All Answers 358187 Answers

Always remember PEMDAS, and strictly follow the order.

To solve this, I started experimenting the profit of books going down from the price of $26 AND going up from the price of 0$

I realized that 0 and 26 had the same profit, the same as 1 and 25. This pattern continues on with the profit always going on a increase.

Because of this trend, we can find the median number between 1 and 25, which is 13.

So the book store should make its books cost $13 to maximize its revenue.

SO SORRY FOR THE INCONVINCE!! I figured it out. Srry. 

\(2^5−(2^3)(2^3)\)

\(=32−(2^3)(2^3)\)

\(=32−8(2^3)\)

\(=32−(8)(8)\)

\(=32−64\)

\(=−32\)

This mean that the answer is -32. Hope this is what you wanted!!

Thx for the help!! 

(10x1950[7000-{1950x2}])x5.4%=326430000%

(10x1950[7000-{1950x2}])x0.6%=36270000%

Successively calculate the curly bracket, the square bracket, the round bracket.

  !

-5 x -2 = 10 is the max value of b

f(x)=204+3e/0.2x

I think the guest accidentally pressed shift and didn't change it...

ANSWER:(I hope...)

\(204 + \frac{3e}{0.2*3} \\=204+\frac{3e}{0.6} \\=204+13.591409142313.5914091423 \\=217.591409142313.5914091423\)

ANSWER:

217.591409142313.5914091423

no, I think it means to divide and he/she just accidentally pressed shift...

Sequence problem
The sequence \(\{a_n\}\) is defined by
\(a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \dfrac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2\).

The sequence \{b_n\} is defined by
\(b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \dfrac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2\).

Find \(\dfrac {b_{32}}{a_{32}}\).

\(\begin{array}{|rcll|} \hline a_n &=& a_{n - 1} + \dfrac { a_{n - 1}^2 } { a_{n - 2} } \\ \mathbf{ a_n } &=& \mathbf{ a_{n - 1} \left( 1 + \dfrac{a_{n-1}}{a_{n-2}} \right) } \\ \hline a_0 &=& 1 \text{ or } a_0 = 0! \\ a_1 &=& 1 \text{ or } a_1 = 1! \\ \hline a_2 &=& 1!\left(1+\dfrac{1}{1}\right) \\ &=& 1!\cdot 2 \\ &=& \mathbf{2!} \\\\ a_3 &=& 2!\left(1+\dfrac{1!\cdot 2}{1!}\right) \\ &=& 2!\cdot 3 \\ &=& \mathbf{3!} \\\\ a_4 &=& 3!\left(1+\dfrac{2!\cdot 3}{2!}\right) \\ &=& 3!\cdot 4 \\ &=& \mathbf{4!} \\\\ a_5 &=& 4!\left(1+\dfrac{3!\cdot 4}{3!}\right) \\ &=& 4!\cdot 5 \\ &=& \mathbf{5!} \\ \ldots \\ \mathbf{a_n} &=& \mathbf{n!} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline b_n &=& b_{n-1} + \dfrac{ b_{n-1}^2 } { b_{n-2} } \\ \mathbf{ b_n } &=& \mathbf{ b_{n-1} \left( 1 + \dfrac{b_{n-1}}{b_{n-2}} \right) } \\ \hline b_0 &=& 1 \text{ or } b_0 = \dfrac{2!}{2!} \\ b_1 &=& 3 \text{ or } b_1 = \dfrac{3!}{2!} \\ \hline b_2 &=& 3\left(1+\dfrac{3}{1}\right) \\ &=& 3\cdot 4 \\\\ b_3 &=& 3\cdot 4\left(1+\dfrac{\not{3}\cdot 4}{\not{3}}\right) \\ &=& 3\cdot 4\cdot 5 \\\\ b_4 &=& 3\cdot 4\cdot 5\left(1+\dfrac{\not{3}\cdot \not{4}\cdot 5}{\not{3}\cdot\not{4}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6 \\\\ b_5 &=& 3\cdot 4\cdot 5\cdot 6\left(1+\dfrac{\not{3}\cdot \not{4}\cdot \not{5}\cdot 6}{\not{3}\cdot\not{4}\cdot \not{5}}\right) \\ &=& 3\cdot 4\cdot 5\cdot 6\cdot 7 \\ \ldots \\ \mathbf{b_n} &=& \mathbf{\dfrac{(n+2)!}{2!} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{b_n}{a_n} &=& \dfrac{\dfrac{(n+2)!}{2!}} {n!} \\\\ &=& \dfrac{ (n+2)! } {2!n!} \\\\ &=& \dfrac{ n!(n+1)(n+2) } {2!n!} \\\\ &=& \dfrac{ (n+1)(n+2) } {2!} \\\\ \mathbf{ \dfrac{b_n}{a_n} } &=& \mathbf{ \dfrac{(n+1)(n+2)} {2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{b_{32}}{a_{32}} &=& \dfrac{(32+1)(32+2)} {2} \\\\ &=& \dfrac{33\cdot 34}{2} \\\\ &=& 33\cdot 17 \\\\ \mathbf{ \dfrac{b_{32}} {a_{32}} } &=&\mathbf{561} \\ \hline \end{array} \)

