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Compute 

$1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb$

$\text{Let $x = \dfrac{1}{2} \quad|x|<1$ }$

Now we have:  $\displaystyle s= \sum \limits_{n=1}^{\infty} nx^n$

Let the progression to be summed be put equal to s:

$s = x+2x^2+3x^3+4x^4+\cdots + nx^n +\cdots$

It is divided by $x$ and multiplied by $dx$ then

$\dfrac{s\ dx}{x} = dx +2x\ dx+3x^2\ dx+4x^3\ dx+\cdots + nx^{n-1}\ dx +\cdots$

and with the integrals taken this equation is found
$\displaystyle \int \dfrac{s\ dx}{x} = x +x^2+x^3+x^4 +\cdots + x^n + \cdots = \dfrac{x}{1-x} \quad \text{ infinite geometric progression }$

Therefore from the equation:
$\displaystyle \int \dfrac{s\ dx}{x} = \dfrac{x}{1-x}$

on differentiation s is found. For the equation becomes:
$\begin{array}{rcll} \displaystyle \dfrac{s}{x} &=& \dfrac{x}{1-x}\left( \dfrac{1}{x}- \dfrac{(-1)}{1-x} \right) \\\\ &=& \dfrac{1}{1-x} + x(-1)(1-x)^{-2}(-1) \\\\ &=& \dfrac{1}{1-x} + \dfrac{x}{(1-x)^2} \\\\ &=& \dfrac{1-x+x}{(1-x)^2} \\\\ &=& \dfrac{1 }{(1-x)^2} \end{array}$

thus there is produced:
$\displaystyle s = \dfrac{x}{(1-x)^2} \\$

So  $\text{let $x = \dfrac{1}{2} $}:$

$\begin{array}{|rcll|} \hline s &=& \dfrac{ \dfrac{1}{2} } {\left(1-\dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ \dfrac{1}{2} } {\left( \dfrac{1}{2} \right)^2} \\\\ &=& \dfrac{ 1 } { \dfrac{1}{2} } \\\\ \mathbf{s} & \mathbf{=} & \mathbf{2} \\ \hline \end{array}$

finally
$\displaystyle 1 \cdot \dfrac {1}{2} + 2 \cdot \dfrac {1}{4} + 3 \cdot \dfrac {1}{8} + \dots + n \cdot \dfrac {1}{2^n} + \dotsb = 2$

SOLUTION:

$(A)$ Since we know the bottom left and bottom right angles of this trapezoid, we can construct two different triangles with angles $60 ^{\circ}$ and  $90 ^{\circ}$. The trianges have angles  $90​​ ^{\circ}$and $60 ^{\circ}$ and since the angles of a triangle sum up to $180 ^{\circ}$, it is a " $90 ^{\circ}, 60 ^{\circ}, 30^{\circ}$ triange". " $90 ^{\circ}, 60 ^{\circ}, 30^{\circ}$ trianges" have the following properties:

Hypotenuse = $2x$

Larger Triange Side = $x\sqrt{3}$

Smaller Triange Side = $x$

In the trapezoid, the larger triangle side equals the altitude thus the altidute equals $\sqrt{3}$ since the hypotenuse equals $2$ and therefore $(2x = 2), x = 1$. Since the bottom base of the trapezoid merely equals $x + x + 2 = 4$ and we already have the top base, we can find the perimeter. The perimeter equals the sum of both bases and the two other sides of the trapezoid thus the perimeter equals $4 + 2 + 2 + 2 = \boxed{10}$ 

$(B)$ As you have most likely realized, the altitude isn't utilized whatsoever in the first problem, however, it is needed in order to compute the area. The formula of the area of a trapezoid is $(\frac{a+b}{2})h$($a$ and $b$ are the bases of a trapezoid and $h$ is the altitude). Since we already know the bases and altitude of this specific trapezoid, we can quite feasibly compute the area. $(\frac{a+b}{2})h = (\frac{4 + 2}{2})\sqrt{3} = (\frac{6}{2})\sqrt{3} = \boxed{3\sqrt{3}}$

RB - ∃

See here :

https://www.wolframalpha.com/input/?i=1-(1-(1%2F(2.3*10%5E40200)))%5E7.14*10%5E94

