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Simplify as

3a + 4b = 7

3a + 4b  =  k

Identical equations produce infinite  solutions, so  k   = 7

3 / (4 + 2 + 3 + 7)  * 240  =

(3 / 16)  * 240  =

3 * (240 / 16)   =

3 * 15  = 

45 blue

The area of parallelogram  MBND  = (1/2)AB * BC = (1/2)(1)(1)  = 1/2

The shaded area is (1/2) of this =  (1/2)(1/2)  = 1/4

We set the 2 colors as x and y. You can have xxx,yyy,xxy,yyx. Notice that all other combinations of xxy and yyx can be repeated by just rotating the triangle.

The final answer will be \(\fbox{4}\)

x = 2,4,6,8,10

If x = 10,

u,v,w,y,z = 0

1 CHOICE

If x = 8,

     If there is one 2:

     5 CHOICE

     If there is two 1:

     10 CHOICE

15 CHOICE

You keep doing this for all possible combinations, for example x=6, you can do 1111,112,22,13,4

This at first feels like it will be a really nasty equation, but lets just start writing it out.

Suppose the original fraction is \(\frac{a+1}{a}\). The next one is \(\frac{a+2}{a+1}\). The third one is \(\frac{a+3}{a+2}\). Multiplying these, we get \(\frac{a+1}{a}*\frac{a+2}{a+1}*\frac{a+3}{a+2}=\frac{a+3}{a}\). We see these cancel out!

This equals five, so \(\frac{a+3}{a}=5\), 5a = a+3, 4a=3. a = 3/4. The original fraction written is \(\frac{\frac{3}{4}+1}{\frac{3}{4}}=\frac{7}{3}\). BUT NOTE!! A better answer might be better LEAVING IT IN THE FORM \(\frac{\frac{3}{4}+1}{\frac{3}{4}} \) or \(\frac{\frac{7}{4}}{\frac{3}{4}}\), BECAUSE \(\frac{7+1}{3+1} \neq \frac{\frac{7}{4}+1}{\frac{3}{4}+1}\).

Here : https://web2.0calc.com/questions/help-with-counting_71

This will have a domain of  all real numbers if  the discriminant of the quadratic in the  denominator is < 0

So

b^2  - 4*18 < 0

b^2  - 72 < 0

b^2  < 72

b =  8

I'm afraid that's not coreect. 

This is pretty simple partial fraction decomposition. 

Our first indication that partial fraction decomposition works is \((x-2)(x+1)={x}^{2}-x-2\)

\(\frac{x+17}{{x}^{2}-x-2}=\frac{A}{x-2}+\frac{B}{x+1}\), and we make the denominators equal, \(\frac{x+17}{{x}^{2}-x-2}=\frac{A(x+1)}{(x+1)(x-2)}+\frac{B(x-2)}{(x+1)(x-2)}\), and \(x+17=A(x+1)+B(x-2)\). To solve, we plug in some special values, like 2, so 19 = 3A, A = 19/3. Similarly, by plugging in -1, we get B = -16/3. So out answer is A = 19/3, B = -16/3.\(\)

No. of possible identifiable sequences  =   (No. of rolls)!  / ( multiple occurences of any number(s)) ! =

  7!  / 2!  =  2520

The answer  lies  in the  solution to this : https://web2.0calc.com/questions/angle-bisectors_36

The math books can occupy  any of  4 positions

And for each of these positions they can be arranged in 4! = 24 ways

And for each of these the psychology books can be arranged in 3! = 6 ways

So

4 * 24 * 3  =  288 arrangements

The first ball doesn't matter, because no matter what ball they draw, the only condition is the next one needs to be the opposite color

After Rainas draws the first one, there is 11 balls, 6 of the opposite color, so the probability the next ball is the opposite color is \(\frac{6}{11}\). If this happens, there is now 10 balls, 5 of each color, so now there is a \(\frac{1}{2}\) chance of drawing a different colored ball.

So the probability is \(\frac{6}{11}*\frac{1}{2}=\frac{3}{11}\).

Suppose point A is the first point choosen. To satisfy the contition, the second point has to be on the major arc of BB^'\(\). But the question now is how do we find the value of \(\angle{BO{B}^{'}}\)? Well, we know \(\angle{BO{B}^{'}}=\angle{AO{B}^{'}}*2\). There are many ways to find angle AOB^', but I choose to use law of cosines. I won't get into the details, but

 \(\cos(\angle{AO{B}^{'}})=\frac{1+1-\frac{9}{4}}{2}=-\frac{1}{8}\), so \(\angle{AO{B}^{'}}=\arccos(-\frac{1}{8})\), and \(\angle{BO{B}^{'}}=2*\arccos(-\frac{1}{8})\)\(\).

