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-3.2

say that the blue pieces are 7x and the green pieces are x.  we know that 7x=63, solving that, we get x=9.  so there are 23 red pieces, 63 blue pieces, and 9 green pieces.  adding, we get 23+63+9=95

this has already been answered :))

https://web2.0calc.com/questions/find-the-value-of-cfrac-1-1-cfrac-1-2-cfrac-1-1-cfrac-1-2-dotsb

omg thank you so much for your help!!!!! :D

The point  (a, -4/5)  is  on the  graph of sin x

So

 a  is a third quadrant angle  so sin a =  -4/5   and   cos a =  -3/5

And   (a, c)  is on the graph of  sin (2x)

So

sin (2a)  =  2sin a  cos a  =     2 (-4/5) (-3/5)  = 2 (12/25) = 24/25  = c

Likewise

(b , 5/13)  is on sin x

sin b = 5/13

b is a first quadrant angle so cos b =  12/13

So

sin (2b)  = 2 sin b  cos b  =   2 (5/13) (12/13)  =  120 / 169  =  d

this question has already been answered :)

https://web2.0calc.com/questions/help_81031

yeah!! that's so helpful!! thank you so much i get it now :D

tan (A + B) =   [ tan A + tan B ]  / [ 1 - tan A tanB ] =   [ 5/2 + 1/2] / [ 1 - (5/2)(1/2) ] = 3 / (-1/4)  =  -12

We can solve for tan G thusly :

tan (B + G)  = [ tan B + tan G ] / [ 1 - tanB tanG ] 

     3/2 =   [   1/2 + tan G ] / [ 1 - (1/2) tan G ]

     (3/2)  [ 1 - (1/2) tan G ]   =  1/2 + tan G

      (3/2) - (3/4)tanG = 1/2+ tanG

     (3/2) - (1/2)  = tan G + (3/4)tan G

           1 =   ( 1 + 3/4) tan G

           1 =  (7/4) tan G

            tan G  =  4/7

tan ( A + B + G)  =   [ (tan A + tan B + tan G)  - tanA tanB tan G ]  /  [ 1 - ( tanAtanB + tanAtanG + tanBtanG) ]

                                [ (5/2 + 1/2 + 4/7 )  - (5/2)(1/2)(4/7) ] / [ 1 - ( 5/4 + 20/14 + 4/14) ] =

                                [  (3 + 4/7 ) - ( 20/ 28) ] / [ 1 - ( 5/4 + 12/7 ) ]  =

                                [   25/7 - 5/7 ]  / [  -55/28 ]  =

                                [ 20/7 ] / [ -55/28 ] = 

                                  -(20 * 28) / ( 7 *55)  = 

                                   -20 * 4  / 55  =

                                     -80  / 55  =

                                      -16 / 11

I'm not too sure about the formula for the triple sum, but I think that it's correct !!!

tan (A + B) =   [ tan A + tan B ]  / [ 1 - tan A tanB ] =   [ 5/2 + 1/2] / [ 1 - (5/2)(1/2) ] = 3 / (-1/4)  =  -12

tan (B + G)  = [ tan B + tan G ] / [ 1 - tanB tanG ] 

     3/2 =   [   1/2 + tan G ] / [ 1 - (1/2) tan G ]

     (3/2)  [ 1 - (1/2) tan G ]   =  1/2 + tan G

      (3/2) - (3/4)tanG = 1/2+ tanG

     (3/2) - (1/2)  = tan G + (3/4)tan G

           1 =   ( 1 + 3/4) tan G

           1 =  (7/4) tan G

            tan G  =  4/7

tan ( A + B + G)  =   [ (tan A + tan B + tan G)  - tanA tanB tan G ]  /  [ 1 - ( tanAtanB + tanAtanG + tanBtanG) ]

                                [ (5/2 + 1/2 + 4/7 )  - (5/2)(1/2)(4/7) ] / [ 1 - ( 5/4 + 20/14 + 4/14) ] =

                                [  (3 + 4/7 ) - ( 20/ 28) ] / [ 1 - ( 5/4 + 12/7 ) ]  =

                                [   25/7 - 5/7 ]  / [  -55/28 ]  =

                                [ 20/7 ] / [ -55/28 ] = 

                                  -(20 * 28) / ( 7 *55)  = 

                                   -20 * 4  / 55  =

                                     -80  / 55  =

                                      -16 / 11

Just add the tips  and then divide this sum by 7.....that will give you the mean  (average)

I don't believe that an integer  exists to  make this prime

Simplifying  we have

l 6x^2 - 75x  + 15 l

l 3 ( 2x^2  - 25x + 5) l

3 l 2x^2 -25x + 5 l

The only way this  can  be  prime is  if the expression   in the absolute  value  bars  is  either = 1  or  = -1

2x^2 - 25x + 5  = 1

2x^2  - 25x + 4 =  0 

Using the Q formula  x is   not an integer  because  the discrminant = 25^2 - 4(2)*(4)  = 593 which is  not  a perfect square

Likewise

2x^2 - 25x + 5 =   -1

2x^2 - 25x + 6  =  0

The discriminant = 25^2 - 4* 2 * 5  =   585  which is also not a  perfect square

D =  .75A     →   D / .75 =  A  →  D / (3/4)  = A  →  (4/3) D  = A       (1)

(D + 20)  = (5/8) ( A + 40)          (2)

Sub ( 1)  into (2)

D + 20 =  (5/8) [ (4/3)D  + 40]

D + 20 =  (20/24)D + (5/8) (40)

