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-BARSIC- [3]
3 years ago
7

Question: is 1 > 0.99999999...?Prove algebraically.

Mathematics
1 answer:
Crazy boy [7]3 years ago
6 0
No.  We claim that 1=0.\overline{9} and use algebra to prove the statement.

Let x=0.\overline{9}.  Multiply this by ten to get 10x=9.\overline{9}.  Subtract the initial equation to give 9x=9 and divide by 9 to see that x=1.  Substituting into the original equation gives 1=0.\overline{9}, proving the desired statement.
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14x. ....................

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Two Trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that spe
sashaice [31]
Let s = northbound train
Then
2s = southbound train
:
Distance = time * speed
4s + 4(2s) = 600
:
4s + 8s = 600
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12s = 600
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s = 600/12
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s = 50 mph is the northbound train
Then
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Check:
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3 0
3 years ago
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I need help with this question please
Gemiola [76]

Answer:

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Step-by-step explanation:

7 0
2 years ago
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At this rate, would a person your age (18 years old) have contributed a ton of garbage? On average, how long does it take for ea
Sloan [31]

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6 0
3 years ago
Describe the translation of the function from the parent function f(x) = x².
kkurt [141]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
f(x)=&{{  A}}\sqrt{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}}(\mathbb{R})^{{{  B}}x+{{  C}}}+{{  D}}
\\ \quad \\
f(x)=&{{  A}} sin\left({{ B }}x+{{  C}}  \right)+{{  D}}
\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}
\\\\

\end{array}\\

\bf \begin{array}{llll}
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\
\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
\bullet \textit{ vertical shift by }{{  D}}\\
\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
\end{array}


now, notice the template above... now let's see your function \bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}

so.. what do you think was the shift then?
8 0
3 years ago
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