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deff fn [24]
3 years ago
13

A 170-inch board is cut into two pieces. One piece is four times the length of the other. Find the length of the shorter piece.

The shorter piece is ___ inches long. PLS HELP ILL GIVE BRAINLIST!!!!​
Mathematics
1 answer:
matrenka [14]3 years ago
4 0

Answer:

34 in

Step-by-step explanation:

Short piece= x

Longer piece= 4x

The shorter piece and longer piece = 170 in, so 5x= 170

To find "x", divide both sides by 5

5x/5= x

170/5=34

x=34

Hope this helps! :)

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If by this you mean perpendicular bisectors, then they would intersect at one point.
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What are the phase shift and period for the function y = 5cos[9(θ − 30°)] − 2?
svetoff [14.1K]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}
\\\\
f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\
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% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\
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\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\
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\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{  D}}\\
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so, with that template in mind, let's see

\bf y=5cos[9(\theta-30^o)]-2\implies y=5cos\left[9(\theta-\frac{\pi }{6})  \right]-2
\\\\\\
y=5cos(9\theta-\frac{3\pi }{2})-2\\\\
-------------------------------\\\\
\begin{array}{lllcclllll}
y=&5cos(&9\theta&-\frac{3\pi }{2} )&-2\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
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-------------------------------\\\\

\bf \textit{horizontal/phase shift}\implies \cfrac{C}{B}\implies \cfrac{-\frac{3\pi }{2}}{9}\implies \cfrac{-\frac{3\pi }{2}}{\frac{9}{1}}
\\\\\\
-\cfrac{3\pi }{2}\cdot \cfrac{1}{9}\implies -\cfrac{\pi }{6}\\\\
-------------------------------\\\\
period\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{9}
6 0
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</span>
8 0
3 years ago
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