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wariber [46]
2 years ago
12

1/2x - 3/4y=-2  Solve for Y

Mathematics
1 answer:
sertanlavr [38]2 years ago
5 0
\frac{1}{2}x-\frac{3}{4}y=-2\ \ \ |Add\ \frac{3}{4}y\\\\
\frac{1}{2}x=-2+\frac{3}{4}y\ \ \ |Add\ 2\\\\
\frac{1}{2}x+2=\frac{3}{4}y\ \ |*\frac{4}{3}\\\\
y=\frac{4}{3}(\frac{1}{2}x+2)\\\\
\boxed{y=\frac{2}{3}x+\frac{8}{3}}
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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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the sum is -7/10
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and their product is -1/125
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from the 2nd equation we can take the cube root of both sides to get
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note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
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subsituting -1/5 for ar
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which simplifies to
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multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
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
r=\frac{5+3}{4}=\frac{8}{4}=2 or r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}

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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Step-by-step explanation:

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