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Gnom [1K]
3 years ago
12

Solve for the perimeter and area of a rectangle that has a length of 4in. and a width of 2 in

Mathematics
2 answers:
Vinil7 [7]3 years ago
7 0
The perimeter is either 10 (if the rectangle has two sides that are 4in and one side that is 2in) or 8 (if the rectangle has two sides that are 2in and one side that is 4in)
ioda3 years ago
5 0
P=4+4+2+2
=8+4
=12in

A=l×w
=4×2
=8in^2
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v(t)=\displaystyle\int a(t)\,\mathrm dt
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s(t)=\displaystyle\int v(t)\,\mathrm dt
s(t)=\displaystyle\int(-10\cos t+3\sin t+C_1)\,\mathrm dt
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With the boundary conditions s(0)=0 and s(2\pi)=12, we have

\begin{cases}0=-3+C_2&s(0)=0\\12=-3+2\pi C_1+C_2&s(2\pi)=12\end{cases}\implies C_1=\dfrac6\pi,C_2=3

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<h3>How to solve algebraic word problems?</h3>

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Read more about algebra word problems at; brainly.com/question/13818690

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