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In answering this question, I'm assuming that all the tiles used are "whole"...no "partial" tiles are used.

If the room is 16m^2, this means that each side of the room is 4m = 400cm. And if 256 tiles were used of type P, there must have been 16 on each side. Then each one would have been .25cm on each side. To see this, note that (.25cm)x16 = 400cm

Now, if whole tiles are used, and they have to be more than 15 cm on a side.....we need to find all the side lengths between 15 and 25 that divide 400 evenly. Only two fit the bill.....a side length of 16cm and a side lenth of 20cm.

If a side length of 16cm is used  (400/16)^2 = 625 tiles of type R must be used.

If a side length of 20cm is used (400/20)^2 = 400 tiles  of type R must be used.

x^5 = 200

We need to take the fifth root of both sides (not divide by 5...)

Doing this, we have

x =  ⁵√200 = about 2.885

3.41 x 10^-4

We just move the decimal back 4 places to the left

= .000341

a) From E, draw a perpendicular to AD  and a perpendicular to DC. The length of each of these will be the radius of the circle centered at E. Then, DE is just the hypotenuse of a isoceles right triangle and its length is just 4√2 cm.

By similar reasoning, BF = 6√2cm

b) So, the length of the diagonal will be 4√2 + 4 + 6 + 6√2  = 10 + 10√2

c) The length of the side of the square will be  [10 + 10√2] / √2.....and the area will be

([10 + 10√2] / √2)^2 = [100 + 200√2 + 200] / 2 =  [300 + 200√2] / 2 =

[150 + 100√2]cm^2

d) The area of the two circles = pi(4^2 + 6^2] =pi[16 + 36] = 52pi cm^2

e) The % area occupied by the two circles will be [52pi / (150 + 100√2) ] = about 56% of the total area of the square

To find that x = 6, use the pythagorean theorm, $${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$

$${\mathtt{a}} = {\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}$$, $${\mathtt{b}} = {\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}$$

$${\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}} = {{\mathtt{c}}}^{{\mathtt{2}}}$$ replace $${{\mathtt{c}}}^{{\mathtt{2}}}$$ with $${{\mathtt{x}}}^{{\mathtt{2}}}$$

$${\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{3.027\: \!735\: \!832\: \!268\: \!483\: \!3}}$$

The answer x = 6 is not correct.

To find the area of the triangle us the formula for area of a triangle, $${\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{l}}{\mathtt{\,\times\,}}{\mathtt{w}}$$

$${\mathtt{l}} = {\mathtt{length}}$$, $${\mathtt{w}} = {\mathtt{width}}$$

$${\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)$$

$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right) = {\mathtt{2.291\: \!796\: \!067\: \!500\: \!630\: \!9}}$$

The area of the triangle is approximately 2.30

To find the perimeter of the triangle, add up all the sides

$${\mathtt{p}} = \left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}{\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}}$$

$$\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right){\mathtt{\,\small\textbf+\,}}{\sqrt{{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\left({\sqrt{{\mathtt{15}}}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}} = {\mathtt{7.309\: \!600\: \!909\: \!545\: \!562\: \!5}}$$

The perimeter of the triangle is approximately 7.31

Of course......"permutes"  permutes to  "tee-rumps" which is what one has when he is holding all the cards....

Not to mention......all the medals.....in some combination, of course.......

My answer assumes that the total paid to the players and other workers is $2000 and $1600, respectively.....

Of course.....Anonymous has the correct idea if the amounts are per player and per worker.......we wouldn't have enough info to give an exact answer....

√[5 /(9a^2)] = √5 / (3a)

√(120 / 121)  = √120 / 11  = √(4 * 30) / 11  = (√4 * √30) / 11 = (2/11)√30

Notice that 10% of 280 is just 28

So....30% must be three times as much....

I'll let you take it from here....

I'm assuming this is

[2√3+9√2] /  6

We can split this up into

(2√3 / 6 )  + ( 9√2) / 6 =

(√3/3) + (3/2)√2

The only way that x = 6 is if the other two sides are √15 - √3  and √15 + √3

Proof

√[(√15 - √3)^2 + (√15 + √3)^2 ] =

√(15 - 2√45 + 3) + (15 +2√45 + 3) =

√(30 + 6) = √36 = 6

Assuming the above is true....the area =

(1/2)[√15 - √3][√15 +√3] =

(1/2)[15 - 3] = (1/2)(12) = 6 sq units

And the perimeter is just [√15 - √3] + [√15 +√3] + 6  =  6 + 2√15

12 -5y = 3y -12  add 5y and 12 to both sides

24 = 8y       divide both sides by 8

3 = y

3 / 2  = 1 +1/2  apples each

Just using normal "FOIL"....we have...

(2√3+5)(√3-√5)

2√3 *√3 = 2*3  = 6

2√3 *(-√5)  = -2√15

5 * √3  = 5√3

5 * (-√5)  = -5√5

And combining  terms, we have

6- 2√15 + 5√3- 5√5

Take the slope between the first two.....and then the slope between the first and last.....if these are equal, the points are on the same line.

The number of horizontal dots will be one greater that the number of horizontal toothpicks.

So.....this is 11.

And the number of rows of dots will be one more than the number of vertical toothpicks.

And this is 21

So the number of dots is just 11 x 21  = 231.

The number of horizontal toothicks we need in each row is just one less than the number of dots in that row = 10......and we have 21 rows of these - the same as the number of rows of dots.....so 10 x 21 = 210 horizontal toothpicks.

