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I just looked through the options and only one worked.

A+B<6

so   1<=A<=4           1<=B<=4         2<=C <= 5

I really do not know but this is my best guess.

the variance is down the bottom of the 2nd diagram.

oh yea  Sorry   

https://web2.0calc.com/questions/geometry_41255

It's been answered.

Angle A = 60 degrees

If

$x = \sqrt{2006}$ and $y = \sqrt{2007}$,
what is the simplified value of

$(x + y)^2 + (x - y)^2$?

$\begin{array}{|rcll|} \hline (x + y)^2 + (x - y)^2 &=& x^2+2xy+y^2+x^2-2xy+y^2 \\ &=& x^2+y^2+x^2+y^2 \\ &=& 2x^2+2y^2 \\ &=& 2(x^2+y^2) \quad | \quad x^2=2006,~y^2=2007 \\ &=& 2(2006+2007) \\ &=& 2*2013 \\ \mathbf{ (x + y)^2 + (x - y)^2 } &=& \mathbf{4026} \\ \hline \end{array}$

LCM of 12 and 16 is 48   

What is the number of degrees

in the acute angle formed by the hands of a clock at 6:44?

Formula: $\boxed{\Delta\alpha^\circ = 330 * t^h}$

$\begin{array}{|rcll|} \hline \Delta\alpha^\circ &=& 330 * t^h \\\\ \Delta\alpha^\circ &=& 330 * (6+\dfrac{44}{60}) \\\\ \Delta\alpha^\circ &=& 330 * 6.7333333333^h \\ \Delta\alpha^\circ &=& 2222^\circ \\ \Delta\alpha^\circ &=& 62^\circ +6*360^\circ \\ \mathbf{\Delta\alpha^\circ} &=& \mathbf{62^\circ} \\ \hline \end{array}$

3 g / cm^3   *  1 kg / 1000 g  *   100* 100 * 100 cm^3 / m^3   =

3 * 1 000 000 / 1000 =  3000 kg / m³

A = 1, B = 2, and C = 2, so B + C = 4.

Angle A = 120 degrees.

Since it's an isoceles right triangle, the altitude to the hypotenuse is exactly half of the length of the hypotenuse.

The area of a triangle is bh/2, so it would be 4*8/2=16 

L1    3x+4y = - 14    is tha same as     y = - 3/4 x - 14/4       slope, m = - 3/4      perpindicular slope is  - 1/m = 4/3

Your new line L2  is thus   y = 4/3 x + b    sub in the point given to calculate 'b'

                                         7 = 4/3 (-5) + b

                                              so b = 13 2/3

y = 4/3 x + 13 2/3          Guest messed up with the slope ...it is not negative in the perpindicular line....

                   you can finish......   

Yep... Melody is behind her time on this one....   LOOOOOL

Agee!   catmg method  was pretty slick !  

Melody....   I summed  the   6 oclock time + 44 minutes for the hour hand....

You will need a calculator  ...here are three ways to find ?

sin = opp/hyp        arc sin ( 20/52)= ....... degrees

or

cos = adj/hyp         arc cos (48/52) = .......    degrees

or

tan = opp/adj        arc tan (20/48) = .......   degrees

8 out of 4 ball is in left side

Now dividing 50 by (8+4) = 12

We get 4.166, Taking only integer part 4

In 4 turns, he will put 32 balls in right side , 16 balls in left side

In fifth turn , he has 2 ball only

Which he will again put in right box

Therefore total balls in right box = 32+2 = 34 balls

But at 6:44 the hour hand is not at 6 so this not an exact answer.

That is a nice easy way too EP Thanks for showing us.

Your way is better than doing them all individually like in my demo. 

Personally, I like catmgs way best but it is really good to see lots of alternative methods.

Thank you @ElectricPavlov!

From the 12:00 position the hour hand moves   1/2 degree per minute

   the minute hand moves  6 degrees per minute

at 6 oclock + 44 minutes  the hour hand is at 180 degrees + 44 (1/2) = 202 degrees

   the minute hand at :44 is at     44 * 6 = 264 degrees

264 - 202 = 62 degrees between them

$\frac{9!}{2!3!4!}$

A  X  X  X  X       Four ways for B to be to the right

X A X X X           Three ways

X X A X X            Two ways

X X X A X              One way     total of 10 ways     

                                      the other three slots can be filled in   3 * 2 *1   ways  = 6 ways for EACH of the 10 possibles

      10 * 6 = 60 ways