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3 years ago

Find the reference angle of 14 pie over 11

Mathematics

1 answer:

saw5 [17]3 years ago

8 0

Answer:

Step-by-step explanation:

reference angle=(14 π)/11-π=(14-11)π/11=3π/11

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A trimaran boat, with a top speed of 55 knots, won the tournament cup in 2010. Its top speed is 6 knots more than 3.5 times the

slavikrds [6]

Answer:

14 knots

Step-by-step explanation:

We know the speed (55 knots) is 6 more than 3.5 times the winner of the monohull in the 2007 tournament.

So let's set the variable for the speed of the monohull to m. Now, let's set up an equation:

$55 = 6 + (3.5 * m)$

We can solve this by first subtracting 6 from both sides to get:

$49 = 3.5 * m$

Dividing both sides by 3.5 we get:

$14 = m$

That means that the top speed of the monohull in the 2007 tournament was 14 knots

4 0

2 years ago

HELP!!!! I need help

ahrayia [7]

Answer:

1 Tenths = 10 Hundredths

10.74 - 10.60 = 0.14

0.14 = 14 hundredths

Thus, you need 10 ones, 6 tenths and 14 hundredths to make 10.74.

3 0

2 years ago

QUICK PLS I ONLY HAVE 7 MINS LEFT

Vikentia [17]

Answer:

first one

Step-by-step explanation:

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2 years ago

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PLEASE HELP!!!!! WILL GIVE BRAINLIEST AND GOOD RATING!! NO LINKS!

natta225 [31]

Answer:

Answer is B hope this helps :)

Step-by-step explanation:

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2 years ago

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For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\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}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\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{function period or frequency}\\
\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 in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago