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Notice that he never scored higher than 90 on any test.......but......he could have scored 90 on four of them....so....we might have something like this......

[90 + 90 + 90 + x + 90] / 5   = 82     multiply both sides by 5  and simplify the left side

360 + x  =  410     subtract 360 from both sides

x = 50      and this would be his lowest score

(^x/₃) + (^3x/₄) = 2

Well first let's make these fractions have the same denominator  

(^4x/₁₂) + (^9x/₁₂) = 2

Now combine the fractions together

^13x/₁₂ = 2

Multiply both sides by 12 to remove the fraction

13x = 12 * 2

13x = 24

Divide both sides by 13 so "x" is the subject

x = ²⁴/₁₃

x = 1.846 (to 3 d.p.)

(18x^2) /( 55x)  * (121x^3) / (9x^4)     .....notice that we can rearrange this as....

(18/9)* ( 121/55) * (x^2 * x^3) / (x * x^4)  =

2 * [ ( 11*11) / (5*11) ]  *[ (x^5)/(x^5)]     ...notice that the last terms "cancel"  and we're left with...

2 * (11*11)/(5*11)     and each  "11" in both numerator and denominator "cancels"....so we have...

2 * (11/5) = 22/5

Let x be the time it takes in hours for B to clear the land......then....the amount of work that B can do in one hr = 1/x

Let x + 2 be the time in hours for A to clear the land..and the amount of work done in one hour by A is         1/(x +2)

(Amt done in one hour by each) * 22 hours = 1 whole job done.....or, mathematically......

[1/x  + 1/(x+2)] *22   =  1       simplify  by getting a common denominator on the left

[(x + 2 + x)] *22 / [(x(x + 2)]  = 1   multiply both sides by the reciprocal of .... 22 / [(x(x + 2)]       

2x + 2   =  [x(x + 2)] / 22        multiply both sides by 22....expand the right side

44x + 44  = x^2 + 2x     simplify some more

x^2 - 42x - 44  = 0    and using the onsite calculator we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{42}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{44}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{21}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{485}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{485}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{21}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{1.022\: \!715\: \!545\: \!545\: \!240\: \!5}}\\
{\mathtt{x}} = {\mathtt{43.022\: \!715\: \!545\: \!545\: \!240\: \!5}}\\
\end{array} \right\}$$

So...it would take "B" about 43 hours to clear the land working alone......[rounded to the nearest tenth]

a) A square of a prime number,

Reason: primes are the only numbers to have 2 divisors (1 and itself).

To get an odd-number of primes, the value must be a square number (Such as 25 has the divisors 5*5. Since these are both just "5" this is only going to provide 1 extra divisor, giving an odd number of divisors)

So, the only way to remain as square number which is indivisible by anything besides it's square root (and 1 and itself, of course), it's got to be a product of primes.

For the lowest possible on a single test to still result in an average value that is particularly high, then we give the other tests the highest possible value and see what we can get:

89 is the highest possible value, thus:

82 = (4*89 + fourth test) / 5

82 = (356 + fourth test) / 5

410 = 356 + fourth test

fourth test = 410 - 356

fourth test = 54

If y + 1  = 2015     ....then y = 2014

And we are asked to find w such that......

2(2014)^2 + 2(2014)  = 2015w   ......   rewrite as

2(2014)^2 + 2(2014) = (2014 + 1)w    ....  simplify

2(2014)^2 + 2(2014) = 2014w + w .....  and notice that......if we let w = 2*2014, we have

2(2014)^2 + 2(2014)  = 2014(2*2014) + 2(2014) 

2(2014)^2 + 2(2014)  = 2(2014)^2 + 2(2014).....so w = 2*2014 = 4028

There are probably easier ways to do this.....but....here's my reasoning.......

The "filling" scoop has to operate an odd number of times, because 46,899 *  an odd will produce an odd. And the "removing"  scoop always removes an even number of candies. So, the difference between an odd and an even is always an odd......