D + 20  = (5/6)D + 25

D - (5/6)D =  25  -20

(1/6)D =  5

D = 5 ( 6/1)  =  30

David had $30 at first

thank you!!! that was really helpful!

angle BCA =  30°

Using the Law of Sines

sin BCA / AB  = sin ABC / 10

sin (30°) / 4 =  sin ABC /10

(1/2) / 4  = sin ABC /10

(1/8)  = sin ABC  /10

(10/8) = sin ABC

(5/4)  = sin ABC        { impossible}

We have the following equations  where (x,y)  is the  center of the  circle and r^2  = the radius^2

(x + 7)^2  + y^2  =  r^2

(x - 1)^2 + (y - 4)^2 = r^2

(x - 7)^2 + (y - 2)^2  =  r^2

Simplify these

x^2 + 14x + 49 + y^2   = r^2             (1)

x^2  - 2x + 1 + y^2 - 8y + 16 = r^2    (2)

x^2 - 14x + 49 + y^2 - 4y + 4 = r^2    (3)

Subtract (1) from (2)

-16x - 48 -8y + 16  = 0

-16x -8y - 32  = 0         divide through by -8

2x + y + 4  =  0

2x + y  =  -4   →  y = -2x - 4       (4)

Subtract ( 3)  from (2)

12x - 48 - 4y + 12  = 0    divide through by 4

3x - 12 - y + 3   =  0

y = 3x - 9     (5)

Set (4) = (5)

-2x - 4   = 3x  -9

5 = 5x

x = 1

y = 3(1) - 9  =  -6

The center   is  ( 1, -6)

Using (1)

(1 + 7)^2   + (-6)^2  = r^2

8^2  + 36  =  r^2

100 = r^2

The equation for  the circle is

(x - 1)^2  + ( y + 6)^2 =  100

x^2  - 2x + 1  + y^2 + 12y + 36  = 100

x^2 + y^2 - 2x + 12y - 63  =   0

To find the largest real number \( c \) such that \( 1 \) is in the range of \( f(x) = x^2 - 5x + c - 3x + 8 \), we first simplify the function to: \[ f(x) = x^2 - 8x + (c + 8) \] For \( 1 \) to be in the range of \( f(x) \), there must be some value(s) of \( x \) for which \( f(x) = 1 \). This can be written as: \[ x^2 - 8x + (c + 7) = 0 \] For this equation to have at least one real root, the discriminant of the quadratic equation must be non-negative: \[ \Delta = b^2 - 4ac \] \[ \Delta = (-8)^2 - 4 * 1 * (c + 7) \] \[ \Delta = 64 - 4c - 28 \] \[ \Delta = 36 - 4c \] For \( \Delta \) to be non-negative, \( 36 - 4c \geq 0 \), or \( c \leq 9 \). Therefore, the largest real number \( c \) for which \( 1 \) is in the range of \( f(x) \) is c = 9.

For problem 2, the answer is 1 + sqrt(2)*i.

For problem 3, we can use DeMoivre's theorem to simplify (1+sqrt(3i))^6. DeMoivre's theorem states that for any complex number z = r(cosθ + isinθ), where r is the modulus and θ is the argument, we have

(z)^n = r^n (cos nθ + isin nθ)

In this case, z = 1+sqrt(3i), so r = 1 and θ = arctan(sqrt(3)) = pi/3. Plugging these values into DeMoivre's theorem, we get

(1+sqrt(3i))^6 = 1^6 (cos 6θ + isin 6θ) = 1 (cos (6 * pi/3) + isin (6 * pi/3)) = 1 (-1 + 0i) = -1

Therefore, (1+sqrt(3i))^6 = -1.

try punching it into a calculator! :)

\(5c+cd = 5(\frac{1}{5}) + (\frac{1}{5})(15)\)

2^(4728246)  -  1

The smallest prime factors of: 2^(4728246)  -  1 are:

3 x 3 x 7 ==63

OH MY GOD i almost completely forgot about that part.... i kept thinking i was doing something wrong lol

thank you so so so much!!!!

18 P 3 = 18! / 15! = 4896 ways

18 choices for gold    17 choices for silve 16 choices for bronaze =   18 x 17 x 16 = 4896 ways 

Note that the number of Americans is irrelavent.

Find the domain of the function \(f(x) = \sqrt{6-x-x^2-2x^2}\) .

Hello Guest!

\(f(x) = \sqrt{6-x-x^2-2x^2}\\ -3x^2-x+6=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {1 \pm \sqrt{1+4\cdot 3\cdot 6} \over -2\cdot 3}\\ \color{blue}x\in \{-1.591,1.257\}\)

The domain of the function \(f(x) = \sqrt{6-x-x^2-2x^2}\)  goes from -1.591 to 1.257.

  !

Hello heartSTORM907!

There is no result for the distance BC.

\(\overline{BC}\in \{\}\\ \sum(\ \overline{BC}\ )=0\ as\ specified\ in\ the\ task.\)

  !.

Hi Guest,

If they have an arithmetic mean of 18, this means: \(\dfrac{x+y}{2}=18\)

So, \(x+y=36\)

If they have a geometric mean of \(\sqrt{47}\), this means: \(\sqrt{xy}=\sqrt{47}\)

Then, \(xy=47\)

Now, we want to find \(x^2+y^2\)

We can rewrite it as follows: \((x+y)^2-2xy\)  [Check this by expanding it]

So: \(x^2+y^2=(x+y)^2-2xy=(36)^2-2(47)=1296-94=1202\)

I hope this helps.