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sumfor(n, 0, 511,2^(n + 2)* (n + 2) = 2.745919064 E+157

Note: Each term of this sequence can be represented by this "closed form": 2^(n + 2)* (n + 2). I don't know how to maniplulate it algebraically to find "n", but  maybe Alan, heureka, Max, Melody and others can sum it up as geometric sequence. The first term is: (2^(n+1))*(n+1) and the common ratio:  ((2 *(n + 2))/(n + 1)) 

Why don't you list all the ones that you can find and invite people to find more?

OH wow thank you!! I understand now becuase of your note at the top so I can rember to use that in similar problems. Thank you so much!

I got the same answer as you did.    

x-2, x, x+2

\((x+2)^2=x^2+60\\ x^2+4x+4=x^2+60\\ 4x+4=60\\ x+1=15\\ x=14\)

so like guest said,  the numbers are  12,14, and 16

Convert r = 2 / (7sinθ-cosθ) to rectangular form.

Enter your answer in slope-intercept form

Yes it is so

I do not like this question much but I think this is what is meant.

-5*5     *      -4*4       .....      -1*1

-25      *         -16    *   (by extension)   -9*-4*-1     =    - 25*16*9*4*1  =   -14400

If you had looked properly at guest's answer maybe you could have found their little error.   

7x^2+27xy-4y^2​

7*-4=-28

I need to numbers that add to 27 and mult to -28     

They are 28 and -1

\(7x^2+28xy-1xy-4y^2​\\ =7x(x+4y)-y(x+4y)\\ =(7x-y)(x+4y)\)

If

\(f(c)=\dfrac{3}{2c-3}\), find \(\dfrac{kn^2}{lm} \)

when \(f^{-1}(c)\times c \times f(c)\) equals the simplified fraction \(\dfrac{kc+l}{mc+n}\),
where \(k\), \(l\), \(m\), and \(n\) are integers.

My attempt:

\(\begin{array}{|rcll|} \hline f(c) &=& \dfrac{3}{2c-3} \quad | \quad c=f^{-1}(c) \\\\ f\Big(f^{-1}(c)\Big) &=& \dfrac{3}{2f^{-1}(c)-3} \quad | \quad f\Big(f^{-1}(c)\Big) = c \\\\ c &=& \dfrac{3}{2f^{-1}(c)-3} \\\\ c\Big(2f^{-1}(c)-3\Big) &=& 3 \\\\ 2cf^{-1}(c)-3c &=& 3 \\\\ 2cf^{-1}(c) &=& 3c+3 \\\\ cf^{-1}(c) &=& \dfrac{3c+3}{2} \\\\ \mathbf{f^{-1}(c)\times c} &=& \mathbf{\dfrac{3c+3}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline f^{-1}(c)\times c \times f(c) &=& \dfrac{(3c+3)}{2}\times \dfrac{3}{(2c-3)} \\\\ f^{-1}(c)\times c \times f(c) &=& \dfrac{3(3c+3)}{2(2c-3)} \\\\ \mathbf{f^{-1}(c)\times c \times f(c)} &=& \mathbf{\dfrac{9c+9}{4c-6}} \\ \hline && \boxed{k=9,\ l=9,\ m=4,\ n=-6} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{9(-6)^2}{9*4} \\\\ \dfrac{kn^2}{lm} &=& \dfrac{36}{4} \\\\ \mathbf{\dfrac{kn^2}{lm}} &=& \mathbf{9} \\ \hline \end{array}\)

The larger of two numbers is 33 more than the smaller.
When added together, the sum of the larger number and five times the smaller number is 129.
What are the two numbers?

\(\text{Let large number $=l$} \\ \text{Let small number $=s$}\)

\(\begin{array}{|lrcll|} \hline (1) & l &=& s +33 \\ (2) & l+5s &=& 129 \\ & s+33 + 5s &=& 129 \\ & 6s &=& 129 - 33 \\ & 6s &=& 96 \quad |\quad : 6 \\ & \mathbf{s} &=& \mathbf{16} \\\\ & l &=& s+33\\ & l &=& 16+33 \\ & \mathbf{l} &=& \mathbf{49} \\ \hline \end{array}\)

The two numbers are 16 and 49

Compute 

\((1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \dotsm (1 + \tan 45^\circ)\)

Formula:

\(\begin{array}{|rcll|} \hline \tan(x-y) &=& \dfrac{\tan x-\tan y}{1+\tan x \tan y }\\\\ \tan(45^\circ) &=& 1 \\ \hline \end{array}\)