Thanks to both of you.

It is good to see you again Will  

You can limit the number of formulas that you need to memorize by knowing and understanding where the formulas come from.

There are still a lot that it is better just to memorize of course but I know a lot less formulas than most other people of my level.

Most times this does not go against me.

For instance, the  distance and midpoint formulas look aweful but if you understand where they come from you do not need to know the formulas at all!

The midpoint formula is just (the average of xs, the average of ys) which is totally logical.

And the distance formula is just pythagoras's theorem.

If 

\(x^2>0.5\\ then\\ x<-\sqrt{0.5}\qquad or \qquad x>\sqrt{0.5}\\ \text{the distance from $\sqrt{0.5}$ to 1 = $1-\sqrt{0.5} $ }\\ \text{the distance from -1 to $-\sqrt{0.5}$ also = $1-\sqrt{0.5} $ }\\~\\ \text{So the given probability will be }\\ \frac{2(1-\sqrt{0.5})}{1--1}=\frac{2(1-\sqrt{0.5})}{2}=1-\sqrt{0.5}\approx 0.29 \)

a - 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, ...
a(n) =1(n - 1) .......This is called "rising factorial"

b - 1, 3, 12, 60, 360, 2520, 20160, 20160, 181440, 1814400, 19958400, 239500800, 3113510400, ...
a(n) =1(n + 1) / 2 ............This is called "rising factorial"

b(32) / a(32) =[1(32+1)! / 2] / 1[32 - 1]! =[33!/2] / 31! = 528

Nothing is mentioned about how many tiles make up the perimeter, where presumablly vertical walls restrict where the coin can land,

so I'll assume the question is equivalent to finding the probability that the coin lands entirely in a single tile.

Let's look at the case where the coin doesn't land entirely in the tile.  The coin center must land in a strip around the edge of the tile of width

r=1cm.  Thus in order to land entirely in the tile the coin center must land in a 10-2=8cm square.

Thus the probability is the ratio of this area over the entire tile area or p = 64/100 = 16/25

There are really too many formulas...

:I 

Try rephrasing your question guest...

Just repost a better version of it...

Substitute in 3 for a   and 1 for b

3(3) + 4(1) = 9+4 = 13

Maybe you mean f(x) = 204 + 3e^(.2x)  ?

204+3*e^(.2*3) = 209.4663564011715267

Sub in the values to the function:

a * 4^2 +5  = a * 6^2 +7

16a+5 = 36a+7

a = - 1/10   = - 0.1

Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c".

\((x-2)^2=4a(y-4)\\ sub\; in\; (1,1)\\ 1=4a*-3\\ a=\frac{-1}{12}\\ so\\ (x-2)^2=\frac{-(y-4)}{3}\\ \)

Now rearrange it to get the form you want.

I checked it with Desmos, which is something you should always do too.

 sqrt(30)+sqrt(50) = 12.5482933869171364     which is between   m and n    12 and 13

Yea sorry

There were 2 answers and they were different.

It often takes less effort to answer a question then it does to check the answers that are already there.

I do appologize though, your answer was a very good one and it was rude of me  to not acknowledge it. 

To be honest I did not even look at your answer, other than to recognise that 30 corresponded to my answer. 

I did give you a point in acknowledgment but I do not consider that that was enough.

All I can do is offer you my appology.

Why don't you join up and then you will get a reputation for being a reliable, good quality, answerer.   

It is worth presevering with. 

Desmos really is not that hard and you can use it to help you learn and to check your answers.

You do have to know what equation to put in of course.

If you want to move a graph left (neg direction) by 3 units you replace x with x+3       (it will move in the neg direction)

If you want to move a graph right (pos direction) by 4 units you replace x with x-4

If you want to move a graph up 5 units you replace y with y-5

If you want to move a graph down 6 units you replace y with y+6

It is that simple!