The number of vertical toothpicks we have in one column is just one less than the number of vertical dots in that column = 20.  And we have 11 of these columns......the same as the number of  dots in each row.  So 20 x 11  = 220 vertical toothpicks

The total number of toothpicks is just 210 + 220= 430.

Let's take the second part of this in relation to the dots. The total number of dots is just the dots in one horizontal row x  the number of rows of dots.

Then, the number of horizontal toothpicks is just (the number of horizontal dots in one row - 1) x the number of rows of dots.

And the number of vertical toothicks is just (the number of vertical dots in one column - 1) x the number of columns of dots

I hope that helps.....

In order to answer the question, more information is needed: how many players are reciving $2000 each and how many workers are recieving $1,600 each.  Once that information is given, you can add up the cost of what is spent on the players and workers, add the cost of the bats given away to the players, subtract the money made in concessions and then divide by the cost of the tickets to get the number of people who will need to buy a ticket to break even.

Lets say 10 players are paid $2000 each and 10 workers are paid $1600 each.  

To put into an equaion, it would be ($2000)(10) + ($1600)(10) + $3X = $5X.  

Since then numer of bats and the number of people are the same, bats and number of people can be represented by the same variable.

Now solve for x.  

1. Multiply $2000 and 10 togother and $1600 and 10 togother.  $20000 + $16000 + $3X = $5X.

2. Add $20000 and $16000 togother.  $36000 +$3X = $5X.

3. Subtract 3X to both sides.  $36000 = $2X

4. Divide $2 to both sides. X = 18000 people and bats

Assuming that 10 players and 10 workers are getting paid, 18,000 people will need to buy a ticket to break even.

Call N the number of tickets sold...this also equals the number of bats given away...and since revenue must = expenses, we have

5N + 1400 = 2000 + 1600 + 3N   simplify

2N  = 2200    divide by 2 on each side

N = 1100 tickets

Since the pro distance is twice the little league distance, 1/2 of the pro distance must equal the little league distance....

400(1/2)  = 200ft

(.026)+(.40*X)+(.05118) is not an equation.  In oredr to have an equation, there needs to be an equals sign.  Once there is an equals sign, then you can solve for x.

There is two ways to convert a decimal number into a fraction. 

First Way

Example 1

Lets say you have a decimal number 0.23

To change the decimal number 0.23 into a fraction

!.  Write the decimal with the number 1 underneath to form a fraction.  0.23/1

2.  Multiply the top and button by 100.  23/100

3.  Simplify the fraction.  23/100

Example 2

Lets say you have a decimal number 0.802

To change the decimal number 0.802 into a fraction

1. Write the decimal with the number 1 underneath to form a fraction. 0.802/1

2. Multiply the top and bottom by 1000. 802/1000

3. Simplify the fraction. 401/500

Notice that between the two examples, steps one and three are the same; however, sttep two is a little different. The reason for that depeds on how mnay digets are in the decimal number.  If there is one digit in the decimal number, multiply both top and bottom numbers by 10, if there are two digits in the decimal number, multiply both top and bottom numbers by 100, if thre are three digits in the decimal number, multiply both top; and bottom numbers by 1000, and so on.

Second Way

Example 1

Lets say you have a decimal number 0.3333333333333333333...

1. Put the decimal into algebraic terms. Let x = 0.3333333333333333333...

2.Move the decimal so the first recurring number is before the decimal point.  In this case that means that you need to multiply both sides by 10. 10x = 3.333333333333333333...

3.Remove the decimal numbers by subtracting x from 10x.  9x = 3

4. Solve for x.  x = 3/9

5. Simplify the fraction. 1/3

Example 2

Lets say you have a decimal number 0.1231212121212121...

1. Put the decimal into algebraic terms. Let x = 0.1231212121212121...

2.Move the decimal so the first recurring number is before the decimal point. In this case that means that you need to multiply both sides by 100000. 100000x = 12312.12121212121...

3.Remove the decimal numbers by subtracting x from 100000x. 99999x = 12312

4. Solve for x. x = 12312/99999

5. Simplify the fraction. 1368/11111

Notice that the first way are for decimal numbers where the decimal number does not repeat forever and the second way are for decimal numbers that do repeat forever.

The only decimal numbers that you cannot put into a fraction are decimal numbers that go on forever but do not repeat.  An example of a number that goes one forever but does not reopeat is $${\mathtt{\pi}}$$.  No fraction will equal $${\mathtt{\pi}}$$.

OK. The link now functions.

There is something about the inflection on permutation that is reminiscent of someone. I’m sure I’ll figure it out, soon.

Anyway, your answer may be a permutation, but it’s wrong. This answer should be appened to the multiple answers for this question that's on another post. It needs three wrong answers and one correct answer, like a multiple choice question.

Your incorrect answer will be the first one discarded -- at least by those who can find their butts in the dark with a torch, a GPS, and an anatomy guide. No buts about it!

72  =  (1/2)(18)(12)sin(A)

But:  (1/2)(18)(12)  =  108

So:  72  =  108sin(A)

Divide both sides by 108:

  72/108  =  sin(A)

Find the inverse sine:

   sin^-1(72/108)  =  41.8°

(9+10=21) (step 1) First we find the two numbers whose product is equal 9 (3x3) and we find the two numbers whose product is equalvilant to 10 (5x2). (step 2) Now we put the eqations next each other 5x2 3x3 and multiply them starting from the middle outwards (2x3=6) (5x3=15). (step three) Now we add those answers (15+6) and the answer comes out to be 21.