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Alenkinab [10]
3 years ago
14

Find a16 of the sequence 1,6,11,16

Mathematics
1 answer:
babymother [125]3 years ago
6 0

\bf 1~~,~~\stackrel{1+5}{6}~~,~~\stackrel{6+5}{11}~~,~~\stackrel{11+5}{16}...\qquad \qquad \stackrel{\textit{common difference}}{d=5} \\\\[-0.35em] ~\dotfill\\\\ n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} a_n=n^{th}\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ d=\textit{common difference}\\ \cline{1-1} a_1=1\\ d=5\\ n=16 \end{cases} \\\\\\ a_{16}=1+(16-1)5\implies a_{16}=1+(15)5\implies a_{16}=76

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Leto [7]

Area inside the semi-circle and outside the triangle is  (91.125π - 120) in²

Solution:

Base of the triangle = 10 in

Height of the triangle = 24 in

Area of the triangle = \frac{1}{2} bh

                                $=\frac{1}{2} \times 10\times24

Area of the triangle = 120 in²

Using Pythagoras theorem,

\text{Hypotenuse}^2=\text{base}^2+\text{height}^2

\text{Hypotenuse}^2=10^2+24^2

\text{Hypotenuse}^2=100+576

\text{Hypotenuse}^2=676

Taking square root on both sides, we get

Hypotenuse = 23 inch = diameter

Radius = 23 ÷ 2 = 11.5 in

Area of the semi-circle = \frac{1}{2}\pi r^2

                                      $=\frac{1}{2} \pi \times (13.5)^2

Area of the semi-circle = 91.125π in²

Area of the shaded portion = (91.125π - 120)  in²

Area inside the semi-circle and outside the triangle is  (91.125π - 120)  in².

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Sana makatulong po sa inyo;)

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