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Thank you sooo much CPhill these really helped me. One question though. For the second question, is that formula for asymptotes applicable to all hyperbolas or just this one specifically?

BTW....here's a resource that I find helpful :

https://www.purplemath.com/modules/index.htm

Last one

x^2         y^2

___  +   ____  =   1   

25          16

The " + "  between the terms signals an ellipse

Like the hyperbola before, this is centered  at the origin = (0, 0)

The major axis is along x  and the minor along y

a^2  = 25       b^2  = 16

a = 5     b  =  4

The vertices here are located at   ( 5, 0) , (-5, 0), (0, 4) and (0, -4)

The foci are located on the major axis and  are given by  ( ±√[ a^2 - b^2 ] , 0) =

( ±√[ 25 - 16 ] , 0 )   =   (± √9 , 0 )   =  (3, 0) and (-3, 0)

Here's a graph :  https://www.desmos.com/calculator/j579zix3zu

Second one

x^2          y^2

___   -     ____   =  1

 25             1

We have the form

(x - h)^2           (y -k)^2

______   -      ________   =       1         

    a^2                  b^2

This is a hyperbola intersecting the x axis  with a center  at (h, k)  = (0, 0)

a = 5      and b  = 1

Tve vertices   lie    along the x axis  and  are given by  (a, 0)  and (-a, 0)  = (5, 0) and (-5, 0) 

The foci  are given by   ( √[ a^2 + b^2], 0 )    and (- √[a^2 + b^2], 0 )   =

( √ [25 + 1] , 0 )   =  (√26, 0)       and  ( - √26, 0)

The asypmtotes are given by :

y = ± (b/a) (x - h) + k     =   

y  = ± (1/5) ( x - 0) + 0    = 

y =  ± (1/5)x 

Here's a graph:  https://www.desmos.com/calculator/aabdp29eok

a=3 is the only answer I see.

First one

-16x  = y^2      this is a parabola  opening to the left.....we have the form

4p ( x - h)  =  (y - k)^2

4p ( x - 0)  = (y - 0)^2

The vertex is at  (h, k)  = (0,0)

We can find "p" [ the distance between the focus and the vertex ] as

4p = - 16       divide both sides by 4

p = - 4

The focus is given by

(0 + p, 0)  =  (0 + (-4), 0)  =  (-4, 0)

The directrix is given by

x =  - (p) 

x= - (-4)

x  = 4

Here's a graph :  https://www.desmos.com/calculator/cdaauapqz1

This sum is obviously telescoping :)

\(\begin{array} {rll} \phantom{=}&\displaystyle\sum^{9999}_{n=1}\dfrac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\color{red}{\sqrt[4]{n+1}-\sqrt[4]{n}}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)\color{red}{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{n+1 - n}\\ =&\displaystyle \sum^{9999}_{n=1}\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)\\ =&\displaystyle (\color{blue}{\sqrt[4]{2}}\color{black} - \sqrt[4]1) + (\color{green}{\sqrt[4]{3}}\color{black} - \color{blue}\sqrt[4]2\color{black}) + (\sqrt[4]4 - \color{green}\sqrt[4]3\color{black}) + \cdots+(\sqrt[4]{10000}-\sqrt[4]{9999})\\ =&\displaystyle \sqrt[4]{10000} - \sqrt[4]1\\ =&\displaystyle 10 - 1\\ =& 9 \end{array}\)

sum_(n=1)^9999 1/((sqrt(n) + sqrt(n + 1)) (n^(1/4) + (n + 1)^(1/4)))≈9.000000000000000000000000000000000000000

\(\dfrac{1}{n^2+n}=\dfrac{1}{n(n+1)} = \dfrac{A}{n}+\dfrac{B}{n+1}\\~\\ A+B=0,~A=1,~B=-1\\~\\ \dfrac{1}{n^2+n} = \dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\sum \limits_{n=1}^{1000}~\left(\dfrac 1 n - \dfrac{1}{n+1}\right) = \\~\\ \left(1 - \dfrac 1 2\right) + \left(\dfrac 1 2-\dfrac 1 3\right) + \left(\dfrac 1 3 - \dfrac 1 4\right) + \dots + \left(\dfrac{1}{1000}-\dfrac{1}{1001}\right) = \\~\\ 1-\dfrac{1}{1001} = \dfrac{1000}{1001}\)

Nice, sinclairdragon.....you approached it from the other "end"  !!!!

