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LenKa [72]
3 years ago
12

At a large nursery, a border for a rectangular garden is being built. Designers want the border’s length to be twice as long as

its width. A maximum of 150 ft of fencing is available for the border. Write and solve an inequality that describes possible widths of the garden.
Mathematics
1 answer:
sasho [114]3 years ago
7 0

Let

x--------> the border’s length

y--------> the border’s width

P--------> perimeter of the border

we know that

x=5+y------> equation 1

P=2*[x+y]-----> P=2x+2y

P <=180 ft

(2x+2y) <= 180-------> equation 2

substitute the equation 1 in equation 2

2*[5+y]+2y <= 180

10+2y+2y <= 180

4y <= 180-10

4y <=170

y <=42.5 ft

so

the maximum value of the width is 42.5 ft

for y=42.5 ft

x=42.5+5------> x=47.5 ft

the answer is

the width of the border is less than or equal to 42.5 ft

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\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

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\boxed{\sf (a+b)^2=a^2+2ab+b^2}

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\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

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\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

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\underline{\bf More\: simplification:-}

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\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

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