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Hey man, can you tell me what you actually mean?

Yay!

Multiply 39,000 by 0.05, then take that away from 39,000.

After that, you get 37,050.

Take 37,050, and multiply it by 0.05, then add it to 37,050.

You get 38902.5

So yeah.

Yay!

so the common difference is 3 right ?

A) geometric sequence, and the sum is just \(-\frac{512+256+128+\dots+2+1}{512} = \boxed{-\frac{1023}{512}}\).

(notice that 1023 is one smaller than 1024, which is 2^10)

B) Idk if my solution is even close to optimal, but here it is:

Let the common difference be x. Then, 

\((35-6x)+(35-5x)+(35-4x)+(35-3x)+(35-2x)+(35-x)+(35)+(35+x)=220\\ 8(35) - 20x=220\\ -20x=-60\\ \boxed{x=3}\)

The common difference is therefore equal to 3

The first term is just \(35-6(3)=\boxed{17}\)

varvaax lmao I'm not that good :)

hahah its okay mate..i hope his answer will make sense 

textot is going to run over my math with a dump truck, so beware

hey thanks, Varxaax.. i appreciate your time mate 

A.) Geometric, sum of all terms is gotta be either -infinite, or all of that added up, I don't have the patience to work that out.

B.) No.

C.) 1+2+4+8+16+32+64+128+256=511, so n=9

D.) So uhh, I'm confused, but I think I get it. It's 4550, because 650+600+550+450...+150+100+50+0=4550

Someone smarter than me can do B.) because I'm lazy.

Yay!

Gotta give us a question, mate. G'day.

Yay!

Hey there!

First use pythagoras to gain the length of \(WX\)

\(WX=\sqrt{(\sqrt{13})^2-3^3}=2\)

Now we can take the tangent of \(V\)

\(tan(V)=\frac{\sqrt{(\sqrt{13})^2-3^3}}{3}=\frac{2}{3}\)

Hope this helped :)

Edit: Had \(13^2\) instead of \(\sqrt{13}^2\) oops

The surface area of the prism is 132, and I'm too lazy to put the equation.

The volume is 60, and once again, I'm too lazy to put the equation.

Yay!

Hey there!

Since \(sin\) is opposite over hypotenuse, we can use thid to find \(BD\):

\(AB=hyp=24\), \(sin(A)=2/3\), using this ratio we know that \(BD=16\)

We can also use the same stategy to find \(BC\)

\(BD=opp=16\). \(sin(C)=1/3\), using this ration we know that \(BC=48\)

Finally we need to manipulate Pythagoras to give us \(DC\)

\(DC=\sqrt{(BC)^2-(BD)^2}=\sqrt{48^2-16^2}\approx45.3\)

Hope this helped :)

The tangent lines are the same length

13x-16 = 4x+ 11

x = 3

4x+11 is then   23 long      pythag theorem  FD = sqrt (12^2 + 23^2) =25.9 units

One other way without finding the height

A  = (1/2) (2.6) (5.6) sin (70)  ≈  6.84096 yd^2

Thank you! Also if anyone needs the exact form instead of the rounded form its \(32\sqrt{2}\)

tan 21 = 17 000 / d

d = 17000/ tan21 = 44286.5 ft

sin A = 2/3   =   BD/24     meaning   BD = 16

sin C = 1/3   =   BD / BC  = 16/BC       meaning BC = 48  

48^2 - 16^2 = DC ^2     ( Pythag Theorem)

DC = 45.25 units

Here's what I get

1  2  5   5   6     identifiable sequencwa  =  5 !  / 2 !  =   60

2  2  3  5   5     identifiable sequences  = 5 !  / (2! * 2!)  =  30

1 3  4  5   5      identifiable sequences  =  5 ! / 2 !  =   60

=  150  sequences  

The top/bottom   are composed of  two right triangles  with legs of  10 and 24

Their  total surface  area =   10 *24  = 240 cm^2

We  have  3 rectangles 

One 24 x 50   =  1200 cm^2

One  25 x 50   =   1250 cm^2

One  10 x 50   = 500 cm^2

So....total surface area  =   [ 240  + 1200 + 1250  + 500 ]  cm^2    = 3190 cm^2

Total circumference   =  pi * d    =  pi ( 2 * 12.6) = 25.2 pi    cm

   this arc is a fraction of the 360 degree circumference      100 / 360

100/360 *  25.2 pi = ...........   cm

Find  h  (height)   as   2.6 sin 70 = 2.44 yd

  then area = 1/2 base * height      = 1/2 * 5.6 * 2.44   yd²

Highest   86     lowest   41        86-41 = 45     Amplitude is 1/2 of this = 22.5

22.5 sin

The graph starts at 62 and heads downward  so it is  - sin     (reflected)

shifted up     so midline is   (86+41)/2 = 63.5 up

- 22.5 sin (          )     + 63.5

Month zero = 62   so there is a very small shift of the graph to the left of .127

 -22.5 sin (  (x+.127)   )   + 63.5

Period is a year = 12 months       2pi / s  =  12      s =   pi/6

- 22.5 sin ( pi/6 (x+.127) + 63.5    =    temp                    where x = month          (this graph is not exact, but a 'best fit' for the data)  

f(x)  =  12sec (2x)

Here's  the graph  : https://www.desmos.com/calculator/7ivgku7b28

We  have  this

(x - 2)^2   + ( y + 1)^2  =  16

(x - 2)^2  +  ( y - 5)^2  =   10   subtract  the  second equation form the first

(y + 1)^2  - ( y - 5)^2   =  6        simplify

y^2  + 2y  + 1   -  y^2 + 10y  - 25    = 6

12y  - 24  =  6

12y  =  30  

y  = 30 /12   =  2.5 = 5/2

Using  either equation   we  can find x  as follows

(x - 2)^2 + ( 2.5 + 1)^2   = 16

(x - 2)^2  +  12.25 =   16

(x - 2)^2  =  16 -12.25

(x  -2)^2  =  15/4         take both roots

x - 2  =  sqrt (15)/2                              x   -  2  = -sqrt (15) /2

x =  sqrt (15)  / 2  +  2                        x  =  2 - sqrt (15) /2

x=  [ 4  +  sqrt (15)  ]   /2                    x=  [4 -sqrt (15)]  / 2

A = (  [4 + sqrt (15) ] / 2 , 5/2 )

B = ( [4 - sqrt (15) ]   / 2 , 5/2 )

My question is.....  What is AB  supposed to represent ????

Is that point given   (-7,\(\sqrt{23}\) )     ??    I will assume it is

Find hypoytenuse length of the terminal ray  ( or magnitude)

  = sqrt (  (-7)^2  + (sqrt23)^2 )     = sqrt  (49 + 23) = sqrt 72

cos   = -7 / sqrt (72)    = - 7/ (6sqrt 2)