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When the numbers \($\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[5]{5}$ \) are listed in order from least to greatest, which number is in the middle?

Compare  

√2   ³√3 , ⁵√5       raise each  to the 30th power

(2^(1/2))^30        (3^(1/3))^30    (5^(1/5))^30

2^15                  3^10                  5^6

32768              59049                15625

So

√2    is the middle number

That depends, where to you want to put the brackets?

Chris, DBA is 124^o  not  125^o 

A straight line is 180^o so 180^o – 124^o  =  56^o  =  angle ABC 

A triangle contains 180^o and so add ABC + BAC to see how much you need left over for BCA  

ABC + BAC  =  56^o + 95^o  =  151^o   

Since the whole triangle contains 180^o then BCA has to be 180^o – 151^o  =  29^o 

.

I don't see how this problem is flawed?

The numbers seem to work out.

f(x)  = 4x^4 - 25x^2 -5x - 13

Test 1  using synthetic division

1  [ 4    0    -25     -5        -13 ]

            4       4

    ______________________

      4    4   -21       stop

Test 2

2 [  4   0     -25       - 5    -13  ]

           8     16

    ________________________

      4   8      -9      stop

Test 3

3  [ 4     0       -25       -5       -13  ]

             12      36      33        84

   ___________________________

     4     12      11      28        71            all are positive......3   is the upper bound

What is the value of \(\sum \limits_{n=1}^\infty \left(\tan^{-1}(\sqrt{n})-\tan^{-1}(\sqrt{n+1})\right)\)?

\(I\ assume \ \boxed{\tan^{-1}(x) = \arctan(x)} \)

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^\infty \left(\arctan(\sqrt{n})-\arctan(\sqrt{n+1})\right) \\\\ &=& \overbrace{ \arctan(\sqrt{1})-\arctan(\sqrt{2}) }^{n=1} \\ && + \quad\qquad \qquad \quad \overbrace{ \arctan(\sqrt{2})-\arctan(\sqrt{3}) }^{n=2} \\ && + \qquad\qquad \qquad \qquad \qquad \quad\quad \overbrace{ \arctan(\sqrt{3})-\arctan(\sqrt{4}) }^{n=3} \\ && \cdots \\ && + \overbrace{ \arctan(\sqrt{\infty})-\arctan(\sqrt{\infty+1}) }^{n=\infty} \\ \hline \end{array} \)

\(\text{shorten} \ldots \)

\(\begin{array}{|rcll|} \hline && -\arctan(\sqrt{2})+\arctan(\sqrt{2}) \\ && -\arctan(\sqrt{3})+\arctan(\sqrt{3}) \\ && -\arctan(\sqrt{4})+\arctan(\sqrt{4}) \\ && \cdots \\ && -\arctan(\sqrt{\infty}) + \arctan(\sqrt{\infty}) \\ &=& 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{n=1}^\infty \left(\arctan(\sqrt{n})-\arctan(\sqrt{n+1})\right)} \\\\ &=& \arctan(\sqrt{1})-\arctan(\sqrt{\infty+1}) \\ &=& \arctan(1)-\arctan( \infty ) \\\\ &=& \dfrac{\pi}{4} - \dfrac{\pi}{2} \\\\ &=& -\dfrac{\pi}{4} \\\\ &=& \mathbf{-0.7853981634} \\ \hline \end{array}\)

x=2n, y=12-5n, z=3n-4 for any n

tangent is cyclic

so the value of tangent is cyclic

so its converges to invinity

it converges to cot(1)-cot(sqrt(m+1))

Thanks!

