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There's a shortcut:::

Important:   The ratio of  QO  to  NO is  2 : 1        ( The Median Theorem )

NR = 1/2 * 14 = 7            ( QN ⊥ PR )

NO = 1/3 * 12 = 4

OR = sqrt (7² + 4²) = √65 

Plane B flies   30 min + 10 min     + 20 km    @ 240 km/60 min

40 min * 240 km/60min + 20 km = 180 km           .....as guest found ! 

If plane A flew 30 minutes, plane B also flew 30 minutes.

30min = 0.5h

240km/h * 0.5h = 120km

10min = 1/6h

240km/h * 1/6h = 40km

120km + 40km + 20km = 180km

Please proof read your question....it really states that the more your income, the more your take home pay.  

80km

The greatest common factor is 11:

when you alternatedly add and subtract the digits, you will always end up with 0, which is why 11 is the GCF.

A Simple question! Give each child 5 /4 of a muffin!

Hint:   connect points T, A, and U to the center of a circle. 

You should be able to solve it now.

To do this, we see that 2^20 is also known as 20 2's. So, we start from 1, and keep on going until we get a total of more than 20 2's:

Number.         How may 2s

1                    0

2                    1

3                    0

4                   2

5                    0

6                  1

7                   0

8                  3

9                   0

10                1

11                  0

12                 2

13                0

...

keep on going until the sum of the collumn "How many 2s" is greater than 20!

You could also use an alternate solution by using the diagonal formula for a rhombus area. That formula is multiply the diagonals together and divide by 2.

The great thing about this problem is that the angles are 120, 120, 60, 60, which means that you can divide this rhombus into two equilateral triangles. 

Using the area formula for euqilateral triangles ,\(\frac{\sqrt{3}}{4} a^{2}\), we see that  the areas of each triangle is \(9\sqrt 3\). Multiply that by 2, and you get your answer: \(\boxed{18\sqrt 3}\)

a-b = 40     so   b = a-40

then     a      +   a-40     + 50      = 180        solve for 'a'

Similar triangles:

QD = 19.2/32  * 15 

then

QD + 15 = BD

Add 5 to both sides

2|x-3| = 12

|x-3| =6

Now you have to evaluate    x-3 = 6       and   x-3 = -6

                                                 x = 9                x = -3           Just as W1000 found !

Top part  :    5 x 10 x 12  cm³

Bottom part      5 x 12 x10 cm³         Add together......

In this problem, the value of the product wxyz is not given, so I don't think I can solve it in the same way. I'm also more interested in how to solve the problem than just the answers. Thanks for linking the other question and for your help - I enjoyed reading heureka's solution!

QN = 12                PR = 14       PN = 7

F is a foot of the altitude from M to PN

Find angle QPN using QN and PN

Find QP and MP using the Pythagorean theorem

Find MF and PF using angle QPN and MP

Find angle MRF using MF and FR

And finally, find  OR = 5*sqrt(2).

For this, we can use half angle formulas. Since 45 degrees is a common angle on the unit circle, we can easily see that it's sine and cosine is \(\frac{\sqrt 2}{2}\). Using the half angle formula for cosines, which is: \(\cos \frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}\), we can get our answer. 

Here is what you should do:

Plug in \(\frac{\sqrt 2}{2}\) for cos A and solve.

Good luck! 

If ATQ = 41, and AUR = 63, then find QXR, in degrees.

Sorry, I realized I can't use the angle sign...

Wow! Great job on my challenge! That is correct, and your way to do solve it was perfect!

Not sure if I'm correct, but here goes:

2|x-3|=12

|x-3|=6

x = 9 or -3

Use tangent to find angle AQB 

90^o – AQB = AQR 

CQR = AQR 

90^o – CQR = CQD  

Use tangent to find length of QD  

BD = 15 + QD  

_.

Because if I don;t repost than the question gets drowned out by other questions and nobdy answers them.