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vova2212 [387]
3 years ago
7

Find the sum of the geometric series:

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0
\bf \qquad \qquad \textit{sum of a finite geometric sequence}
\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}
\end{cases}\\\\
-------------------------------

\bf \sum\limits_{i=1}^{5}~\stackrel{a_1}{375}\stackrel{r}{\left( \frac{-1}{5} \right)}^{i-1}\qquad 
\begin{cases}
a_1=375\\
r=-\frac{1}{5}
\end{cases}\implies S_5=375\left( \cfrac{1-\left(  \frac{-1}{5}\right)^5}{1-\left(  \frac{-1}{5}\right)} \right)


\bf S_5=375\left( \cfrac{1-\left( \frac{-1}{3125} \right)}{1+\frac{1}{5}} \right)\implies S_5=375\left( \cfrac{1+\frac{1}{3125}}{1+\frac{1}{5}} \right)
\\\\\\
S_5=375\left( \cfrac{\frac{3126}{3125}}{\frac{6}{5}} \right)\implies S_5=375\left(\cfrac{3126}{3125}\cdot \cfrac{5}{6}\right)\implies S_5=375\left(\cfrac{521}{625}\right)
\\\\\\
S_5=\cfrac{1563}{5}\implies S_5=312\frac{3}{5}
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a sailmaker is designing a mainsail for a newly constructed sailboat. the mainsail foot is 13 feet long, the luff is 10 feet lon
Vikentia [17]
  • (A) The mainsail foot is 13 feet long, the luff is 10 feet long, the leach is 6.93 feet long, and the mainsail area is 64.85 ft².
  • (B) The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff forms a 50° angle with the leach at the head, requiring the length of the leach and the area of the mainsail.

<h3>What exactly is sines law?</h3>
  • The law of sines refers to the ratio of a triangle's side length to the sine of the opposite angle, which is the same for all sides.
  • SinA/a = SinB/b = SinC/c
  • Where A, B, and C represent the triangle's angle and a,b, and c represent the opposite sides of the triangle's angle.

So, Part A:

By employing the sines law,

  • SinA/10 = SinB/13
  • SinA/10 = Sin50/13
  • A = 36.10
  • C = 180-(180-36.10)
  • C = 93.90

So,

  • Sin36.10 / a = Sin50/ b = Sin93.90/c
  • Sin36.10 / 10 = Sin50/ 13 = Sin93.90/c
  • c = Sin93.90 × 13 ÷ Sin50
  • c = 16.93 ft

Now, Part B:

Area of the mainsail:

  • = 1/2 × abSinC
  • = 1/2 × 10 × 13 × Sin93.90
  • =  64.85

Therefore, the answers to both parts are:

  • (A) The mainsail foot is 13 feet long, the luff is 10 feet long, the leach is 6.93 feet long, and the mainsail area is 64.85 ft².
  • (B) The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff forms a 50° angle with the leach at the head, requiring the length of the leach and the area of the mainsail.

Know more about sines law here:

brainly.com/question/27174058

#SPJ4

The complete question is given below:

A sailmaker is designing a mainsail for a newly constructed sailboat. The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff makes a 50° angle at the head with the leach.

Part A: Determine the length of the leach.

Part B: Find the area of the mainsail.

PLS HELP :) 50 POINTS!!!

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1 year ago
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