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olchik [2.2K]
3 years ago
7

Find three consecutive odd numbers whose sum is 303

Mathematics
2 answers:
CaHeK987 [17]3 years ago
8 0
n;\ n+2;\ n+4-three\ consecutive\ odd\ numbers\\\\n+(n+2)+(n+4)=303\\\\3n+6=303\\\\3n=303-6\\\\3n=297\ \ \ \ /:3\\\\n=99\\\\Answer:99;\ 101;\ 103.
Roman55 [17]3 years ago
5 0
99 + 101 + 103 = 303
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we are given

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vagabundo [1.1K]

Answer:

Considering the given equation y = log_{3}x\\

And the ordered pairs in the format (x, y)

I don't know if it is log of base 3 or 10, but I will assume it is 3.

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x=\frac{1}{3}

y=a_{0}

y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1

So the ordered pair will be (\frac{1}{3}, -1 )

For (1, a_{1} )

x=1

y=a_{1}

y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0

So the ordered pair will be (1, 0 )

For (3, a_{2} )

x=3

y=a_{2}

y = log_{3}x\\y = log_{3}3\\y = 1

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For (9, a_{3} )

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y=a_{3}

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So the ordered pair will be (9, 2 )

For (27, a_{4} )

x=27

y=a_{4}

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So the ordered pair will be (27, 3 )

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x=81

y=a_{5}

y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4

So the ordered pair will be (81, 4 )

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4 years ago
Find a point on the graph of the function y=f(x)+3
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In the general case, it is (x, y+3), where y = f(x).
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