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lets call the ribeye R and the salmon S. (R is the # of ribeyes. btw, what is a ribeye?)

We have the equations 11R + 18S = 554.59 and 30R + 9S = 582.58. We can multiply the second equation by 2 to get 60R + 18S = 1165.16.

We can form the system of equations. 

          60R + 18S = 1165.16

    -     11R + 18S = 554.59

_______________________

          49R           =  610.57     --> R = 12.46.

Now susbtitute R = 12.46 into the first equation (11R + 18S = 554.59). We have 11(12.46) + 18S = 554.59 --> 18S = 417.53 --> S = 23.2. 

The ribeye costs $12.46 and the salmon costs $23.20. 

- PM

\(A=\{\text{exactly 3 oranges}\}\\ B=\{\text{exactly 6 apples}\}\\ P[A \cup B] = P[A]+P[B] - P[A \cap B]\)

\(\text{The total number of possible arrangements is found from Stars and Bars }n=8,~k=3\\ N=\dbinom{n+k-1}{k-1} = \dbinom{8+3-1}{3-1} = 45\)

\( \text{To determine }P[A] \text{ we note that first we pick 3 oranges}\\ \text{We then fill the remaining 5 fruits from oranges or bananas.}\\ \text{There are }\dbinom{5+2-1}{2-1} = 6 \text{ ways of doing this, so}\\ P[A]=\dfrac{6}{45} = \dfrac{2}{15}\)

\(\text{Likewise }\\ P[B] = \dfrac{\dbinom{2+2-1}{2-1}}{45}=\dfrac{1}{15}\)

\(\text{Finally we see that }A \cap B = \emptyset \text{ so } P[A \cap B] = 0\\ P[A \cup B] = \dfrac{2}{15}+\dfrac{1}{15} = \dfrac{3}{15} = \dfrac{1}{5}\)

I agree with you.  yes it has helped thanks 

it is either b or c due to the fact that (0, 3) is on the graph, and since the frequency is 3, the period is 1/3, and since it is 2pi/(the coefficient of x inside the sin function), then only \(\boxed{B}\) satiesfies it, because the coefficient is 6pi, and you get 2pi/6pi or 1/3. 

HOPE THIS HELPED!

due to the symmetry of the problem the probability that Bo rolls greater than Amaya equals the probability that Amaya rolls greater than Bo.

The probability that they roll the same is 6/36=1/6

So the probability that they do not roll the same is 1-1/6 = 5/6

And thus the probability Amaya rolls greater than Bo is half of this or (5/6)/2 = 5/12

15 / 36   I think   5/12

What might be going on is that 3 of the axes of symmetry do not intersect any of the dots.

The website in question might not consider these axes of symmetry.

140 x 3 = 420

log_1/2 (3x + 1)^1/3 =  - 2

In exponential form we have

(1/2)^-2  =  (3x + 1)^1/3

2²   = (3x + 1)^1/3

4  =   (3x + 1)^1/3     cube both sides

64 =   3x + 1        subtract 1 from both sides

63 =  3x        divide both sides by 3

21 = x

The sum of the elements in any row of Pascal's Triangle  = 2^n.....where n is the row number

So....the base 10  log of 2^n   = n log₁₀ 2

So

f(n) / log₁₀ 2   =

n log₁₀ 2 / log₁₀ 2  =

n

Yes I have a lot of fun using Geogebra too. 

I am still finding new functions in there. It can really do a lot of things. 

For anyone that is interested,it is a free download.

Notice that (2,3) will be the mid-point of both diameter endpoints

So..let (x, y) be the other diameter endpoint....and we have

(x   +  - 1) / 2 = 2             and        (y + -1) / 2 = 3

          multiply both sides by 2

x - 1     = 4                                 y - 1   =  6

                      add 1 to both sides

x =  5                                        y  =  7

So the other endpoint  is   (5, 7)

Cross-multiply and we get that

2xr - 2x^2 = 7     rearrange as

2x^2 - 2rx + 7 =  0

Single real solutions will occur when the dicriminant = 0

So

(2r)^2 - ( 4) (7) (2)  = 0

4r^2  -  56 = 0     

4r^2 =  56    divide both sides by 4

r^2 = 14      take both roots

r =  ±√14

And the product of these r's  =   (√14) ( - √14)  =    -14

OK....no prob...!!!

THANK YOU SO MUCH :)

26 possible letters  x  10 possible digits  =

26 x 10   =

260 possible plates ....  

There isn't anything to  solve.....we just want to evaluate this :   tan (acos(3/x))

Let acos(3/x) = theta

So.....theta is an angle whose cosine = 3 / x

So.... using the  Pythagorean Theorem......y =  sqrt (x^2 - 3^2) = sqrt (x^2 - 9)

So..... tan (acos(3/x)) =  tan (theta)  =  y / x    =   sqrt (x^2 - 9) / 3

Thanks, Melody...!!!!

.....it took me long enough to make it in GeoGebra......this utility is still kinda' funky to use...IMHO....

Nice, asdf    >>>>>!!!!!

Just a rough pic for reference......definitely not exact!!!!

                     A

     E                            B

  D

                        C                                     

Since ABC = 128°.......then major arc CA is twice 128° = 256°

Therefore.......arc ABC =  360° - 256° = 104°

But.....since AB = BC....then each of these chords must span an arc of 52°

And since  AB = BC = CD = DE.....these equal chords each span an arc of 52°

So....arc EA  must be   360 - 4(52)  = 152°

And angle ACE is an inscrbed angle subtending EA....so it's measure is (1/2) measure arc EA = 76°

i dont know how to solve this, but i think a way to make it easier is to draw all the diagonals