\(\small{ \begin{array}{|rcll|} \hline && \mathbf{(1 + \tan 1^\circ)(1 + \tan 2^\circ)(1 + \tan 3^\circ) \dotsm (1 + \tan 22^\circ)(1 + \tan 23^\circ) \dotsm (1 + \tan 43^\circ)(1 + \tan 44^\circ)(1 + \tan 45^\circ)} \\\\ &=& (1 + \tan 1^\circ)(1 + \tan 44^\circ) \times (1 + \tan 2^\circ)(1 + \tan 43^\circ) \times (1 + \tan 3^\circ)(1 + \tan 42^\circ) \\ && \vdots \\ && \times (1 + \tan 22^\circ)(1 + \tan 23^\circ) \times (1 + \tan 45^\circ) \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \tan (45^\circ-1^\circ)\Big) \times (1 + \tan 2^\circ)\Big(1 + \tan (45^\circ-2^\circ)\Big) \times (1 + \tan 3^\circ)\Big(1 + \tan (45^\circ-3^\circ)\Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \tan (45^\circ-22^\circ)\Big) \times (1 + \tan 45^\circ) \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 1^\circ} {1+\tan 45^\circ\tan 1^\circ} \Big) \times (1 + \tan 2^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 2^\circ} {1+\tan 45^\circ\tan 2^\circ} \Big) \times (1 + \tan 3^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 3^\circ} {1+\tan 45^\circ\tan 3^\circ} \Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \dfrac{\tan 45^\circ-\tan 22^\circ} {1+\tan 45^\circ\tan 22^\circ}\Big) \times (1 + \tan 45^\circ) \quad | \quad \mathbf{\tan 45^\circ = 1} \\\\ &=& (1 + \tan 1^\circ)\Big(1 + \dfrac{1-\tan 1^\circ} {1+\tan 1^\circ} \Big) \times (1 + \tan 2^\circ)\Big(1 + \dfrac{1-\tan 2^\circ} {1+\tan 2^\circ} \Big) \times (1 + \tan 3^\circ)\Big(1 + \dfrac{1-\tan 3^\circ} {1+\tan 3^\circ} \Big) \\ && \vdots \\ && \times (1 + \tan 22^\circ)\Big(1 + \dfrac{1-\tan 22^\circ} {1+\tan 22^\circ}\Big) \times (1 + 1) \\\\ &=& (1 + \tan 1^\circ)*\dfrac{(1+\tan 1^\circ+1-\tan 1^\circ)} {(1+\tan 1^\circ)} \times (1 + \tan 2^\circ)*\dfrac{(1+\tan 2^\circ+1-\tan 2^\circ)} {(1+\tan 2^\circ)} \times (1 + \tan 3^\circ)*\dfrac{(1+\tan 3^\circ+1-\tan 3^\circ)} {(1+\tan 3^\circ)} \\ && \vdots \\ && \times (1 + \tan 22^\circ)*\dfrac{(1+\tan 22^\circ+1-\tan 22^\circ)} {(1+\tan 22^\circ)} \times 2 \\\\ &=& 2\times 2\times 2 \\ && \vdots \\ && \times 2 \times 2 \\\\ &=& 2^{22}\times 2 \\\\ &=& 2^{23} \\\\ &=& \mathbf{8388608} \\ \hline \end{array} }\)

\(\frac{kn^2}{lm} = 8\)

The sum of all possible values of g(8) is 8 + (-1) = 7.

3. b/a = -4.

4. The parabola is y = 1/2*x^2 - 3x + 7/2, so abc = -21/4.

\(f^{-1}(x) = \dfrac{-2x + 5}{x + 1}\)

a/c = -2

The polynomial is \(x^4 - 4x^3 + 4x^2 - x + 5\)

\(f^{-1}(f^{-1}(9)) = \boxed{1}\)

By the Sine Law, XY = 4*sqrt(3).

Isn't that-8?

Important:::  If the value of scale factor k is negative, the dilation takes place in the opposite direction from the center of dilation on the same straight line containing the center and the pre-image point.

A / [xyz] = 0.407407407 = 11/27  

How about you tell me what    \(\lfloor -7.2 \rfloor \)     is and why?

The sum of the interior angles of a rhombus is 360^o so add up all the angles. 

(2)(2w – 62) + (2)(w + 71) = 360 

4w - 124 + 2w + 142 = 360 

6w + 18 = 360 

6w = 342  

w = 57 

So plug in 57 in place of w and find that     2w – 62  =   52^o  each of the acute angles  

                                                         and      w + 71  =  128^o  each of the obtuse angles   

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If the two triangles are Similar, then the sine of D is the same as the sine of G, which is the opposite side over the hypotenuse,

that is:   77 / 85 = 0.9058824  (and angles G and D are 65^o but that's just an oh by the way)    

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It is because the other 2 posts have been blocked.

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