Thank you so much dude 

Question 2

This says to  R1  and from this subtract  (1/2)R2    [and put the result in R2 ]

(1/2) R2  = (1/2)  [  -6   -3   12 ]  =  [ -3  -3/2     6]

Subtract this from R1....so we have

[ 4 - -3      3 - -3/2     0 - 6 ]   =  [ 7    9/2    - 6 ]  

So   the new matrix is

4    3       0

7   9/2    -6

First one says to take 2R2  and subtract R4     [ and I assume we put the result in R4 ]

So   2R2  =    2 [ 0   4 ] =   [ 0   8 ]

And from this subtract R4  =   [ 0 - -3 , 8 -  -4 ]   =  [ 3, 12]

So  the new matrix is

-1   2

0    4

6    3

3   12

Amount of coupon  = $15

The discount  off the price  =  30%......to see this.....the price is .7P  = 70% of the normal price, P...so....the discount must be =  .30   =  30%

1. There are two values of k for which the cubic polynomial 2x^3 - 9x^2 + 12x - k has a double root. What is the sum of those values?

Let  the  double root  be " a "   and the other root be " b"

And from Vieta, we have that

a + a + b =  2a + b   = 9/2    →   b  =  9/2 - 2a      (1)

a^2 + ab + ab  = a^2 + 2ab  =  6       (2)

a^2b =  k/2  →  2a^2b  = k      (3)

Sub (1)  into (2)  and we have that

a^2 + 2a ( 9/2 - 2a)  = 6

a^2 + 9a - 4a^2  = 6

-3a^2 + 9a  - 6  = 0

3a^2 - 9a + 6  = 0

a^2 - 3a + 2  = 0

(a - 2) (a - 1)  = 0

So    a = 2    or     a  =  1

If a = 1, then b =  (9/2) - 2a  =  9/2 -2(1)  =  5/2

And k  = 2(1)^2 (5/2)  =  5

And if a = 2, then b  = (9/2) -2(2)  = (9/2) - 4   =  1/2

And k  =  2(2)^2 ( 1/2)  =  4

So the sum  =  4 + 5  =  9  

oh okay ,thank you for checking that. was question 2 correct?

Ohhhhk....

I am just gonna back my way out of this charging formula....

Good job, tertre   !!!!

Thanks, CU   .....that  cubic "formula" is nasty......!!!!

We have more variables than equations....so.....we will have  infinite solutions

Subtracting the first equation from the second  we get that  3y + 5z  =  16  → 

 3y = [ 16 - 5z]  →     y =  [ 16 - 5z ] / 3 

And using the first equation

x + y + z =  8

x + [ 16 - 5z]/3 + z  = 8       multiply through by 3

3x + 16 -5z + 3z = 24

3x - 2z = 8

3x = 8 - 2z

x = [ 8-2z] /3

So....the solutions for  {x, y , z }  =   {  [8-2z]/3 , [16-5z]/3 , z }

For example...if z = 0,  then x = 8/3  and y = 16/3

So....check that these solutions work :

x + y + z  = 8  ???

8/3 + 16/3 + 0  = 8 ???

24/3 = 8  ???

True

And

x + 4y + 6z = 24  ???

(8/3) + 4 [16/3]  + 0  = 24??

8/3 + 64/3  = 24 ???

72/3  = 24   ???

True

So.....   one possible solution is   { x, y , z}  =  (8/3, 16/3 , 0 }

sorry fixed

I recommend you NOT to use the cubic formula, its kind of long...

3. The fourth-degree polynomial P(x) satisfies P(1)=1, P(2)=2, P(3)=3, P(4)=4, and P(5)=125.What is P(6)?

A  fourth degree polynomial  has the form  ax^4 + bx^3 + cx^2 + dx + e

We have the following equations

a + b + c + d + e  =  1

16a + 8b + 4c + 2d + e  = 2

81a + 27b + 9c + 3d + e  = 3

256a + 64b + 16c + 4d + e  = 4

625a + 125b + 25c + 5d + e  = 125

This system is a little lengthy to solve....Wolfram Alpha gives the following solutions

a = 5  b = -50  c  = 175    d = -249   e  = 120

So.... P(6)  =   5(6)^4 - 50(6)^3 + 175(6)^2 - 249(6) + 120   =  606 

CPhill: Did CU make make a typo here? 1/2x + 1/2y = 180 - 111 = 69 degrees

1/2x + 1/2y =69 

x       +      y  =138

180  -  138   = 42 degrees - angle A ??