = {\frac{{\mathtt{20}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{\mathtt{4}}}} = {\mathtt{5}}$$

I hope I did that right...

Yes, thank you ^_^

That is cute - and very derpy.   :)     (AND EFFECTIVE - I am not criticizing)

$$6\sqrt[5]{32m^3}\times5\sqrt[5]{1024m^2}\\\\
=6*5\sqrt[5]{32*1024m^3m^2}\\\\
=6\sqrt[5]{32*1024m^5}\\\\
=6\sqrt[5]{32*1024}\;\sqrt[5]{m^5}\\\\
=6\sqrt[5]{32*1024}\;\sqrt[5]{m^5}\\\\
=6\sqrt[5]{32*1024}\;m\\\\$$

$${\sqrt[{{\mathtt{{\mathtt{5}}}}}]{\left({\mathtt{32}}{\mathtt{\,\times\,}}{\mathtt{1\,024}}\right)}} = {\mathtt{8}}$$

$$\\=6\sqrt[5]{32*1024}\;m\\\\
=6*8\;m\\\\
=48m$$

Does that help?

It would also be very useful for you to know that fifth root is the same as a power of 1/5

$$\sqrt[5]{32}=32^{1/5}$$

If you are going to put this into any calc it could be done as

32^(1/5)=

In the web2 calc you could also type in sqrt(32,5) and press the equals.  This is just another way to do it.

What does that mean?

N0  no one is here - look under the bed, we might be there. 

I'll try to help you the best I can, but I need a question first

$$\frac{-3}{2}-\frac{\sqrt2}{2}=\frac{-3-\sqrt2}{2}$$

This is an irrational number.  Your calc is not going to give you a fraction answer.  I am a bit confused  

A little derpy, not going to lie, but:

I couldnt figure out how to put the exponents in the square like that, so I just drew them on 

Well you cannot divide by 0 so 0 is one of the numbers.

1+1/x can also not equal zero so x cannot be -1

Also if you simplify this down you get to a point where  

$$f(x)=\frac{1}{\frac{2x+1}{x+1}}$$

You cant divide by 0 so  2x+1 not eqal 0

x not equal -1/2

So the domain does not include  0,-1/2 or -1

the sum of these is -3/2

I have not checked any of my working. :)

Haha, melody nice one. That is hilarious!! 

Maybe you should give us an example of your homework so that we can gauge the difficulty level.

Yes. A negative exponent will flip the number with the exponent to the denominator.

Basically:$${{\mathtt{8}}}^{-{\mathtt{1}}} = {\frac{{\mathtt{1}}}{{\mathtt{8}}}}$$ 

Another example would be: $${{\mathtt{13}}}^{-{\mathtt{32}}} = {\frac{{\mathtt{1}}}{{{\mathtt{13}}}^{{\mathtt{32}}}}}$$

From there it's simple math with exponents to solve it. 

I'm with Happy on this one but Happy's comment is correct - It definitely did not say who the marbles were given to.

So the question made no sense and there are at least 2 'correct' answers.  LOL

Ok  MrKmney but how would you convert 5/6 ?