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pshichka [43]
2 years ago
8

What is the value of z for the equation 1/4z = –7/8 + 1/8z? –3 –7 3 7

Mathematics
1 answer:
DaniilM [7]2 years ago
6 0
The first step for solving this equation is to determine the defined range.
\frac{1}{4z} = - \frac{7}{8} +  \frac{1}{8z}, z ≠ 0
Move the expression\frac{1}{8z} to the left side of the equation and change its sign.
\frac{1}{4z} - \frac{1}{8z} = - \frac{7}{8}
Now we need to write all numerators above the least common denominator of 8z. This will change the equation to the following:
\frac{1}{8z} = - \frac{7}{8}
Simplify the equation using cross multiplication.
8 = -56z
Switch the sides of the equation.
-56z = 8
Divide both sides of the equation by -56.
z =  -\frac{1}{7}, z ≠ 0
Lastly,, check if the solution is in the defined range to get your final answer.
z = -\frac{1}{7}
Let me know if you have any further questions.
:)
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8 0
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nlexa [21]

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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
2 years ago
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kvv77 [185]
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jenyasd209 [6]
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6 0
2 years ago
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