Let's say the flling scoop operates once and the removing scoop operates 3 times......then the number of candies left in the vat is just  46899 mod (3*13576) =  6171

Now...let's say that the filling scoop operates 3 times = 140697 candies added to the vat ....and that the removing scoop operates 10 times = 135760 candies removed from the vat

Then 140697 mod 135760 = 4937 ....  and 6171 - 4937 = 1234

Lastly....let the filling scoop operate 5 times = 234495 candies added to the vat, and let the "removing" scoop operate 17 times = 230792 candies removed from the vat

Then 234495 mod 230792 = 3703   =  6171 x 2(1234)

Notice the patern.....for every 2n - 1 times the "filling" scoop operates and for every 3 + 7(n-1) times the "removing" scoop operates, the mod remainder after the first time (where n = 1) = (6171) ... and this is reduced by 1234 for each successive integer value of "n"

So.....  6171/1234 = 5

This means that, when the "filling" scoop operates n = 5 more times after the first operation, i.e., 2(6) - 1 = 11 total times, the number of candies put into the vat is 11*46899  = 515889.  And the removing scoop will have operated 3 + 7(6-1)  = 38 times  in which 38 * 13576 = 515888 candies have been removed from the vat.....so..... 515889 added - 515888 removed = 1 remaining candy in the vat.....

[I'm pretty sure that there is some sort of algorithm for solving this......but....I'm not familiar with what it might be....!!! ]

x = 29 + (18*y)

x * y = 6354

First we can insert either y or x into the other equation. For this I shall insert y's equivalent.

x * y = 6354

∴ y = ⁶³⁵⁴/_x

Insert that into the below equation like so:

x = 29 + (18*y)

∴ x = 29 + (18 * ⁶³⁵⁴/_x)

Now let's simplify and re-arrange:

x = 29 + (¹¹⁴³⁷²/_x)

Since we don't like that x being on the bottom of the fraction, let's multiply it by "x". of course everything else must be to keep this equation valid:

[x] * x = [29 + (¹¹⁴³⁷²/_x)] * x

x² = 29x + 114372

0 = -x² + 29x + 114372

We have now gotten ourselves a quadratic equation, so use the "quadratic forumula" to find all possible values of "x":

x = (-29 ± sqrt(29² - 4*-1*114372)) / (2 * -1)

x = (-29 ± sqrt(841 + 457488)) / -2

x = (-29 ± sqrt(458329)) / -2

x = (-29 ± 677) / -2

The two results are...

x = (-29 + 677) / -2 = -324

x = 353

Using these two values of x, let's find their associated values of y:

y = ⁶³⁵⁴/_x

y = ⁶³⁵⁴/_-324 = -19.61 _{[1 recurring]}

y = ⁶³⁵⁴/₃₅₃ = 18

So the two results are:

x = -324 and y = -19.61 _{[1 recurring]}

x = 353 and y = 18

2^x = 2²⁰ - 2¹⁹

Another way of reading this is:

2^x = 2²⁰ - 2^20-1

Which equates to:

2^x = 2²⁰ - (2²⁰ / 2)

Now, we get the situation of "x - (x/2) = (x/2)". This means 2^x is a half of 2²⁰, which is another way of saying:

2^x = 2¹⁹

x = 19

What you would do is find the multiples of each number first (besides 0). For 600, there is 600, 1200, 1800, 2400, 3000, 3600, 4000, etc.

And for 108: 108, 216, 324, 432, 540, 648, etc.

The lowest number that is included in both sets would be 5,400, therefore that is the LCM.

66666⁴ / 22222⁴

= (66666 / 22222)⁴

= (3)⁴

= 9²

= 81

Saving = old cost - new cost

Saving = ($4 * ¹⁰⁰⁰/₃₀ per month for a year) - ($4 * ¹⁰⁰⁰/₅₀ per month for a year)

Saving = (⁴⁰⁰⁰/₃₀ * 12) - (⁴⁰⁰⁰/₅₀ * 12)

Saving = (1600) - (960)

Saving = $640 a year

F = (9/5)C  + 32   (1)     

C = (F - 32)(5/9)   (2)       ....so.....setting (1) = (2), we have

(9/5)C + 32  = (F- 32)(5/9)      multiply both sides by  (9/5)

(81/25)C  + 288/5  = F - 32     add 32 to both sides

(81/25)C + 288/5 + 32  = F

(81/25)C + 448/5 = F    (3)       and putting (3) into (1), we have

(81/25)C + 448/5  =  (9/5)C  + 32     simplify

(81/25)C - (9/5)C  =  32 - 448/5      ......     (9/5) = (45/25)   and  32  = 160/5

(36/25)C  = [160 - 448]/5

(36/25)C  = [-288/5]    

(36/25)C = -1440/25       multiply through by 25

36C  = -1440       divide both sides by 36

C = -40      and F = (9/5)(-40) + 32  =  -72 + 32 =  -40

So....they are equal at -40

I hope I understood the question correctly:

We have 2 wins, and 8 loses. They then win x amount more and end on equal loses and wins

So:

2 + x = 8

x = 6

10 meets + 6 = 16 total meets.

