All Answers 358290 Answers

DEG  +  GEF   =  90

2x  +   5x -  22  =  90

7x   =  90  + 22

7x   =  112

x  =   112  /  7   =    16° 

2.

  B                    C

      31         85

  E  47              D

Let   BD  and   CE  intersect  at  P

We  have  triangles  BPC   and EPD

Angle  CDB =  angle  CBP  =  31

Angle  BCE  = angle BCP  =  85

So  angle  BPC =  180  - 31  -   85  =    64  =  angle  EPD

So  in triangle   EPD 

Angle CED  = angle  PED

Angle  BDE =   180  - angle  EPD   -angle  PED

Angle  BDE  = 180   - 64  -  47  =  69°  

These  are  vertical angles......they  are equal....so....

2x  +  7 =   4x  -  14

7 + 14   =  4x  -  2x

21  =  2x

x =   21 / 2   =   10.5°

Thank you for helping me!! I really appreciate it!

:)

Using euclidean's algorithm, we see that

\(\gcd(n^2, n-12)\\=\gcd(n^2-n(n-12), n-12)\\=\gcd(12n, n-12)\\=\gcd(12n-12(n-12), n-12)\\=\gcd(144,n-12)\)

Therefore, 144 must be divisible by n-12 in order for n^2 to be divisible by n-12.

\(144=2^43^2\), so it has (4+1)(2+1) = 15 factors (which are \(2^03^0, 2^13^0, 2^23^0, 2^33^0, 2^43^0, 2^03^1, 2^13^1, 2^23^1, 2^33^1, 2^43^1, 2^03^2, 2^13^2, 2^23^2, 2^33^2, 2^43^2\)), which means that if you add 12 to any of the numbers mentioned above, you will get a number n that satisfies the condition above.

Can you finish it?

2 - 4 - 8

Evaluate the sum \(\dfrac{1}{2} + \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots\)

\(\begin{array}{|rcll|} \hline s &=& \dfrac{1}{2} &+ \dfrac{3}{2^2} + \dfrac{5}{2^3} + \dfrac{7}{2^4} + \ldots \\ s &=& 1*2^{-1} &+ 3*2^{-2} + 5*2^{-3} + 7*2^{-4} + 9*2^{-5} + \ldots \\ 2^{-1}s &=& &1*2^{-2} + 3*2^{-3} + 5*2^{-4} + 7*2^{-5} + \ldots \\ \hline s-2^{-1}s &=& 1*2^{-1} &+ 2 (\underbrace{2^{-2} + 2^{-3} + 2^{-4} + \ldots}_{=2^{-2}*\dfrac{1}{1-2^{-1}} } ) \\ s(1-2^{-1}) &=& 1*2^{-1} &+ 2*2^{-2}*\dfrac{1}{1-2^{-1}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*\dfrac{1}{\dfrac{1}{2}} \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&+\dfrac{1}{2}*2 \\ s*\dfrac{1}{2} &=& \dfrac{1}{2}&*(1+2) \\ s &=& &1+2 \\ \mathbf{s} &=& \mathbf{3} \\ \hline \end{array}\)

//////////// Oops! 

You define the function K(x) (the blue graph) so that it is the mirror image, in the line y=x (the dotted graph), of f(x) as defined on x<=2 (the red graph). All you have to do is change the place of x and y in y = - 3x + 8 and solve for y to get

                                                                                      \(k(x) =-\frac{1}{3}x+\frac{8}{3}\),    x>2.

This has been asked and answered before.  See https://web2.0calc.com/questions/please-help-me-im-so-lost-and-this-is-my-last-question for example.

Search the site for 'marching band' to see other replies.

