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sweet-ann [11.9K]
3 years ago
10

Work out the difference between 5/9and4/9

Mathematics
1 answer:
tatiyna3 years ago
7 0

Answer:

1/9

Step-by-step explanation:

Subtract them

5/9-4/9=1/9 Answer

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Minchanka [31]
Does this mean:
1x3=3
4-1=3
2+1=3
5-2=3
6/2=3
?
8 0
3 years ago
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What are the period and phase shift for f(x) = 3 tan(4x + π)?
Y_Kistochka [10]
\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}}
\\ \quad \\

\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{  A}}|\\\\
\bullet \textit{ horizontal or "phase" 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{function period}\\
\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\
\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}


now.. with that template above in mind, let's see yours


\bf \begin{array}{lllllll}
f(x)=&3tan(&4x+&\pi )\\
f(x)=&3tan(&4x+&1\pi )&+0\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}

7 0
3 years ago
Find the area of the trapezoid.
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the trapezoid is actually on its side
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b=12.2
h=12.6

A=\frac{10.2+12.2}{2} *12.6\\
=\frac{22.4}{2} *12.6\\
=11.2*12.6\\
=141.12\\
3 0
3 years ago
PLEASE HELP now I really need to know​
Leona [35]

Answer:

Step-by-step explanation:

-2.5 + 5n - 1.5n

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Option B is the correct answer

6 0
2 years ago
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ollegr [7]

Answer:36.67 i think

Step-by-step explanation:

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