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Hint: out of 6 total cards, 3 of them are red. For everyone to get at least one red card, what has to happen?

--> Make a list of what you know and see if it matches the 40% clue given in the problem.

Whoops, I didn't notice that. Thanks for catching it!

Correcting the mistake I think it would be \(a=\sqrt[3]{-8}\)

Ohh that was helpful. Thank you!

That's a good step, but the problem with your reasoning is that -7 by itself doesn't have a^3 + 7 in the denominator, so you would have to convert -7 to (-7(a^3 + 7))/(a^3 + 7) before solving.

Let us write this in a slightly different  manner  noting that y/y   = 1    and y not equal to  0

y / 3   +  y / y    =  (y + 3)   /  y

Now

y / 3     =   (y + 3) / y    -  y/y

y / 3    =     ( y + 3 - y)  / y

y / 3  =   3 / y            cross-multiply

y^2  = 9      take both roots

y = +3   or   -3

Hope that helps  !!!

Set the first two equations equal

ax + 3   =  x + 5

Let x  = 2

2a  + 3  =   2 + 5

2a  + 3   = 7

2a = 7 - 3

2a  = 4

a = 4/2  = 2

Set the second and third equations equal

x + 5  =  8x + b

Let x  = - 2

-2 + 5  =  8(-2) + b

3  = -16  + b

b = 16 + 3  = 19

So   the sytem that makes these continuous is

y= 2x + 3     ( x > 2)

y = x + 5       [ -2, 2 ]

y  = 8x  + 19    ( x < -2 )

a + b =   2 + 19   =  21

See the graph here : https://www.desmos.com/calculator/ywwaz4mykn

I won't mess with your expertise :)

but I understood the problem a little differently than how I think you did...?

No prob  !!!

x/ a  + y/b = 1

Multiply both sides by ab

xb   +  ya   =  ab

Since this passes through   (1,11)  and ( 4,8)  we have this system

  1b +11a  = ab     ⇒  -4b -44a  = -4ab   (1)

  4b  + 8a   =  ab            (2)

Add (1)  and (2)

-36a   =  -3ab          since a  cannot =  0  ......divide by a

-36  = -3b

-36/ -3  =  b  = 12

Using (2)

4(12)  +  8a  =  a (12)

48  +  8a  =  12a

48  = 12a - 8a   

48 = 4a

48/4 = a  = 12

16/4 = a  =  4

Thanks for your help!

Hint: in any one of the equations, manipulate it to leave an equation that gives one variable in terms of another. Plug this into a different equation. What do you get?

For example, since p+r=8, we know that p=8-r (subtract r from both sides). Plugging this in to p+2q=0 gives 8-r+2q=0.

Doing this will eventually eliminate the variables so you're left with a simpler system.

In each of the equations, manipulate it to get just one variable on one side. Plug the expression back into the expressions.

For example, we know that p/q=5/6. Multiplying boths ides by q, this gives \(p=\frac{5q}{6}\). Plugging this back into \(\frac{p-q}{q-r}\) will give \(\frac{\frac{5q}{6}-q}{q-r}.\)

I think there might be a slight mistake above

I think we want  A  = 3L

So

mn ( m^2 - n^2 )   =  3  [ 2m (m + n) ]

mn ( m + n) ( m - n)  =  6m ( m + n)              since ( m + n)  > 0    we have that

mn ( m - n)  =  6m

n ( m - n)  =  6

(m - n)  = 6 / n

Possibilities  (with a little help from WolframAlpha )

m    n                     2mn         m^2 - n^2        m^2 + n^2          Perimeter   Area

5     2                       20              21                   29                     70            210     Yes     

5    3                        30              16                   34                     80            240     Yes

7    1                        14              48                   50                    112           336     Yes   

7    6                        84              13                   85                    182           546     Yes

Added areas    =  1332

Pythagorean triples are the right angled triangles with integer leg lengths:

\( \frac{1}{(t-1)^2-(t+1)^2}\)

Let's  simplify  the denominator   as    [ t^2 - 2t + 1 ]  - [ t^2 + 2t + 1 ]   =   -4t

Note  that we can't have  0   as the denomiator, so  the only value  that  makes 4t   =  0    is when t = 0

Then  the domain for t is   ( -infinity , 0)  U  ( 0  , infinity )

"The resulting product is less than 500."

Let a and b represent the two primes chosen. The inequality would be \(ab<500\).

You may have already figured that one out, but try using the same method to come up with an inequality for the first part.

Ohh haha I see that now... thanks for correcting me!

Thanks, dolphinia !!!!!

Hint: we already know one of the side lengths of the triangle; it's 8. Since it's an equilateral traingle, all of the sides have length 8. Use this along with the area of a triangle formula.