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"The sum of the first n positive even integers is 12432. compute the value of n."

https://web2.0calc.com/questions/help_41375 

this link will help :).

2x^2 + 2y^2 + 4x + 12y + 6 = 0     Divide by 2

x^2 + 2x       + y^2 + 6y    = -6        'Complete the square' for x and y

(x+1)²           + (y+3)²       = -6 + 2 + 9  = 5              5 is radius²      radius = sqrt 5

                          center is (-1,-3)

Floor means the greatest integer smaller that sqrt(23), we know sqrt(16) = 4 and sqrt(25) = 5, so the number is between 4 and 5. So the greatest integer smaller than sqrt(23) is 4. 

I hope this helped. :))

=^._.^=

Using  the Remainder Theorem, this  means  that

A (5)^21  + B(5)^17  + C(5)^9   +  3   =  -2 

A(5)^21  +  B (5)^17  + C(5)^9   =   -5

And  when divided  by x  +  5 ,  this means that

A(-5)^21  + B(-5)^17  + C (-5)^9  +  3   =

-A (5)^21  -  B(5)^17  - C(5)^9  +  3     =

-   ( A(5)^21  + B(5)^17  + C(5)^9 )  +  3   =

-  ( -5)   +    3    =

5  +  3     =

8

Let multiplicative factor be x Then amount of walnuts to buy = 1(x), or x Amount of almonds to buy = 2(x), or 2x Amount of raisins to buy = 3(x), or 3x Cost of x lbs of walnuts: 12x Cost of 2x lbs of almonds: 2x(9), or 18x Cost of 3x lbs of raisins: 3x(5), or 15x We then have: 12x + 18x + 15x = 15 45x = 15 x, or multiplicative factor 15/45 , or 1/3 Amount of walnuts to be purchased: x, or lb Amount of almonds to be purchased: 2x, or 2/3  lb Amount of raisins to be purchased: 3x, or , or 1 lb Amount of trail mix that can be made with $15 2 or lbs

  Slope will be the derivative of the parabola at any given point....or the slope of the tangent line at that point....= 0 for this parabloa.

The slope  at the vertex of the parabola (which is what you are seeking) = 0 

...of course there are numerous solutions for just ONE of the substances equations......but we are trying to mix substance A B and C to get the final substance of  10% A  50% B  and 40 % C     .....I don't believe there is a way to mix all three to get the final ....

   In other words....you only solved ONE of the equations......we need solve the SYSTEM of equations to find x y and z  that solves all THREE of the equations......

4.795831...

The  bounds on integration   are  from  x   -1  to  x =  1     (these are  the x intersections of the  graphs)

The base  of  each  triangle  =  ( 1 - x^2) =  the side of each triangle

The   height of  each  triangle   =  side *  sqrt (3) / 2

Thus.......the  area  for  each triangle   =  (1/2) base * ( height)   =

(1/2) ( 1  - x^2) (1  - x^2) *sqrt (3)  /2  = 

(sqrt (3)  / 4)  ( 1  - x^2) ( 1  - x^2)  =   (sqrt (3)/ 4)   ( 1  - x^2)^2    =  (sqrt (3)  / 4) (x^4  - 2x^2  + 1)

And summing the  areas of  all the  triangles  from x  = -1   to  x = 1  will  give us the volume of S

Since  the  volume is symmetric   on each side of  the  y axis,  we can  actually  just  integrate  this

                           1

(2 * sqrt (3)  / 4)  ∫     x^4  -  2x^2   +  1    dx   =

                           0

                                                             1

(sqrt (3) / 2 )  [ x^5 / 5   -  (2/3)x^3 +  x ]       =

                                                             0

(sqrt (3)   / 2 )    [  1/5 - 2/3  +  1  ]  = 

(sqrt (3) / 2)  [ 6/5 - 2/3 ] =

(sqrt (3)   / 2 )  [  18 - 10]   /15   =

(sqrt (3)   / 2  )  [ 8  / 15]  =

(4/15)sqrt (3)   ≈  .4619 units ^3   =  volume of  S

                    0

\(11*101+110 = 1221\)

The  number  of possible sets  that could  be drawn  from choosing any  5  of  the  10 numbers =    10 C 5    =  252

And  the  number  of possible  sets  drawn from ( 4,5,6,7,8,9)  = 6C5  =  6

So

P( that all  5  numbers  drawn are > 3)  =   6 / 252   =   1 /  42

Excellent, Melody   !!!!

