I need help on this math question

Mathematics

1 answer:

sveta [45]3 years ago

5 0

I remember foing this i think it is x>453

I am not 100 percebt sure

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If an instrument measures in units of 10, then any value between which two values would be measured as 80?

enyata [817]

Answer:

The answer is 75 and 85.

Step-by-step explanation:

If the instrument measures in units of 10, then the measurement could be 5 units on either side of 80 (80 − 5 = 75, 80 + 5 = 85). Any value between 75 and 85 would be measured as 80.

4 0

3 years ago

Complete the table by converting each decimal to a fraction.

AlekseyPX

if you've read those links already, you'd know what we're doing here.

we'll move the repeating part to the left-side of the dot, by multiplying by "1" and as many zeros as needed, or 10 at some power pretty much.

on 0.13 we need 100 to get 13.13.... and on 0.1234, we need 10000 to get 1234.1234....

$\bf 0.\overline{13}~\hspace{10em}x=0.\overline{13}
\\\\[-0.35em]
~\dotfill\\\\
\begin{array}{|lll|ll}
\cline{1-3}
&&\\
100\cdot 0.\overline{13}& = & 13.\overline{13}\\
100\cdot x&& 13 + 0.\overline{13}\\
100x&&13+x
\\&&\\
\cline{1-3}
\end{array}\implies \begin{array}{llll}
100x=13+x\implies 99x=13
\\\\
x=\cfrac{13}{99}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
0.\overline{1234}~\hspace{10em}x=0.\overline{1234}
\\\\[-0.35em]
~\dotfill$

$\bf \begin{array}{|ccc|ll}
\cline{1-3}
&&\\
10000\cdot 0.\overline{1234}&=&1234.\overline{1234}\\
10000\cdot x&=&1234+0.\overline{1234}\\
10000x&=&1234+x
\\&&\\
\cline{1-3}
\end{array}\implies \begin{array}{llll}
10000x=1234+x
\\\\
9999x=1234\implies x=\cfrac{1234}{9999}
\end{array}$