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tatyana61 [14]
3 years ago
11

The perimeter of a rectangle table is 680 cm if the length of the table is 24 cm more than the race the width of the table find

the dimensions of the table
Mathematics
1 answer:
postnew [5]3 years ago
5 0

Answer:

Length is 182 cm and width is 158 cm.

Step-by-step explanation:

Given:

Perimeter of rectangle table = 680 cm.

Let the width of the table be x.

Now according to given length of the table is 24 cm more than the race the width

hence length = 24+x.

But perimeter of rectangle = 2(length+width)

Substituting the values we get;

2(24+x+x)=680\\24+2x=\frac{680}{2}\\\\24+2x=340\\2x=340-24\\2x=316\\\\x= \frac{316}{2}=158 \ cm

Hence width = 158 cm

Now, length = 24 cm + width = 24 cm + 158 cm = 182 cm

Hence the dimension of table are 182 cm and 158 cm.

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so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

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so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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use (PEMDAS)
P parentheses ()
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There is division.
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amm1812

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