amygdaleon305

I think there might be an error in the question... that its \({VK \over VJ}=7\) in the second line pls note. 

Now, let's assume this is the line       

                                        ____•__________•_________•____

                                       J       V                   W                U       K

and, \({UJ \over UK}=7\)

⇒     \(UJ=7UK\)

       \({VK \over VJ}=7\)

⇒    \(VK = 7VJ\) 

Now, \(JK=VJ+VK\)                 And, \(JK= UJ+UK\)

                 \(=8VJ\)                                            \(=8UK\)

From this we infer that... \(VJ=UK\)

Also it is given that, W is midpoint of UV

\(VW=UW\)

∴ \({WJ \over WK} = {VW+VJ \over UW+UK}\)

            \(={VW+VJ \over VW+VJ}\)

            \(=1\)

The required ratio is 1. 

No prob! :)

You're very welcome! :D 

Given set of equations are 

           \(y=x^2-3x+4\)                       ...(1)

and     \(x+y=4\)     ⇒ \(y=4-x\)        ...(2)

From eq. (1) and (2)

∴ \(x^2-3x+4=4-x\)

   \(x^2-2x=0\)

   \(x(x-2)=0\)

   \(x=0\)  or  \(2\) 

For x=0 in eq (1)                     For x=2 in eq(1)

⇒ y = 4                                   ⇒ y = 2

∴ The solutions are (0,4) and (2,2). 

~Looking forward 

Equation is   \(x+2={\sqrt{3x+10}}\)

a) First of all, squaring both sides

\((x+2)^2=3x+10\)

\(x^2+4x+4=3x+10\)

\(x^2+x-6=0\)

⇒ \(x^2+3x-2x-6=0\)

\((x+3)(x-2)=0\)

∴ x = -3 or 2

b) Here we can't have \({\sqrt{3x+10}}<0\)

∵ Any value  \(x<{-10 \over 3}\)  is extraneous. 

∴ Squaring both sides in  \({\sqrt {3x+10}}=x\)

\(3x+10=x^2\)

\(x^2-3x-10=0\)

⇒ \((x-5)(x+2)=0 \)

\(x=5\) or \(-2\)

1st case : Putting x=5 in eq.                          2nd case: Putting x=-2 in eq. 

\({\sqrt{3(5)+10}} = 5\)                                       \({\sqrt{3(-2)+10}}=(-2)^2\)

⇒ \(5=5\)                                                     ⇒ \(2≠-2\)

∴ x = 5 is the extraneous solution. 

c) An extraneous solution is a solution that's evolved from the process of solving of the equation but its not a valid solution to the equation. You know, they are those "extra answers" that we get while solving the problem and they may not necessarily be valid

~Roll the credits 

Equation of hyperbola is 

     \({(x-h)^2 \over a^2}-{(y-k)^2 \over b^2} = 1\)

∴ Center of hyperbola is (h,k)

Now, from figure 

(h,k) = (5,4) 

⇒ h = 5

⇒ k = 4

a = 3 

b = 2 

∴ \(h+k+a+b=5+4+3+2\)

                             \(=14\)

~ Hope this helps :) 

See Beto's equation...   \(3m=3m+5\)

                                    ⇒\(3m-3m=5\)

But                                  \(0≠5\)

Now Mark's equation...   \(6-n=-n+2\)

                                   ⇒\(-n+n=2-6\)

But                                   \(0≠-4\)

∴Beto's and Mark's equations have no solutions. 

Let the common ratio be \(x\) then, 

No. of white beads = \(3x\)

No. of black beads = \(5x\)

Now, according to question

⇒\({3x+22 \over 5x}={4 \over 3}\)

⇒\(9x+66=20x\)

⇒\(11x=66\)

⇒\(x=6\)

∴ There were 5*6 = 30 black beads.

Modulus = \({ \sqrt{2^2+6^2}}\)

              = \({\sqrt{4+36}}\)

              = \({ \sqrt{40}}\)

              = \({2 \sqrt{10}}\)