X=length
y=width
Area (rectangle)=length x width
we suggest this system of equations:
y=x/5
xy=9/20
solve this system of equations by substitution method:
x(x/5)=9/20
x²/5=9/20
Least common multiple=20
4x²=9
x²=9/4
x=⁺₋√(9/4)
we have two solutions:
x₁=-3/2 it does not validate
x₂=3/2 ⇒y=x/5=(3/2)/5=3/10

The dimensions of the lake are:
lengh=3/2 miles
width=3/10 miles

to check:
Area=3/2 miles x 3/10 miles=9/20 miles².
width is 1/5 the length of the lake ⇒ 3/10 miles=(3/2) /5 miles

7 0

3 years ago

What is 2\20 divided by 2

Valentin [98]

$\frac{2}{20} \div 2 \\ \\ \frac{1}{10} \div 2 \\ \\ \frac{1}{10} \times \frac{1}{2} \\ \\ \frac{1 \times 1}{10 \times 2} \\ \\ \frac{1}{20} \\ \\$

The final result is: 1/20 or 0.05.

3 0

3 years ago

.<br> Evaluate the series 1 + 0.1 + 0.01 + . . .

EastWind [94]

We can employ a simple repeated decimal trick:

$x=1.111\ldots$

$0.1x=0.111\ldots$

$\implies x-0.1x=1\implies0.9x=1\implies x=\dfrac1{0.9}=\dfrac{10}9$

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Alternatively, we can compute the partial sum of the series.

$\displaystyle S_n=\sum_{k=0}^n\dfrac1{10^k}$

$S_n=1+0.1+0.01+\cdots+\dfrac1{10^n}$

$0.1S_n=0.1+0.01+0.001+\cdots+\dfrac1{10^{n+1}}$

$\implies S_n-0.1S_n=0.9S_n=1-\dfrac1{10^{n+1}}$

$\implies S_n=\dfrac{10}9-\dfrac9{10^n}$

As $n\to\infty$ , the second term vanishes and we're left with $\dfrac{10}9$ . Notice that this is really just a more formal version of the earlier trick.