Let \(S={1\over 2}+{3\over 2^2}+{5\over2^3}+...+{2n-1\over 2^n}\)                   ...(1)

⇒ \({S\over 2}={1\over 2^2}+{3\over 2^3}+{5\over2^4}+...+{2n-2\over 2^n}+{2n-1\over 2^{n+1}}\)         ...(2)

Subtracting eq(2) from (1), 

\({S\over2} = {1\over 2}+{2\over 2^2}+{2\over 2^3}+{2\over 2^4}+...+{1\over 2^n}-{2n-1\over 2^{n+1}}\)

     \(={1\over 2}+({1\over 2}+{1\over 4}+{1\over 8}+....{1\over 2^n})-{2n-1\over 2^{n+1}}\)

     \(={1\over2}+[{{1\over2}(1-{1\over2^n})\over 1-{1\over2}}]-{2n-1\over 2^{n+1}}\)

     \(={1\over2}+1-{1\over 2^n}-{2n-1\over2^{n+1}}\)

     \(={3\over 2}-[{2+2n-1\over 2^{n+1}}]\)

\({S\over 2} = {3\over 2}-{2n+1\over 2^{n+1}}\)

⇒\(S = 3-{2n+1\over 2^n}\)

sin  (3pi/2  - a)  =    sin(3pi/2)cosa  + cos(3pi/2)sina  =  -cosa

So

sin⁴  ( 3pi/2  -a)  =   (-cos a)^4  =   (cos a)^4

sin ( 3pi  + a)   =   sin (3pi) cos a   -  cos(3pi)sina   =   sina

So

sin ⁴ ( 3pi  + a)   =  ( sin a)^4

sin ( pi/2 + a)  =  sin (pi/2)cos a   -  sin a cos (pi/2)   =  cos a

So

sin ⁶  (pi/2  + a)    =   (cos a)^6

sin (5pi  -a)  =  sin (5pi)cosa   + sin a cos (5pi)  =  -sina

So

sin ⁶ ( 5pi -a)   =  (-sin a)^6  =   (sin a)^6

So

3   [  (cos a)^4   +  (sin a)^4  ]  -  2  [ (cos a)^6  + (sin a)^6  ]   =

Special  identites

cos ^4 a    =    (3 + 4cos (2a) + cos (4a))  / 8

sin^4  a   =     (3  - 4cos (2a)  + cos (4a) )  / 8

So.....the first expression   evaluates  to

(3/8)  ( 6  + 2cos (4a) )  = 

(9/4) + (3/4)cos(4a)

Also

cos^6 a   =  (1/32)  ( 15cos (2a) + 6cos (4a) + cos (6a) + 10 )

sin ^6 a    = (1/32)   (-15cos (2a)  + 6cos (4a)   - cos (6a) + 10 )

So......the  second expression  evaluates  to  

-(2/32)   ( 12cos (4a)  +  20)  =

-(1/16)   ( 12cos (4a) +   20)  =

-(3/4)cos (4a)  -  5/4

So

[ (9/4)  + (3/4)cos (4a)  ]  -  [ (3/4)cos (4a)  + 5/4)  ]   =

9/4    -  5/4   =

4/4    = 

1

I did not know that! Thank you for letting me know!

Sequence A

2, 4, 8, 16, 32, 64, 128, 256

Sequence B

10, 30, 50, 70, 90, 110, 130, 150, 170, 190, 210, 230, 250, 270, 290,

130 - 128 = 2 is the closest we can get. 

=^._.^=

y + x + y - x = 180

2x = y - x

Can you take it from here?

=^._.^=

7y + 5y = 180

5y + 3y + x = 180

Can you take it from here?

=^._.^=

a no like boi, do your own work. lmao  

Posting too many questions isn't allowed. You know it won't be answered

If 3x=k (mod 9) has a solution, then 3x - k = 9m, for some integer m. That is k = 3x - 9m = 3(x-3m), which implies that k is a multiple of 3. But k is the remainder upon division of any of the numbers in the set {3, 6, 9, 12, 15} by 9, so k =0, 3, 6 are the only values k can take, if the given equation has a solution. But in general division of a integer by nine can result in 0, 1, 2, 3, ..., 8 left over as remainders; so 1, 2, 4, 5, 7, 8 are the values k can take if the equation has no solutions.

Oh okay Ginger I guess I might've overthought it again... thanks for clarifying 

 f(f(f(f(1+i))))

f(1+i) = 2i

f(2i) = -4

f(-4) = -8

f(-8) = -16

=^._.^=

You are correct.....

they could be separated like this : 

 |     |     |     |     |     |     |     |     |     |   

T               T               T         T                   

                 or other ways.....   let me think about how to derive the answer for a little while......If I come up with a solution I will let ya know !

THANX !  ~ EP

Hi! That sounds like a great idea!