All Answers 356462 Answers

Using OJ start:

30 cans in total. 60% coke, 40% pepsi.

that means 18 cans is coke and 12 cans is pepsi.

  75% of x is now Pepsi     .75x = 12

                                            x = 16 cans total

there WAS 30   30 -16 = 14 cans of Coke were consumed by diabeticNoah !

let the dimensions be x and y          DRAW PICTURES OF THE RECTANGLES

x * y = ?

decrease each of the sides by 1.  Now we know that

(x-1)(y-1) = ?

expand this and work out the value of  x+y

If you increase the sides by 1 then you have an area of   (x+1)(y+1)

expand this then substitute the known values.

Now you have your answer.

Give it a go and get back with what you have done.   

You can use the quadratic equation to solve this.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Think about it.  When will this have exactly one solution?

30 cans in total. 60% coke, 40% pepsi.

that means 18 cans is coke and 12 cans is pepsi.

After drinking some amount of cans of coke, now 25% is coke and 75% is pepsi.

the remaining amount has to be divisible by 4, so trying is probably faster than using equations.

if remaining 28, coke 16- no good

if remaining 24, coke 12- no good

remaining 20, coke 8- no good, but close

remaining 16, coke 4- that's our answer, and that means Noah drink 14 cans of coke.

We find a pattern: 6, 12, 18, 24, 30, 36, 42 and the rest of it.

It repeats every five numbers. 6, 36, 66, 96 or 12, 42, 72, 102.

We add 6+2+8+4+0 and multiply that by 20, but now we also include 606, so we subtract 6 from that.

20*20 is 400, minus 6 is 394.

So the units digit is 4

The following should help:

If you want to assume that the children are not identical then you tie the boys together and not you have 4 'children'

they can be arranged in a line in 4! ways.  But either one of the boys can be to the left so we need to double this answer.

4!*2 = 48 ways

Guest is probably telling you that you need to do your own homework.

I have deleted most of your questions.  

Please stop using this as a cheat site. 

People here want to help others, they do not want to do all your homework for you.

Good, I hope it is.

Do not dump all your homework here and expect the rest of the world to do it for you.

Its okay, its a really hard question... it may take some time to get solved 

If
\(z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)
find \(\lfloor z \rfloor\).

\(\begin{array}{|rcll|} \hline z &=& \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \\\\ z &=& \dfrac{ (\sqrt{3}-1)^2 - 2 (\sqrt{2}-1)^2 } { \sqrt{3}-1 - 2 (\sqrt{2}-1) } \\\\ z &=& \dfrac{ 3-2\sqrt{3}+1 -2(2-2\sqrt{2}+1) } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& \dfrac{ -2\sqrt{3}+4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& \dfrac{ -2(\sqrt{3}-2\sqrt{2}+1) } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& -2 \\\\ \lfloor z \rfloor &=& -2 \\ \hline \end{array}\)

Let \(P = 4^{1/4} \cdot 16^{1/16} \cdot 64^{1/64} \cdot 256^{1/256} \dotsm\)
Then \(P\) can be expressed in the form \(\sqrt[a]{b}\),
where \(a\) and \(b\) are positive integers.
Find the smallest possible value of \(a+b\)

\(\begin{array}{|rcll|} \hline P &=& 4^{1/4} \cdot 16^{1/16} \cdot 64^{1/64} \cdot 256^{1/256} \dotsm \\ P &=& 2^{\frac{1}{2^1}} \cdot 2^{\frac{2}{2^3}} \cdot 2^{\frac{3}{2^5}} \cdot 2^{\frac{4}{2^7}} \dotsm 2^{\left(\frac{n}{2^{2n-1}} \right)} \dotsm \\ P &=& 2^{\frac{1}{2^1}+\frac{2}{2^3}+\frac{3}{2^5}+\frac{4}{2^7}+\dotsm+\frac{n}{2^{2n-1}}+\dotsm } \\ P &=& 2^{S_{\infty}} \\ \hline S_n &=& \frac{1}{2^1}+\frac{2}{2^3}+\frac{3}{2^5}+\frac{4}{2^7}+\dotsm+\frac{n}{2^{2n-1}} \\ \frac{1}{2^2}S_n &=& \frac{1}{2^3}+\frac{2}{2^5}+\frac{3}{2^7}+\dotsm+\frac{n-1}{2^{2n-1}} +\frac{n}{2^{2n+1}}\\ \hline S_n-\frac{1}{2^2}S_n &=& \frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}}-\frac{n}{2^{2n+1}} \\ \frac{3}{4}S_n &=& s-\frac{n}{2^{2n+1}}\qquad (1) \\ s&=& \frac{1}{2^1}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}} \\ \frac{1}{2^2}s &=& \frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dotsm+\frac{1}{2^{2n-1}}+\frac{1}{2^{2n+1}} \\ \hline s-\frac{1}{2^2}s &=&\frac{1}{2^1} - \frac{1}{2^{2n+1}} \\ \frac{3}{4}s &=&\frac{1}{2^1} - \frac{1}{2^{2n+1}} \\\\ s &=& \frac{4}{3}* \left( \frac{1}{2^1} - \frac{1}{2^{2n+1}} \right) \\ \hline \frac{3}{4}S_n &=& s-\frac{n}{2^{2n+1}}\qquad (1) \\ \frac{3}{4}S_n &=& \frac{4}{3}* \left( \frac{1}{2^1} - \frac{1}{2^{2n+1}} \right)-\frac{n}{2^{2n+1}} \\ \frac{3}{4}S_{\infty} &=& \frac{4}{3}* \left( \frac{1}{2^1} - 0 \right)-0 \\ \frac{3}{4}S_{\infty} &=& \frac{2}{3} \\ S_{\infty} &=& \frac{4}{3}*\frac{2}{3} \\ S_{\infty} &=& \frac{8}{9} \\ \hline P &=& 2^{S_{\infty}} \\ P &=& 2^{\frac{8}{9}} \\ P &=& \sqrt[9]{2^8} \\ P &=& \sqrt[9]{256} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+b &=& 9+256 \\ \mathbf{a+b} &=& \mathbf{265} \\ \hline \end{array}\)

By long division, 3(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2) = 3x^2 + 12 - 15/(x + 2) + 4/(x - 1), so g(x) = -3x^2 - 12.

simplify to get 36*sqrt(2) - 8*sqrt(2) and that equals 28*sqrt(2). If we are trying to find n without any numbers outside the square root, n is 28*28*2 = 1568

Correct......   what do YOU find for that answer?    I'll let you answer that......  Thanx !    ~EP

Verical asymptopes occur when the denominator would be = 0

x^2 + 5x + 2 = 0     Quadratic formula shows x = - 5/2 +- sqrt (17)/2         sum = -5

bbggg

gbbgg

ggbbg

gggbb         four ways for the two boys......but the two boys could switch places making 8 ways for the boys to be in line

                        then the other three spots can be filled by the girls   3 x 2 x 1    6 ways

8 x 6 = 48 ways    (this assumes the boys and girls are distiguishable)

EC = 3.33333333333333333333333333333333333333333333333333333333333333333

It's another square; so, n = 4