Hi OldTimer, it is nice to see you again.

I got the same as you until the end bit.

For stationary points

\( \frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{n+1}{n}-1) }\\ \frac{2}{n}x^{(\frac{2}{n}-1)}=\frac{n+1}{n}x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1)}=(n+1)x^{(\frac{1}{n}) }\\ 2x^{(\frac{2}{n}-1-\frac{1}{n})}=(n+1) \\ x^{(\frac{1-n}{n})}=\frac{n+1}{2} \\ x=\left(\frac{n+1}{2}\right)^{\frac{n}{1-n}} \\\)

You should do more to show it is a maximum and not some other kind of stationary point though.

This desmos graph shows that it is correct:

x^2/49   +  y^2 / b^2    =    1

The   center is  (0, 0)

c   =  distance  from  the  center  to either focal point  =  5

a  = sqrt (49)   = 7  =  length of semi-major axis

b  =  sqrt (a^2   - c^2 )   =   sqrt ( 7^2   - 5^2 )   = sqrt ( 49  -  25 )  =  sqrt  (24)   =  2sqrt (6)

Equation is

x^2/49   + y^2/24     = 1

See the graph  here  :  https://www.desmos.com/calculator/3l1f0moddl

See my answer  here, WDIPT  :   https://web2.0calc.com/questions/help-now_15

The first few  terms are

1,1/2,1/3, 1/4, etc.

The  common difference is      

1/ (n + 1)    - 1 / n  = 

(n  -  (n + 1))                 -1

___________   =   ________

   n ( n + 1)               n^2  + n

a_n+1  =    1  / n    -     1  / ( n^2 + n)     =     ( n^2  + n  -  n)  / ( n * (n^2 +n))   = 

 n^2 / ( n^2 (n + 1))  = 

1/ n + 1

I can confirm that it is correct!

The forumla to find the sum of the geometric series is...

A/(1-r)

where A is the first term...

r = the common ratio

you can plus everything in and you'll get.....

2/(1-3/5)

2/2/5

2 x 5/2

10/2

5

The sum of the infinite geometric series is 5

p.s correct me if im wrong

Hi,

The graphs of x^2-2x+y^2+10y=0. Find the radius of this circle.

We can start this off by completing the square to get it into the general formula,

[(x-1)^2-1] + [(y+5)^2-25] = 0

(x-1)^2 + (y+5)^2 = 26

the formula for the graph of a circle is...

(x-h)^2 + (x-k)^2 = r^2

r^2 = radius^2

r^2 = 26

r = sqrt(26)

p.s plz tell me if I did anything wrong. 

One of the roots of 2x^3-9x^2+13x+k=0 is \(x_1=2\).Find the other 2 roots. Explain how you got the answer. 

Hello whatdoiputhere!

\((2x ^ 3-9x ^ 2 + 13x + k) \)  \(:(x-2)=\)  \(2x^2-5x+3\)

 \(\underline{2x^3-4x^2}\)

          \(-5x^2+13x\)

          \(\underline{-5x^2+10x}\)

                           \(3x+k\)

                           \(\underline{3x-6}\)

                                     \(0\)

\(k=-6\)

\(2x^2-5x+3=0\\ x^2-2.5x+1.5=0\\ x=1.25\pm \sqrt{1.25^2-1.5}\\ \color{blue}x=1.25\pm0.25\)

\(x\in\{1,1.5,2\}\)

  !

C

All the answers simplify to x ≥ 2, x ≠ 3 besides C. C is not possible because if a number is less than 2, it cannot be greater than 3.