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If  9x^2+24x+a is the square of a binomial, then the binomial has the form  for some number , because . So, we compare  to . Expanding  gives 

(3x+b)^2=(3x)^2+2(3x)(b)+b^2=9x+6bx+b^2

Equating the linear term of this to the linear term of 9x^2+24x+a, we have 6bx=24x, so b=4. Equating the constant term of 9x^2+6bx+b^2 to that of 9x^2+24x+a gives us a=b^2=16.

1 : 2.414        is the ratio of their radii

Ratio:>>         A_insc  :  A_sec  = 11 : 16   or  11/16

Thank you, Melody!

Thanks very much Heureka :)

As follows:

I'm sure you can calculate A+B+C

A circle is inscribed in a quarter-circle sector. It is tangent to the arc of the sector and the two perpendicular radii of the sector. What is the ratio of the area of the inscribed circle to the area of the sector?

Hello Guest!

\(a=r+r\sqrt{2}\\ a=r(1+\sqrt{2})\)          Radius of the sector

\(A_{sec}=\pi r^2(1+\sqrt{2})^2/4\)

\(A_{insc}=\pi r^2\)

\(\frac{A_{insc}}{A_{sec}}=\frac{4}{(1+\sqrt{2})^2}\)

\(\frac{A_{insc}}{A_{sec}}=\frac{2}{1+\sqrt{2}}=1:2,414\)

  !

Simpify \(\dfrac{\sin 25^\circ \sin 35^\circ \sin 85^\circ}{\sin 75^\circ}\)

Formula:

\(\begin{array}{|rcll|} \hline \sin(x)\sin(y) &=& \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\\\ \sin(x)\cos(y) &=& \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\\\ \cos(60^\circ) &=& \dfrac12 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{\sin(25^\circ) \sin(35^\circ) \sin(85^\circ)}{\sin(75^\circ)} } \\\\ && \qquad \boxed{ \sin(x)\sin(y) = \dfrac{1}{2}\Big( \cos(x-y)-\cos(x+y) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(35^\circ-25^\circ)-\cos(35^\circ+25^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\cos(60^\circ) \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\Big( \cos(10^\circ)-\dfrac{1}{2} \Big) \\ \sin(35^\circ)\sin(25^\circ) = \dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4} } \\\\ &=& \dfrac{\Big(\dfrac{1}{2}\cos(10^\circ)-\dfrac{1}{4}\Big) \sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{2}*\cos(10^\circ)\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ && \qquad \boxed{ \sin(x)\cos(y) = \dfrac{1}{2}\Big( \sin(x-y)+\sin(x+y) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(85^\circ-10^\circ)+\sin(85^\circ+10^\circ) \Big) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(95^\circ) \Big) \quad | \quad \sin(95^\circ)=\sin(180^\circ-95^\circ)=\sin(85^\circ) \\ \sin(85^\circ)\cos(10^\circ) = \dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big) } \\\\ &=& \dfrac{\dfrac{1}{2}*\dfrac{1}{2}\Big( \sin(75^\circ)+\sin(85^\circ) \Big)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)+\dfrac{1}{4}*\sin(85^\circ)-\dfrac{1}{4}*\sin(85^\circ)}{\sin(75^\circ)} \\\\ &=& \dfrac{\dfrac{1}{4}*\sin(75^\circ)}{\sin(75^\circ)} \\\\ &=& \mathbf{\dfrac{1}{4}} \\ \hline \end{array}\)

A + B + C = 4 + 3 + (-2) = 5.

-5x -49 >= 113      add 49 to both sides of the equation

-5x >= 162             multiply both sides by -1     >>>>>>>> this reverses the > sign !!!!

5x <= -162           divide by 5

x <= - 162/5

x = 4y-3     from second equation    sub this in for 'x' in the first equation

5 (4y-3) + 2y = -15

20y - 15 + 2y  = -15

y = 0      then     x = -3

Find the radius of the circle.....  then the sqaure has sides = 2 * r     and you can calculate the area

2x^2 = -2y^2 + 12x - 4y + 20

2x^2 - 12x    + 2y^2 + 4y  = 20

x^2 - 6x + y^2 + 2y = 10

(x-3)^2    + (y-1)^2   =  10 + 9 + 1       20 = r^2 of the circle    

Square   2*r  x  2*r =   4 r^2       4(20) = 80  units²  

There's not enough information!!! 

1st    Any circle with a diameter of 24 units or longer can have a chord of 24 units.

2nd   We don't know for a fact that a diameter of a larger circle is 4X the diameter of a smaller circle.

13/x = x/3

39 = x^2

x = sqrt39

Hmmm... it seams that your LaTeX has not loaded correctly. You dont need the dollar signs.

Anyways, I am assuming that you mean \(\sqrt{a}\cdot\sqrt{a+6}\cdot\sqrt{b}\cdot\sqrt{b+6}\), and I also assume that you mean a=7, and b=91, right? From there, all you have to do is plug in 7 for a, and 91 for b. This will give \(\sqrt{7}\cdot\sqrt{13}\cdot\sqrt{91}\cdot\sqrt{97}\) , which is \(\sqrt{803257}\). Sadly, this is not a nice number, but we end up with \(91\sqrt{97}\)

ah right, oops! Clear triangles always confuse me...

3 miles / 10 m/hr = .3 h   for car    = 18 minutes

3 m / 65 m/hr  x 60 min/hr   + 5 x 3  min    =   17.78 min by train

7/20 x 100% = 35 %

THANK YOU!

Let x =  .3636....

100x = 36 . 3636

100x - x  = 99x = 36 .

     x = 36/99 = 4/11          4 + 11 = 15

You will have to re-fuel each car  FOUR times

Car A    100 miles / 185 m/hr   = 1945.94 seconds   + 4 x 30 = 2065.95 seconds to finish

Car B     100 miles/ 200 m/hr  =   1800  seconds      + 4 x 90 = 2160 seconds to finish   

      Which car would you like?    

I guess it's r =? 

(1 + r)² = (2 - r)² + 1²  

We know angles A and C are right angles=90 degrees. However, we don't know that angles B and D are 90 degrees. They could very well be 89 and 91 degrees!
Therefore, it is prudent to extend the diagonal from B to D. Now, we can use Pythagoras's Theorem on triangle BCD. BD^2 =5^2 + 4^2 =41 and BD =Sqrt(41)
Again, we can use Pythagoras's Theorem on triangle BAD. Sqrt(41)^2 - sqrt(5)^2 =AD^2. AD^2 =41 - 5 =36.
AD = 6

Here is the graph:

You've got it all wrong! 

Two right triangles are  ABD  and  BCD

First, find a diagonal BD             BD = sqrt( 4² + 5² )

Then, using BD and AB find side AD       AD = sqrt( BD² - AB² )