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Slope of (2x + 3y = 4k) = -2/3

Slope of (x - 2ky = 7) = -1/(-2k) = 1/2k

\(\dfrac{1}{2k} = \dfrac{-2}{3}\\ -4k = 3\\ k = \dfrac{-3}{4}\)

Consider the slope:

\(\dfrac{5-3}{t-0} = \dfrac{3-0}{0-(-8)}\\ 16 = 3t\\ t = \dfrac{16}{3}\)

she has been walking 22 min

.....

use this link to find the answer

Thank you Travisio

Maya has $15 to spend. She spent $5.50, including tax, to buy a notebook

$9.50 left

15-5.50=19/2=9.5

9.50-7.75=7/4=$1.75

She can spen $1 and 75¢ on the cookies

1.75/.25=7

She can buy up to 7 packs of cookies and still have $7 and 75¢ left

Travisio

I got a riddle 

I'm a number thats odd

But I'm a number of three

I'm more than 20

I'm less that 27

And half of one 

And the other

I'm stuck in the middle

What am I?

\(Travisio\)

GA, you were raised in a cathloic enviroment?

I would agreee that waiting is the best option unless there is something wrong with your connection mostly because (If it's sayign that) your not reciving  the iinformation formt he server back to you or AKA the requested infromation to download. I'm not informed on solving this sort of techinacla diffuclty so I would go with waitng some sites will have a server error for a degree of time longest i have ever had it was about a 2 - 3 days.

Thanks! Maybe I'll do that

I don't really know but maybe you can try again in a day or 2 and see if it makes any difference.

Maybe someone else has got a better suggestion....

Thank you very much, Melody ;P

So we know that our given varible is B and that it eqaul's 3, so we substatuite that into our equation 

\((3 +7)({3}^{2}-10)\)

Two thing's we want to remember is PEMDAS and that (3 + 7) and (3² - 10) are to multiply on one another we can tell this because both equations are next to each other to indicate multiplication there closed in parenthasses because there both seperate values and they want you to solve for each value before multiplying. 

So \((3 + 7) = 3 + 7 = 10\)

we end up with a sum of   \((10)\) 

For the second part: \(({3}^{2}-10)\)

We need to remeebr that we need to take care of exponet's first in pemdas, because it is the first one avaible the parenthassis just symbolizes that this is one value. 

So 3²  =  9 and 9 - 10  =  -1 

Our final equation should look like this: \((10)(-1)\)and we know this mean's 10 * -1 and all you have to do is 

1. Multiply the factor's. 

2. Get your product. 

3. Dispaly your product as your answer.

2) 

In a certain sequence the first term is \(a_1= 2007\) and the second term is \(a_2 = 2008\).

Furthermore, the values of the remaining terms are chosen so that \( a_n + a_{n + 1} + a_{n + 2} = n\) for all \(n \ge 1\).

Determine \(a_{1000}\).

\(\begin{array}{|lrcll|} \hline (1) & a_n + a_{n + 1} + a_{n + 2} &=& n \\ (2) & a_{n+1} + a_{n + 2} + a_{n + 3} &=& n+1 \\ \hline (2)-(1): & a_{n+1} + a_{n + 2} + a_{n + 3}-(a_n + a_{n + 1} + a_{n + 2}) &=& n+1-n \\ & a_{n+1} + a_{n + 2} + a_{n + 3}-a_n - a_{n + 1} - a_{n + 2} &=& 1 \\ & a_{n + 3}-a_n &=& 1 \\ & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & \mathbf{a_{n + 3}} &=& \mathbf{1+a_n} \\ \hline n= 1: & a_{1 + 3}=\mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \hline n= 4: & a_{4 + 3}=a_{7} &=& 1+a_4 \\ & &=& 1+( 1+a_1) \\ & \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \hline n= 7: & a_{7 + 3}=a_{10} &=& 1+a_7 \\ & &=& 1+( 2+a_1) \\ & \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \hline n= 10: & a_{10 + 3}=a_{13} &=& 1+a_{10} \\ & &=& 1+( 3+a_1) \\ & \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \hline \ldots \\ \hline \end{array}\)

generalized:

\(\begin{array}{|rcll|} \hline \mathbf{a_{4}} &=& \mathbf{1+a_1} \\ \mathbf{ a_{7}} &=& \mathbf{2+a_1} \\ \mathbf{ a_{10}} &=& \mathbf{3+a_1} \\ \mathbf{ a_{13}} &=& \mathbf{4+a_1} \\ \ldots \\ \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline \end{array} \)

\(\mathbf{a_{1000} =\ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{ a_{1+3m}} &=& \mathbf{m+a_1} \\ \hline 1+3m &=& 1000 \\ 3m &=& 999 \\ m &=& 333 \\ a_{1000} &=& 333+a_1 \quad | \quad a_1 =2007 \\ a_{1000} &=& 333 + 2007 \\ \mathbf{a_{1000}} &=& \mathbf{2340} \\ \hline \end{array}\)

 Put 3 in there where B is

B+7   would be   3+7 = 10

B^2-10  would be 3^2-10 =-1     10 * -1 = -10

1)
Find the sum of the infinite series
\(1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots\).

\(\text{Let $r=\dfrac{1}{1998} $}\)

\(\begin{array}{|rcll|} \hline s&=& \mathbf{1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots} \\ s&=& \mathbf{1+2r +3r^2+4r^3+\cdots + nr^{n-1} + \ldots} \\ \hline \end{array} \)

\(\begin{array}{|rcrl|} \hline s_n &=& 1+&2r +3r^2+4r^3+\cdots + nr^{n-1} \quad |\quad \cdot r \\ rs_n &=& & 1r +2r^2+3r^3+\cdots + (n-1)r^{n-1}+nr^n \\ \hline s_n-rs_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} -nr^n \\ \hline \end{array}\\ \begin{array}{|rclrcrl|} \hline s_n(1-r) &=& \underbrace{1+r+r^2+r^3+\ldots + r^{n-1}}_{=S_n(\text{ geometric progression})} -nr^n \\ s_n(1-r) &=& S_n -nr^n & S_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} \quad | \quad \cdot r\\ & & & rS_n &=& &r+r^2+r^3+\ldots + r^{n-1}+r^n \\ \hline & & & S_n-rS_n &=& 1- &r^n \\ & & & S_n(1-r) &=& 1- &r^n \\ & & & S_n &=& \dfrac{1- r^n}{1-r} \\ s_n(1-r) &=& \dfrac{1- r^n}{1-r} -nr^n \\ s_n &=& \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ s &=& \lim \limits_{n\to \infty} s_n \\ s &=& \lim \limits_{n\to \infty} \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ && \boxed{\lim \limits_{n\to \infty} r^n = \lim \limits_{n\to \infty} \left(\dfrac{1}{1998}\right)^n = 0} \\ s &=& \dfrac{1- 0}{(1-r)^2} - 0 \\ s &=& \dfrac{1}{(1-r)^2} \\ s &=& \dfrac{1}{\left(1-\dfrac{1}{1998}\right)^2} \\ s &=& \left(\dfrac{1998}{1997}\right)^2 \\ \mathbf{s} &=& \mathbf{1.00100175300507095\ldots...} \\ \hline \end{array}\)

Yes. Well, Mr. BB, I am a chimp for all seasons, including the cold, cold winter.  

BTW, how’s GetSmart doing?  The last I heard from him, he said he will be volunteering himself for a Christmas-time turkey dinner.  Is he going to do that, or are you just going to eat him yourself?

I hope not. He is the only genetically enhanced turkey in the world (that I know of).  

What is the sum of the following sequence:
\(1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000\) .

