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so what is the answer? please

Yah....didn't see the little dot above the 6 ...I am more accustomed to a bar...though I have seen the dot before...

Hooray !......if I ever need to call Germany....  I would likely just use Wi-Fi calling for free...

See below:

Angle AGD is 80 degrees.

In your statement of the problem, f(4) is undefined.

Since angle(BAC) intercepts the same arc as angle(BPC), angle(BAC) = ½x^o.

Adding the interior angles of quadrilateral(ABPC):  20^o + ½x^o + 20^o + (360-x)^o  =  360^o 

Solving, we get x = 80^o.

The possible values of j are -22 and 13.

a + b + c = 14.

There are N numbers in the sum (N+1) + (N + 2) + ... + (2N)

Their sum is:  N( (N + 1) + 2N ) / 2  =  N(3N + 1) / 2

The average is:  [ N(3N + 1) / 2 ] / N  =  (3N + 1) / 2

Since this average is 20:  (3N + 1) / 2  =  20     --->     3N + 1  =  40

                                                                                            3N  =  39

                                                                                              N  =  13

The inominative interest rate rate is  8% p.a   or   0.08% pa

\(\frac{0.08}{12} = 0.00\dot 6\)       per month

q(x) = ax + b

p(x) = cx + d

p(2) = 2c + d = 3

p(q(x)) = c(ax + b) + d = acx + bc + d = 4x + 7

Now we have a system of equations, where we're looking for q(-1) = -a + b.

Can you take it from here?

=^._.^=

(-k)^2  - 4 (k-5)(5)   = 0

k^2 - 20k + +100 = 0

(k-10)(k-10) = 0       k= 10   or  10       (so really, only ONE solution)

   Are they ALSO asking for the solution to the ORIGINAL equation when k = 10  maybe ?????

The solutions are 2 and -5.

There are 320 possible lineups.

I got ten, but it says that there are multiple solutions.

Hey Melody.... I didn't round...... the diff was my mistake  using  Kaylee's 10 year balance for 40 years  ( SHOULD be  39)  ....should have re-read the Q ! 

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

you'll find there are TWO values of 'x'  that satisfy this quadrtatic     x  =  4 +- sqrt(15)   

     sub each of these x values in the   x + 1/x    equation              <=====     Take it from here !

The 'discriminant' is the     b^2 - 4ac      portion of the Quadratic Formula:

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

a = k-5     b = -k       c = 5 

When the discriminat = 0   there is only one solution to the quadratic        

(-k)^2  - 4 (k-5)(5)   = 0      <======== can you take it from here....solve for k ?

Yeah, I came across this problem too, can someone help me?

No that is not why our results differ.    Your r is the same as mine, you, or rather your calculator, has just rounded off r. 

BUT it would not make THAT mcuh difference.  

You have to use 2 formulas for Karlee.

First the Future value of an annuity due (120 months), and then the compound interest formula  (12*39 months)

Hi proyaop,

The general term is

\(\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\ =\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\ =\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\\)

You want the constant term so     

\(\frac{9-3x}{2}=0\\ 9-3x=0\\ 9=3x\\ x=3\)

So the contatn term is

\(=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\ =\displaystyle {\binom{9}{3}*5(x)^{\frac{9-3*3}{2}}}\\~\\ =\binom{9}{3}*5\\~\\ =84*5\\~\\ =420\)

You do need to check my working for careless mistakes of course. 

LaTex:

\displaystyle {\binom{9}{n}(\sqrt x)^{9-n}(\frac{5}{x})^n}\\~\\
=\displaystyle {\binom{9}{n}(x)^{\frac{9-n}{2}}(\frac{5}{x^n})}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-n-2n}{2}}}\\~\\
=\displaystyle {\binom{9}{n}*5(x)^{\frac{9-3n}{2}}}\\~\\

96 people is 40% of the seats because when 96 people leave it changes a difference of 40%. 40% x 2.5 = 100%, so to get the number of seats you just do 96 x 2.5 = 240.

Thus, there are 240 seats in the theater.