Probably easier ways to do this....but...let the first term  = a

So we have

a , (2a + 3) , (4a + 9) , (8a + 21), (16a + 45), (32a + 93), (64a + 189)

So

64a + 189  = 125

64 a  =  -64       divide both sides by -64

a  =  - 1

So  ....the sequence is

-1, 1, 5, 13, 29, 61, 125

125-3=122

122/2=61

61-3=58

58/2=29
29-3=26
26/2=13
13-3=10
10/2=5
5-3=2
2/2=1
1-3=-2
-2/2=-1

Term 7: 125
Term 6: 61
Term 5: 29
Term 4: 13
Term 3: 5
Term 2: 1
Term 1:-1
The first term is -1

The coordinates of C  =   (2cos (pi/12), 2sin(pi/12)  =  (2cos 15°, 2sin 15°)

cos (15°)  =  [ 1 + √3 ]                   sin (15°)  =  [√3 - 1 ]

                    _______                                      _______

                       2√2                                              2√2

So  C  =  ( [ 1 + √3] /  √2 ,  [ √3 - 1 ] / √2 )

The coordinates of A  =   (4cos (3pi/12) , 4 sin (3pi/12)  =  (4cos (pi/4) , 4sin (pi/ 4) )  =

( 4/√2, 4 / √2 )

The coordinates of B =  (3cos (5pi/12) , 3 sin (5pi/12)  =  (3cos (75°) , 3sin (75°) ) =

( 3 [ √3 - 1] / [2√2] ,  3[ 1 + √3]/ [2√2] )

So   

AB^2   =     ( 3 [ √3 - 1] / [2√2] - 4/√2)^2  +  ( 3[ 1 + √3]/ [2√2] - 4/√2)^2  = 25 - 12√3  ≈ 4.22

AC^2  = (4/√2 -  [ 1 + √3] /  √2)^2  + ( 4/√2 -  [ √3 - 1 ] / √2)^2 =  20 - 8√3  ≈  6.14

BC^2  = ( 3 [ √3 - 1] / [2√2] - [ 1 + √3] /  √2)^2 + (  3[ 1 + √3]/ [2√2] -  [ √3 - 1 ] / √2)^2  = 7

[ Because of all the nasty radicals....I used Wolfram Alpha to compute these ]

Delete someone mode my comment thank you.

List the first few terms

-4, 2, -1, 1/2, -1/4, 1/8........

The common ratio  is  -1/2    and the first term is -4

So....the sum converges to

   -4                    -4

_______  =    _____  =   -4 (2/3)  =  -8/3

1 - (-1/2)           (3/2)

expand   (2x - 3y)^5:

32 x^5 - 240 x^4 y + 720 x^3 y^2 - 1080 x^2 y^3 + 810 x y^4 - 243 y^5

\(\frac{1}{x}+\frac{1}{x+2}=\frac9{40}\\ \frac{x+2+x}{x(x+2)}=\frac{9}{40}\\ \frac{2x+2}{x^2+2x}=\frac{9}{40}\\ 9x^2+18x=80x+80\\ 9x^2-62x-80=0\\ (9x+10)(x−8)=0\\ x_1=\frac{-10}{9}, x_2=8\)

only 8 is an integer. 