Supposing you are calculating in radians, the probability is \(\frac{2*\arccos(-\frac{1}{8})}{2\pi}\), or \(\frac{\arccos(-\frac{1}{8})}{\pi}\), which is approximately 0.53989, or 53.989%.

We see there are quadratic terms and linear terms, so Cauchy Schwarz is perfect!

Applying Cauchy Schwarz inequatlity, we get,

\(({x}^{2}+3{y}^2)(1+\frac{1}{3})\ge{(x+y)}^{2}\). Substituting our values in,

\(18*\frac{4}{3}\ge{(x+y)}^{2}\), or \(24\ge{(x+y)}^{2}\), and because x and y are nonnegative, or \(x+y\ge0\), we have \(\sqrt{24} \ge x+y\).

But we need to satisfy the cauchy schwarz equality condition, to know for sure this can be achieved. This condition is, satisfied when x = r, \(\frac{\sqrt{3}}{3}r=\sqrt{3}y\), or \(y=\frac{r}{3}\), so \({r}^{2}+\frac{{r}^{2}}{3}=18\), or \({r}^{2}=27/2, r = \frac{3\sqrt{6}}{2}\), this puts x and y as \(x = \frac{3\sqrt{6}}{2}, y=\frac{\sqrt{6}}{2}\). So we know that this solution works, so we know the maximum value is \(\sqrt{24}\).\(\)

Let r be the radius of the circumcircles of triangles ABX and ACX. Since the circumcircles have the same radius, we know ∠A=∠C.

Since AX=13 and CX=4, we can use the Law of Sines on triangle ACX to find AC:

\begin{align*} \frac{\sin C}{AC} &= \frac{\sin A}{AX} \\

\sin C &= \frac{AC}{AX} \cdot \sin A \\

\sin C &= \frac{AC}{13} \cdot \sin C \\

AC &= 13 \end{align*}

Similarly, using the Law of Sines on triangle ABX to find AB:

\begin{align*} \frac{\sin B}{AB} &= \frac{\sin A}{AX} \\ \sin B &= \frac{AB}{AX} \cdot \sin A \\ \sin B &= \frac{AB}{13} \cdot \sin A \\ AB &= 13 \end{align*}

Now that we know all side lengths of triangle ABC, we can use Heron's formula to find the area:

\begin{align*} \text{Area of } ABC &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\ &= \sqrt{(13+13+4)/2 \cdot (13+4-13)/2 \cdot (13-4-10)/2 \cdot (4-10-13)/2} \\ &= \sqrt{30 \cdot 0 \cdot -3 \cdot -17} \\ &= \boxed{91} \end{align*}

                C

            6

       Y              6

    12

A              Z                B    

              12

Because BY is a  bisector

BC / CY  = AY / AB

BC / 6 = 12 /12

BC  = 6

And  because CZ is a  bisector

AZ / BZ  = AC / BC

AZ / BZ = 18 / 6

AZ / BZ  = 3

AZ = 3BZ

So     

BZ + AZ   = AB

BZ + 3BZ = 12

4BZ  =12

BZ =  3

xy  = 2/5

y = 2 / (5x)

6x + 3 / (2 / (5x) )  = 

6x +  (15/2)x  = 

(27/2) x

This has no min

x^2 -14x  + y^2 + 16y + 26

Take the derivative with respect to x and  set to 0

2x - 14   = 0

2x = 14

x = 7

Take the derivative with respect to y and  set  to 0

2y + 16 = 0

2y = -16

y = -8

The min is    (7)^2  - 14(7) + (-8)^2 + 16(-8) + 26  =  -87

OP^2         =       AP^2

x^2 + y^2   = (x + 2)^2  + (y -1)^2

x^2 + y^2 = x^2 + 4x + 4  + y^2  - 2y + 1

2y - 1  = 4x + 4

2y    = 4x + 5

y = 2x + 5/2         (1)

OP^2  =    BP^2

x^2 + y*2  =   (x + 3)^2  + (y - 2)^2

x^2 + y^2   =  x^2 + 6x + 9  + y^2  - 4y + 4

4y - 4 = 6x + 9    (2)

Sub (1)  into (2)

4 (2x + 5/2) - 4 =  6x + 9

8x + 6   = 6x + 9

2x = 3

x = 3/2

y  = 2(3/2) + 5/2   =   11/2

Coordinates of center =  (3/2 , 11/2)