The vertex of  f(x)  = (-2, - 5)

So...the smallest "c" we can have is -2

We can find the inverse as

y = ( x + 2) ^2 - 5

y + 5 = (x + 2)^2

√ [ y + 5]  = x + 2

√[y + 5 ] - 2 = x

√[x + 5 ] - 2 = y  = the inverse

Note that the point  (-5, -2)  is on the inverse....as we would expect

See the graph , here :  https://www.desmos.com/calculator/w4bhcvpkkb

Note that if f(x) included any x values in the domain < -2, it would not be one-to-one...and we wouldn't have a true inverse

Consider the curve with equation.$\operatorname{Re}\left( \dfrac{1}{z} \right) = \dfrac{1}{6}$For each complex number in the following list, $1, 4i, 3+3i, 3-3i, 1 - 2i, 2+ 3i, 6$, figure out whether each one is on the curve, then enter "yes" or "no" in the blank corresponding to each option below.

FTFY

sin A / a  =  sin B / b

sin (29.8)  / 28.6  =  sin B / 35.8

35.8 sin (29.8) / 28.6 =  sin B

arcsin  (35.8 sin (29.8) / 28.6)  = B ≈  38.5°

So....C = 180 - 29.8 - 39.5  = 110.7°

And

a/sin A = c / sin C

28.6 / sin 29.8 = c /  sin 110.7

28.6sin 110.7 / sin 29.8  =   = 53.8

So   1st triangle 

A = 29.8  B = 38.5 C = 110.7

a = 28.6  b =35.8   c = 53.8

Subtract  38.5 from 180  = 141.5°   = B

Add this to A  = 171.3 < 180   so we have two triangles

And angle C = 180 - 171.3 = 8.7°

So using the Law of Sines again to find c , we have

a /sin A = c / sin C

28.6 / sin 29.8 = c / sin 8.7

28.6sin 8.7 / sin 29.8 = c = 8.7

So  ...in the second triangle

A = 29.8   B = 141.5  C = 8.7

a = 28.6   b =  35.8    c = 8.7

Since No.1 has been solved by CPhill, I will solve No.2.

Solution No.2:

$\angle ABC$ = $\angle ACB$ which implies that $ABC$ is isosceles. We drop the perpendicular $M$ from $A$ to $BC$; we get that $MC = 9$ and so $MD = 5$ . Now by Pythagoras on $AMD$..., $AM = 12$. Thus $ABC$'s areo equals to $\frac{1}{2}(12*18 = 216) = \boxed{108}$ .

RB - ∃

Solution:

This is very simple. 

$f(x) = 3(x^2 ) - 7$, so $f(4) = 3(4^2) - 7 = (3*16) - 7 = 48 - 7 = 41$.

We know that $g(f(4)) = 9$ so this means that $g(41) = 9$.

$f(-4) = 3 (-4^2) - 7$ and since $-4^2 = -4 * -4 = 16$, this is the same thing as $f(4)$ so $f(-4)$ also equals $41$.

We know that $g(41) = 9$ thus $g(f(-4)) = 9$ as well.

$\boxed{g(f(-4) = \boxed{9}}$

RB - ∃

"The area of a parallelogram can be found by multiplying a side times the height to that side.

Since you will get the same answer using whatever side you want to use:

Area  =  $FH * AD  =  EG * DC$

--->        $FH * 18  =  16 * 27$

--->                $FH  =  24$" ~ geno3141

First, evaluate  f(4)  =  3(4)^2 - 7  =  48 - 7  =  41

Then evaluate f(-4) = 3(-4)^2 - 7 = 48 - 7 = 41

So    g(f(4)) = g(41) = 9

So...it  must also be that  g(f(-4)) = g(41)  = 9

flip it about the x axis, then flip it about the y axis

Thank you! 

sumfor(n, 1, 1000, n/(2^n) = 2

We can use the intersecting chord theorem, here....

We have that 

AP * BP  =  CP * DP

3 * BP  =  8 *  DP          divide both sides by 3, DP

BP / DP  =  8 / 3

Multiply the section of the chord, they should be equal to each other.