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3 years ago

5

the perimeter of a rectangular garden is 84 feet. write and solve the system of equations that can be used to find the dimension

s of the garden if its length, l, is 3 feet longer than its width, w.

1 answer:

scoray [572]3 years ago

6 0

Answer:

The length is 22.5 and the width is 19.5

Step-by-step explanation:

2L + 2W = P

\/

2L + 2W = 84

\/

W + 3 + W + 3 + W + W= 84

\/

4W + 6 = 84

\/

4W = 78

\/

W = 19.5 L = 19.5 + 3 = 22.5

I hope this helps, have a nice day! :D

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The sum of first three terms of a finite geometric series is -7/10 and their product is -1/125. [Hint: Use a/r, a, and ar to rep

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Ooh, fun

geometric sequences can be represented as
$a_n=a(r)^{n-1}$
so the first 3 terms are
$a_1=a$
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from the 2nd equation we can take the cube root of both sides to get
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so now rewrite 1st equation as
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subsituting -1/5 for ar
$\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r$
which simplifies to
$\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}$
multiply both sides by 10r
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add (2r²+2r+2) to both sides
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solve using quadratic formula
for $ax^2+bx+c=0$
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
so
for 2r²-5r+2=0
a=2
b=-5
c=2

$r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}$
$r=\frac{5 \pm \sqrt{25-16}}{4}$
$r=\frac{5 \pm \sqrt{9}}{4}$
$r=\frac{5 \pm 3}{4}$
so
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use them to solve for the value of a
$\frac{-1}{5}=ar$
$\frac{-1}{5r}=a$
try for r=2 and 1/2
$a=\frac{-1}{10}$ or $a=\frac{-2}{5}$

test each
for a=-1/10 and r=2
a+ar+ar²= $\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}$
it works

for a=-2/5 and r=1/2
a+ar+ar²= $\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}$
it works

both have the same terms but one is simplified

the 3 numbers are $\frac{-2}{5}$ , $\frac{-1}{5}$ , and $\frac{-1}{10}$

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