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natita [175]
3 years ago
10

Solve for d: 2d+1/6=6(1/3d+1)

Mathematics
2 answers:
taurus [48]3 years ago
7 0
2d+1/6=2d+6 (I distributed the 6 to what was in the parentheses)
2d-2d=6-1/6 (I subtracted 2d from both sides to get the variables on one side and               1/6 from both sides to get the variables by themselves)
0=35/6 (combined like terms)
0=35/6 is an untrue statement so x doesn't exist

Colt1911 [192]3 years ago
3 0
2d + 1/6 = 6(1/3d + 1)
2d + 1/6 = 2d + 6
2d - 2d = 6 - 1/6
0d = 35/6

Ans: \O
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\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 \\
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\\ \quad \\
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\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}
\\\\
\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}\\\\
\end{array}

\bf \begin{array}{llll}


\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, with that template above in mind, let's see this one

\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}


A=3, B=1,  shrunk by AB or 3 units, about 1/3
C=2,          horizontal shift by C/B or 2/1 or just 2, to the left
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check the picture below

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