How many times does 13 go into 776?

$${\frac{{\mathtt{776}}}{{\mathtt{13}}}} = {\mathtt{59.692\: \!307\: \!692\: \!307\: \!692\: \!3}}$$

59 times. 

So far, we got -59. Now, we need the fraction. 

When you divided 776 by 13, we did not get a number by itself. It came along with a long decimal. This could be converted into a fraction. 

59 times 13 is 767

So, 776-767 is 9 

I think your improper fractions is -59 9/13

Thank you Alan. I understand how that can be.

Let say the big bang happened near where Earth is. Then say the big bang blast the energy and matter out in to the universe as a sphere and say it moves at the speed of light for billions of years until it starts making stars and galaxies and still moves very fast while they are forming. Then it takes billions of years for the stars to make light and then the light travels for billions of years back to where the big bang happen.

So if the big bang happened near earth, and it takes 13 billions years to go out into space, and some billions of years to become stars and 13 billions of years for it to return the light to earth. This makes it seem the universe has to be 26 billions of years plus the billions of more years for the big bang energy to become stars.

If the big bang happen billions of light years from where Earth is now, it also seems the universe would be much more than the 15 billions of years. This would show the matter that made the Milkyway moved away from the big bang center as the matter that make the distant galaxies moved in a opposing direction.

This explains why the galaxies are 13 billions of light years distant after 6.5 billions of years, but not how the light have time to travel from a distant galaxies to the Milkyway and Earth, because this means 6.5 billions of years to go to where they are then 13 billions of years for the light to travel from there to here and that adds to 19.5 billions of years and that dose not count the billions of years for the stars to form.

I have not ask a question here but I explain why this is not easy for me to understand these concepts. The Hubble constant explain some of this, but it does not explain all of why this does not add up in a normal way. If you have more understanding of this or know a website that can explain more I would be appreciative. I have went to some sites on astrophysics but there are not many that much explain time of universe and distance from stars.

Euler's Totient Function briefly explained.......I'm certainly no expert, but, after reading some about this, here is a "quick and dirty" explanation of "why it works".......

Let  p be some prime number.......then the number of positive integers that are not relatively prime to p^k   is just  .... p, 2p, 3p, 4p......p^k-1p

Then, there must be p^k-1 positive integers that are not relatively prime to p^k ......{Notice that the last integer is just p^k itself}.......So, if this is true.....the number of positive integers that are relatively  prime  to p^k must just be  p^k - p^k-1.....or.....expressed another way......

p^k - p^k-1  =     p^k ( 1 - 1/p )........let's look at a simple example.....

Let's find φ(1000).........we are aking...."How many positive integers are relatively prime to 1000??"

Note that 1000  = 2³ * 5³  ...so........  φ(1000)  = φ(2³ * 5³)  = φ(2³) * φ(5³)......then.....applying our "formula".....we have.......

φ(2³)  = 2³(1 - 1/2)  =  8(1/2)  = 4.........note that this is true......the number of positive integers that are relatively prime to 2³ =  (8)  are 1, 3, 5 and 7........seen another way.....p^k - p^k-1  =  2³ - 2^(3-1)  = 8 - 4  = 4

Now......look at φ(5³)......our  "formula"  gives us  5³(1 - 1/5)  = 125(4/5)  = 100.....this means that there are 100 positive integers that are relatively prime to 125.....this is easy to see.......the one's that aren't relatively prime to 125 are.....all multiples of 5 from 1-125  =  25 of them.......that means that 100 positive integers are relatively prime to 125  !!!

So.....the number of positive integers that are relatively prime to 1000 is just 4 * 100 = 400......Again, this is easy to see.......no even positive integers are relatively prime to 1000......this leaves a possible 500 that are......but all odd multiples of 5 aren't relatively prime to 1000, either, and there are 10 in each 100........for a total of 100 in all......so....   500 - 100 = 400 positive integers that are relatively prime to 1000  !!!

Thanks again to Heureka for letting us in on this "trade secret"........this is a pretty neat theorem.....!!!

$${\mathtt{1.3}}{\mathtt{\,\times\,}}{\mathtt{10}} = {\mathtt{13}}$$  nice and clean and no decimals, but it still needs to be equal to 1.3 so we do

$${\frac{{\mathtt{13}}}{{\mathtt{10}}}} = {\mathtt{1.3}}$$and we check if there are still any common divisors... nope none, it's in it's simplest form

You open your eyes! Enter the first number with your keyboard then click on the multiplication button and enter the second number, after that you hit enter.