Using the hockey stick identity:

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& 120 + 720 + 2510 + 6720 + 15120 + 30240 + 55440 +\ldots + 990034950024000 \\ &=& 120 +6*120 + 21*120 + 56*120 + 126*120 + 252*120 + 462*120 +\ldots + 8250291250200*120 \\ &=& 120*(\underbrace{1+ 6 + 21 + 56 + 126 + 252 + 462 +\ldots + 8250291250200}_{\text{long stick of the hockey in Pascals triangle}} ) \\ &=& 120*\left(\dbinom{5}{5}+\dbinom{6}{5}+\dbinom{7}{5}+\dbinom{8}{5}+\dbinom{9}{5}+\dbinom{10}{5}+\dbinom{11}{5}+\ldots +\dbinom{1000}{5} \right) \\ &=& 120*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\ &=& 120* \dbinom{1001}{6} \\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array} \)

See the answer here: https://artofproblemsolving.com/community/c2393h1037309s3_stamp_and_sheet_problem

See the answer here:  https://web2.0calc.com/questions/please-explain-thoroughly-so-i-could-understand-the-procedures

Very nice presentation EmeraldWonder. 

If E is the middle point of the side CA of the triangle ABC and
if R is the area of the triangle, prove that:

\(\cot(AEB) = \dfrac{ BC^2 - BA^2 } { 4R } \)

\(\text{Let $\angle AEB= \epsilon$} \\ \text{Let $AE=EC$} \)

\(\begin{array}{|rcll|} \hline BC^2 &=& AE^2 + BE^2 - 2\cdot AE\cdot BE \cdot \cos(180^\circ-\epsilon ) \\ &=& AE^2 + BE^2 + 2\cdot AE\cdot BE \cdot \cos(\epsilon ) \\\\ BA^2 &=& EC^2 + BE^2 - 2\cdot EC\cdot BE \cdot \cos(\epsilon ) \\\\ BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad & | \quad EC=AE \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline R &=& R_1 + R_2 \quad & | \quad R_1 = \text{area}~\triangle AEB,\quad R_2 = \text{area}~\triangle ECB \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{EC\cdot BE \sin(180^\circ-\epsilon) }{2} \quad & | \quad EC=AE \\ &=& \dfrac{AE\cdot BE\cdot \sin(\epsilon) }{2} + \dfrac{AE\cdot BE \sin(\epsilon) }{2} \\ &=& AE\cdot BE\cdot \sin(\epsilon) \\\\ AE\cdot BE &=& \dfrac{R}{\sin(\epsilon)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline BC^2 - BA^2 &=& 4\cdot AE\cdot BE \cdot \cos(\epsilon ) \quad | \quad AE\cdot BE = \dfrac{R}{\sin(\epsilon)} \\ BC^2 - BA^2 &=& 4\cdot \dfrac{R}{\sin(\epsilon)} \cdot \cos(\epsilon ) \\ BC^2 - BA^2 &=& 4R \cdot \cot(\epsilon ) \\\\ \mathbf{\cot(\epsilon )} &=& \mathbf{\dfrac{BC^2 - BA^2}{4R}} \\ \hline \end{array} \)

If E is the middle point of the side CA of the triangle ABC and if R is the area of the triangle, prove that:

Cot AEB = (BC2 - BA2) / 4R  

Let

x=AB

y=CB

w=EB

z=AE=EC

angle AEB= theta

angle CEB= 180-theta

\(R=0.5zwsin\theta+0.5zwsin(180-\theta)\\ R=0.5zwsin\theta+0.5zwsin(\theta)\\ R=zwsin\theta\\ 4R=4zwsin\theta\\ \)

\(y^2=w^2+z^2-2wzcos(180-\theta)\\ y^2=w^2+z^2+2wzcos(\theta)\\ x^2=w^2+z^2-2wzcos(\theta)\\ subtracting\\ y^2-x^2=4wzcos(\theta)\\ BC^2-BA^2=4wzcos(\theta)\\ \)

\(\therefore\\ RHS =\frac{BC^2-BA^2}{4R}\\ RHS =\frac{4wzcos(\theta)}{4wzsin(\theta)}\\ RHS =cot(\theta)\\ RHS=LHS \qquad QED\)

Ok for some reason they all work now :D