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stealth61 [152]
3 years ago
8

a piece of wire 48 m long is bent into the shape of a rectangle whose length is twice its width. find the length of the rectangl

e.
Mathematics
1 answer:
Drupady [299]3 years ago
3 0

\bf \textit{perimeter of a rectangle}\\\\
P=2(L+w)~~
\begin{cases}
L=length\\
w=width\\[-0.5em]
\hrulefill\\
L=\stackrel{twice~the~width}{2w}\\
P=48
\end{cases}\implies 48=2(2w+w)\implies \cfrac{48}{2}=3w
\\\\\\
24=3w\implies \cfrac{24}{3}=w\implies \boxed{8=w}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
L=2(8)\implies L=16

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Step-by-step explanation:

In the given sequence:

a(3) =  14, a(9)  = -1

The general term of a sequence in Arithmetic Progression is:

a(n) = a + (n-1)d

a(3) = a + (3 -1) d = a + 2 d

and a(9) = a + (9- 1 ) d = a  + 8 d

⇒   a + 2 d =  14      .........  (1)

and a + 8 d  = -1    ...........  (2)

Now, solving the given system of equation, we get:

From (1), a  = 14 - 2 d

Put in (2), we get:

a + 8 d  = -1                ⇒   14 - 2 d + 8d = -1

⇒  14 +   6d   = -1  

or,  6d = -1 -14 = -15

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Then a  = 14 - 2 d = 14 - 2(-2.5)  =14 + 5  = 19, or a  = 19

Now, first four terms of the sequence is:

a = 19

a(2) = a + 4 = 19 - 2.5 = 16.5

a(3) = a + 2d = 19 + 2(-2.5) = 19 - 5 = 14

a(4) = a  + 3d = 19 + 3(-2.5)  = 19  - 7.5  = 11.5

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4 0
3 years ago
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