Author: Anonymous

Given:sin⁡(59.1o)=0.8581cos⁡(59.1o)=0.5135tan⁡(59.1o)=1.6709{“version”:”1.1″,”math”:”\begin{align*} \sin (59.1^o) &= 0.8581 \\ \cos(59.1^o) &= 0.5135 \\ \tan(59.1^o) &= 1.6709 \end{align*}”}determine sin⁡(30.9o)=?{“version”:”1.1″,”math”:”\sin (30.9^o) = ?”}

In which quadrants is y=csc⁡(x){“version”:”1.1″,”math”:”y = \csc (x) “} positive

Given:sin⁡(59.1o)=0.8581cos⁡(59.1o)=0.5135tan⁡(59.1o)=1.6709{“version”:”1.1″,”math”:”\begin{align*} \sin (59.1^o) &= 0.8581 \\ \cos(59.1^o) &= 0.5135 \\ \tan(59.1^o) &= 1.6709 \end{align*}”}determinecos⁡(−59.1o)=?{“version”:”1.1″,”math”:”\cos (-59.1^o) = ?”}

Find the length of an arc on a circle with a central angle of θ=π4{“version”:”1.1″,”math”:”\theta = \frac{\pi}{4}”} and radiusr=12{“version”:”1.1″,”math”:”r=12″}

The average time spent waiting at a stop light, in seconds, for a person living in a city with population, P, is given by the equation: w(P)=0.34ln⁡(P)+15{“version”:”1.1″,”math”:”w(P) = 0.34 \ln (P) + 15″} If Clarksville, TN has a population of 350,000, what is the average wait time at stoplights in Clarksville?

Evaluate the following trig function: csc⁡(π)=?{“version”:”1.1″,”math”:”\csc (\pi)=?”}

Use the above triangle to determine the trig ration fortan⁡A={“version”:”1.1″,”math”:”\tan A = “}

Use the above triangle to determine the trig ration forsec⁡B={“version”:”1.1″,”math”:”\sec B = “}

What is the graph off(x)=3×2x{“version”:”1.1″,”math”:”f(x) = 3 \times 2^x”}

Identify the structure within the (orange) circled area: