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Lana71 [14]
3 years ago
10

35 POINTS! Arrange the steps in the correct sequence: x + y = -2 2x - 3y = -9

Mathematics
1 answer:
lubasha [3.4K]3 years ago
3 0

\left\{\begin{array}{ccc}x+y=-2\\2x-3y=-9\end{array}\right\\\\\text{multiply the first equation by 3}\\3(x+y)=3(-2)\qquad|\text{use distributive property}\\3x+3y=-6\\\\\underline{+\left\{\begin{array}{ccc}3x+3y=-6\\2x-3y=-9\end{array}\right}\qquad\text{add 3x+3y=-6 to 2x-3y=-9, and solve for x}\\.\qquad5x=-15\qquad|:5\\.\qquad x=-3\\\\\text{Substitute the value of x in the first equation (x+y=-2)}\\\\-3+y=-2\qquad|+3\\y=1\\\\\text{The solution for the system of equations is (-3, 1)}

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Answer:

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Evaluate for x=4 and y=-3 write in simplest form<br> 2y/x2
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Hello from MrBillDoesMath!

Answer:

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A business was valued at £80000 at the start of 2013. In 5 years the value of this business raised to £95000. this is equivalent
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the yearly increase of x% assumes is compounding yearly, so let's use that.

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\£95000\\ P=\textit{original amount deposited}\dotfill &\£80000\\ r=rate\to r\%\to \frac{r}{100}\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &5 \end{cases}

95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r

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2 years ago
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