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Fed [463]
4 years ago
5

Last year, Shen opened an investment account with $6600 . At the end of the year, the amount in the account had increased by 7.5

% . How much is this increase in dollars? How much money was in his account at the end of last year?
Mathematics
1 answer:
rewona [7]4 years ago
7 0
6,600 x 7.5% = $495

$495 + 6,600 = 7,095

$495 increase in dollars
$7,095 was in his account at the end of the last year

Hope this helps! =)
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which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the l
victus00 [196]
One equation for this would be

y = \frac{41}{16} x-\frac{55}{8}

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m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}
\\
\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})
\\
\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})
\\
\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

y-y_1=m(x-x_1)
\\
\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})
\\
\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}
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\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}
\\
\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}
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\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}
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\\y=\frac{41}{16}x-\frac{550}{80}
\\
\\y=\frac{41}{16}x-\frac{55}{8}
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