All Answers 358156 Answers

$${\frac{{\mathtt{1}}}{{\mathtt{10}}}} = {\mathtt{0.1}}$$

With PERMUTATIONS   - ORDER COUNTS

There will always be more permutations than combinations (unless the number of elements in the answer is 0 or 1).

with combintations order does not count

Thanks Guys - both these answers have great value. 

I have added this question to reference material.

There is no number multiplied by itself that will give you a negative number. Unless you use imaginary numbers. By definition i^2= -1

6+(8+3^2)-3(11-2(6-(8-6)))+1

Here's some nice PEMDAS here~

Parenthesis

We have:

(8+3^2)

and

(11-2(6-(8-6)))

Inside the parenthesis, we apply PEMDAS again:

(8+3^2) ---> (8+9)

and

(11-2(6-(8-6))) ---> (11-2(6-2))

Again:

(8+9) ---> 17

and

(11-2(6-2)) ---> (11-2(4))

Again:

(11-2(4)) ---> (11-8)

Again:

(11-8) ---> 3

And now, we have done almost everything in parenthesis. Insert the final answers from each part taken out back in the place of the parts in parenthesis.

6+(17)-3(3)+1

There are no exponents to deal with now.

Multiplication and Division

3(3) ---> 9

Addition and Subtraction

6+17-9+1

6 + 17 = 23 - 9 = 14 + 1 = 15

Your final answer is 15

The area will be 180 cm^2

Because you're changing both length and width, you're making each side 3x bigger than it originally was. We can then take the scale to the width (3) and the scale to the length (3) and multiply those together.

(20 cm^2)*(3*3) = 20*9 = 180

To double-check:

4*5 = 20 ---> (4*3)*(5*3) = 12*15 = 180 

2*10 = 20 ---> (2*3)*(10*3) = 6*30 = 180 

Find the molar mass of Iron

55.845 g/mol

Take the 5.5g (what you have) and put it in the numerator

Take the 55.845g/mol (molar mass) and put it in the denominator

$${\frac{{\mathtt{5.5}}}{{\mathtt{55.845}}}} = {\mathtt{0.098\: \!486\: \!883\: \!337\: \!81}}$$

There are 0.098 moles in 5.5g of Fe (iron)

The Pythagorean Theorem allows you to calculate the remaining side of a right triangle, if you are given two of the sides.

$$c^2=a^2+b^2$$

Just remember that c is always the hypotenuse, which is the longest side of a right triangle. A frequent error is to place the information regarding the hypotenuse in either a or b.

Try this.

In the calculator for this site, switch to the Equation Solver. Here you can use the term "i". Remember that i^2=-1

Not necessarily, how did you get that number?

$${\mathtt{60}}{\mathtt{\,\times\,}}\left({\frac{\left({\frac{{\mathtt{7}}}{{\mathtt{9}}}}\right)}{{\mathtt{20}}}}\right) = {\frac{{\mathtt{7}}}{{\mathtt{3}}}} = {\mathtt{2.333\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

Hmmm... Asimov seems pretty intersting. I haven't heard of him, but I just glanced at the book you suggested. This should be pretty fun to read. Than you!

I'm gonna' be honest, Anonymous....I "borrowed" this "formula" from one of our fellow posters, Nauseated, based on his answer to a problem just last week....I haven't had time to digest the rationale behind it, yet.

However.......here is his "non-technical" explanation.....if you care to look at it....

$${\frac{{\mathtt{5}}}{{\mathtt{6}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{9}}}} = {\frac{{\mathtt{11}}}{{\mathtt{18}}}} = {\mathtt{0.611\: \!111\: \!111\: \!111\: \!111\: \!1}}$$

If you would explain one of the factorials it would be greatly appreciated. I don't really understand how they are generated ,say, for the number of routes to the spider.

Thanks, Damian......that was a good problem, too.....a little bit of a challenge.....!!!!

We have n ways to choose 1 person from a department with n people.

Then, from the total remaining  people - 83 - , we just want to choose any 2 of them.

So we have....

25 x  23  x 14 x 7 x 19 x C(83, 2)  = 3,643,421,950 possible teams

I was the original questioner. (Do i say that right?)

I want to really thank you CPhill for your support, and your effort you put into that.

Thank you very much.

~Damian

Using my two previous answers.....

Total committees - Total committtees that include Fred = Total committees that don't include Fred

So we have

120 - 36  =  84 committees that don't include Fred

Note this is the same as C(9,3) = choosing  any 3 of 9 people that don't include Fred

The number of ways to get from (0, 0) to (5, 7) moving in this manner is given by

(7 + 5) ! / (7! 5!) = 12! / (7! 5! )  = 792 ways

From these, we want to subtract the ways to get from (0, 0) to (2, 3)

This is given by 

(3 + 2)! / (3! 2!) = 5! / (3! 2!) =  10 ways

So 792 - 10  = 782 routes

$${\frac{{\mathtt{16.5}}}{{\mathtt{0.264}}}} = {\frac{{\mathtt{125}}}{{\mathtt{2}}}} = {\mathtt{62.5}}$$