Question #dc133

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Question #dc133

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Answer 1 · 2017-10-04T17:25:26

It is not clear that "The length of a rectangle is equal to 2 times the width plus 3 feet" is What?
So here Both the cases are solved.

When l=2b+3

Let the length of rectangle $= l$ $=l$
and Width of rectangle $= b$ $=b$

The length of a rectangle is equal to 2 times the width plus 3 feet.
Hence
$l = 2 b + 3 feet$ $l=2b+3 "feet"$ ...[A]

Area of Rectangle $A = l b = 20 square feet$ $A=lb=20 "square feet"$
Put the value of $l$ $l$ from equation [A]
$(2 b + 3) (b) = 20$ $(2b+3)(b)=20$
$\Rightarrow 2 b^{2} + 3 b - 20 = 0$ $=>2b^2+3b-20=0$
$\Rightarrow b = \frac{- 3 \pm \sqrt{(3^{2}) - 4 (2) (- 20)}}{2 \times 2} = \frac{- 3 \pm \sqrt{169}}{4}$ $=>b=(-3+-sqrt((3^2)-4(2)(-20)))/(2xx2)=(-3+-sqrt(169))/4$
$\Rightarrow b = \frac{- 3 \pm 13}{4} = - 4$ $=>b=(-3+-13)/4=-4$ $or$ $"or"$ $\frac{5}{2}$ $5/2$
Negative value is discarded hence $b = \frac{5}{2}$ $b=5/2$
Hence $l = 2 b + 3$ $l=2b+3$
if $l = 2 (\frac{5}{2}) + 3 = 8$ $l=2(5/2)+3=8$

When l=2(b+3)

If
The length of a rectangle is equal to 2 times the width plus 3 feet.
Hence
$l = 2 (b + 3) feet$ $l=2(b+3) "feet"$ ...[A]

Area of Rectangle $A = l b = 20 square feet$ $A=lb=20 "square feet"$
Put the value of $l$ $l$ from equation [A]
$2 (b + 3) (b) = 20$ $2(b+3)(b)=20$
$\Rightarrow b^{2} + 3 b = 10$ $=>b^2+3b=10$
$\Rightarrow b^{2} + 3 b - 10 = 0$ $=>b^2+3b-10=0$
$\Rightarrow b = \frac{- 3 \pm \sqrt{(3^{2}) - 4 (1) (- 10)}}{2} = \frac{- 3 \pm \sqrt{49}}{4}$ $=>b=(-3+-sqrt((3^2)-4(1)(-10)))/(2)=(-3+-sqrt(49))/4$
$\Rightarrow b = \frac{- 3 \pm 7}{4} = 1$ $=>b=(-3+-7)/4=1$ $or$ $"or"$ $- \frac{5}{2}$ $-5/2$
Negative value is discarded hence $b = 1$ $b=1$
Hence $l = 2 b + 3$ $l=2b+3$
if $l = 2 (1) + 3 = 5$ $l=2(1)+3=5$

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Question #dc133

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When l=2b+3

When l=2(b+3)

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