All Answers 358310 Answers

decibel      db    

Instead of being all wordy:

x = adult tx      revenue from adult tx =    12 x

98-x = senior tx    revenue from senior tx = 8 (98-x)     summed together these equal 1072

12x     +    8 (98-x)  =  1072       solve for x = adult tix     then   98-x = senior tx

If you consider the ellipse whose foci are (4,0) and (8,0) that has major axis endpoints of (0,0) and (12,0), all the points of that ellipse (with the exception of the endpoints of the major axis) will produce a triangle with perimeter 12.

By the definition of what constitutes an ellipse, this ellipse will contan all the points.

There are two isosceles triangles with base (4,0) and (8,0). These have the largest area.

Thanks

please tell me how you got this.

\({-b \pm \sqrt{b^2-4ac} \over 2a} \to \frac{-8\pm \sqrt{8^2-4\cdot \:4\left(-60\right)}}{2\cdot \:4} \to\frac{-8\pm\sqrt{8^2+4\cdot \:4\cdot \:60}}{2\cdot \:4} \to\frac{-8\pm\sqrt{1024}}{2\cdot \:4}\to\frac{-8\pm\sqrt{1024}}{8}\to \frac{-8\pm32}{8}\to\frac{-40}{8}, \frac{24}{8}. \)Thus, the roots are \(\boxed{-5}, \boxed{3}.\) 

I double checked. Your answer is correct! Thanks for the answer.

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Hi there...
 
Problem...:
--------------------
Tickets sold in total = "98"
"Adult-Tickets" = "$12,-" each.
"Senior-Tickets" = "$8,-" each.
--------------------
"How many "Adult-Tickets" were sold?"
--------------------

As a point of refference, here are the extremes...
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((88 "Adult-Tickets")+(2 "Senior-Tickets"))=$1072,-
((131 "Senior-Tickets")+(2 "Adult-Tickets"))=$1072,-
--------------------

I hope that helped a bit?

Kind regards
BizzyX
 
 
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If the probability that I will get at least one of five into the bin is 211/243, 

then the probability that I miss all five is 32/243.

But, 32/243  =  (2/3)⁵.

This means that the probability of missing one is 2/3 and the probability of making one is 1/3.

[miss x miss x miss x miss x miss  =  (2/3) x (2/3) x (2/3) x (2/3) x (2/3)  =  32/243]

So, if you take 6 shots and make at least two:

₆C₂·(2/3)²·(1/3)⁴ + ₆C₃·(2/3)³·(1/3)³ + ₆C₄·(2/3)⁴·(1/3)² + ₆C₅·(2/3)⁵·(1/3)¹ + ₆C₆·(2/3)⁶·(1/3)⁰  = 

All give you a good starting point.

Let  angle b = x

Now you can get every angle there in terms of x.

Start with the most obvious and work from there.

It is not all that hard.

eventually you will have 2 or 3 angles all in terms of x that add to give 180 degrees.

Then you will just have a little equation to solve to get x.

When you finish you should check that it works,

Please no one else give a full answer.

Jenny, you can interact with me if you need/want to.     

If I read you r questin correctly     50 blue beads is 1/3 of the bue beads   so , to start there was 150 blue beads

Blue beads represented 60 % of the original number of beads (because red represented 40%

150 /n  x 100% = 60     n= 250 beads originally

now if she loses 50 beads   she has      200 left

x = stock price at purchase

      down 20 %    means 80% left     .8x    after 20% loss

          then up 30%          (1.3)(.8x) = final value    =1.04 x      final value is UP 4%

That is excellent, if what you say is completely true then you understand, and have memorised, a lot of things now.

Use desmos to look at these 2 graphs, also paste the grapns in your answer (both on the same graph)

y=sinx

y=sin(2x)

What features are the same, what features are different.  What effect does the 2 in front of the x have?

The probablility that I can get a crumpled up piece of paper into the recycling bin is 211/243.  So, if I shoot 6 pieces of paper, then the probablity I can get at least two into the recyccling bin is
\(\left( \frac{211}{243} \right)^6 + 6 \left( \frac{211}{243} \right)^5 \left( \frac{22}{243} \right) + 15 \left( \frac{211}{243} \right)^4 \left( \frac{22}{243} \right)^2 + 25 \left( \frac{211}{243} \right)^3 \left( \frac{22}{243} \right)^3 + 6 \left( \frac{211}{243} \right)^2 \left( \frac{22}{243} \right)^4,\)
because I get two into the recycling bin if I get two, three, four, five or six.  This works out to about 77.91%.

Not POSITIVE, but pretty sure the minimum sampling rate would be    2   x 12 khz = 24 000/sec

put into 512 levels       24000 x 512 = 12.288 Mbs

"In this sense, infinity could be any type of value..."      

Right there is a fundamental error.  Infinity can not be treated as a value. 

Get somebody to play tic tac toe with you.  You'll catch on pretty quick. 

Call the current spped, C

Call the speed of the  boat in still water, R

Effective rate with current  =  R + C

Effective rate against the current  = [ R - C ]

So  we have that  distamce / rate   = time...so...

12/ [ R + C]   = 2             12  /[ R - C]  =  4

12 = 2 [ R + C]                12  = 4 [ R - C]

6  = R + C                        3 = R - C

Add the last two  equations

9 = 2R

4.5 = R

And 

6 = R + C

6 = 4.5 + C

1.5   = C

You can enter the numbers as follows:

1 4 7 10

2 5 8 11

3 6 9 12

Then there are 3! = 6 ways of arranging the numbers in each column, so the number of ways is 6^4 = 1296.

If she picked a blue tile first.....there  are 6 grey ones remaining and  2  blue ones remaining.....so.....

P ( gray  /  blue)  =    6 / (6 + 2)  =  6 / 8   =   3 / 4

Call the windspeed,  W

Call the rate of the plane in still air, R

And

With the wind the effective rate  is R + W

Against the wind the effective rate is R - W

And  we know that   distance / rate   = time  ....so....

2800 / (R + W)  = 5           2800 / (R - W)  = 7

2800 = 5 (R + W)              2800 = 7 (R - W)

2800 / 5  = R + W              2800/7= R - W

560 = R + W                       400 = R - W

Add the last two  equations and we get that

960 = 2R

480  = R

And

560 = R + W

560 = 480 + W

80  = W

Weird, it's not letting me sign in. I apologize that it showed as a guest. I'm working on trying to sign in, but it's fighting me.

Note that

sin (x + pi/6)  =  sin(x)cos(pi/6)  + sin( pi/6) cos (x)

And

sin ( x - pi/6)  = sin (x)cos(pi/6)  -sin (pi/6) cos (x)

So  we have

sin (x + pi/6)  - sin (x - pi/6)  = 1/2 

[ sin(x)cos(pi/6)  + sin( pi/6) cos (x) ]  -  [ sin(x)cos(pi/6)  - sin( pi/6) cos (x) ] =  1/2   

sin(pi/6) cos(x)  + sin (pi/6) cos(x)   = 1/2

(1/2)cos(x)  + (1/2) cos (x)  = 1/2

cos (x)  =  1/2

And this happems at  x =   pi/3    and  x = (5/3)  pi  on  the interval  [0,2pi )