1/x  +  1/(x+2)  =  9/40

Multiply each term by the common denominator of  40(x)(x+2):

[1/x](40)(x)(x+2)  +  [1/(x+2)](40)(x)(x+2)   =  [9/40](40)(x)(x+2) 

40(x+2)  +  (40)(x)  =  9(x)(x+2)

40x + 80  +  40x  =  9x² + 18x

                         0  =  9x²  -  62x  -  80

                         0  =  (9x + 10)(x - 8)                 

Disregarding the answer which is a fraction:

--->     x  =  8;  x + 2  =  10

\(\sum \limits_{n=0}^\infty \alpha^n = \dfrac{1}{1-\alpha},~\forall \alpha \ni |\alpha|<1\\ \text{and does not converge otherwise}\)

\(\sum \limits_{n=1}^\infty ~-4\left(-\dfrac 1 2\right)^{n-1} = -4\left(\sum \limits_{n=0}^\infty~\left(-\dfrac 1 2\right)^n \right)\)

I leave it to you to apply the first line to the second.

The binomial theorem is as follows:

\((a+b)^2=\sum_{k=0}^{n}\binom{n}{k}a^{(n-k)}b^k \)

The formula never helped me much either, but here is the explanation. Looking at any expansion, there are coefficients and exponents. For example, the expansion for \((a+b)^3\) is as follows:

\((a+b)^3=a^3+3a^2b+3ab^2+b^3\)

Just looking at the coefficients, you get 1, 3, 3, and 1. This is the fourth row of the Pascal triangle. 

For the exponents, you get 3, 2, 1, 0 for a; and 0, 1, 2, 3 for b. From this, we can deduce what \((2x-3y)^5\) will be. 

The expansion for just \((x-y)^5\):

The exponents for x will be 5, 4, 3, 2, 1, 0 and the exponents for y will be 0, 1, 2, 3, 4, 5. 

The coefficients will be the sixth row of the Pascal triangle, which is 1, 5, 10, 10, 5, 1

Therefore, the expansion will be \((x-y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)

However, we are not done yet. We need to add the positive/negative signs, alternating. 

\((x-y)^5=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5\)

Using this equation, you can plug in the values to find the given expression.

I hope this helped,

Gavin. 

Exactly! You have something to add in my answer or there are something whitch don't tike you?

Thank you.

\(3\sin\theta\ =\ \sin\theta-\sqrt3\\~\\ 3\sin\theta-\sin\theta\ =\ -\sqrt3\\~\\ 2\sin\theta\ =\ -\sqrt3\\~\\ \sin\theta\ =\ -\frac{\sqrt3}{2}\)

We can look at a unit circle to see what values of  θ  in the interval  [0, 2π]  will satisfy  \(\sin\theta=-\frac{\sqrt3}{2}\) .

Here is a pretty clear unit circle:

I will use distributive property:

\((2\sqrt{3x}-2)(3\sqrt{3x}+5)<=>2\times 3(\sqrt{3x})(\sqrt{3x})+5\times2(\sqrt{3x})-2\times3(\sqrt{3x})-2\times5<=>6\times3x+10(\sqrt{3x})-6(\sqrt{3x})-10<=>18x+4(\sqrt{3x})-10\)

Finally \(18x+4(\sqrt{3x})-10\)

\(0≤θ≤2π \) (1)

\(3sinθ=sinθ -\sqrt{3}<=> 3sinθ-sinθ= -\sqrt{3}<=> 2sinθ= -\sqrt{3} <=>sinθ= -\frac{\sqrt{3}}{2}<=>sinθ=-sin(\frac{π}{3}),sinθ=-sin(\frac{2π}{3})<=>sinθ=sin(-\frac{π}{3}),sinθ=sin(-\frac{2π}{3})\)

With (1) \(sinθ=sin(2π-\frac{π}{3}),sinθ=sin(2π-\frac{2π}{3})<=>sinθ=sin(\frac{5π}{3}),sinθ=sin(\frac{4π}{3}) <=>θ=\frac{5π}{3},θ=\frac{4π}{3}\)

So \(θ=\frac{5π}{3},θ=\frac{4π}{3}\)