The point (-3/5,y) in the third quadrant corresponds to angle θ on the unit circle.

1 answer:

7 0

Sec(theta) = 1 / cos (theta) = hypotenuse / x -coordinate

hypotenuse = 1 (because it is the radius of the unit circle)

sec (theta) = 1 / (-3/5) = - 5/3

cot (theta) = 1 / tan(theta) = x-coordinate / y - coordinate

cot (theta) = -3/5 / y

y^2 + (-3/5)^2 = 1 => y^2 = 1 - 9/25 = 16/25 = y = +/- 4/5

Third quadrant => y = -4/5

=> cot (theta) = (-3/5) / (-4/5) = 3/4

You might be interested in

Which equation is equivalent to the given equation?<br> x² - 6x = 8

Orlov [11]

Answer x^2-6x-8=0

I hope this helps

3 0

2 years ago

Hey so im also looking for the first answer and the choices were “ is not “ and “ is “

dexar [7]

so the investigator found the skid marks were 75 feet long hmmm what speed will that be?

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ d=75 \end{cases}\implies s=\sqrt{30(0.7)(75)}\implies s\approx 39.69~\frac{m}{h}$

nope, the analysis shows that Charlie was going faster than 35 m/h.

now, assuming Charlie was indeed going at 35 m/h, then his skid marks would have been

$s=\sqrt{30fd}~~ \begin{cases} f=\stackrel{friction}{factor}\\ d=\stackrel{skid}{feet}\\[-0.5em] \hrulefill\\ f=\stackrel{dry~day}{0.7}\\ s=35 \end{cases}\implies 35=\sqrt{30(0.7)d} \\\\\\ 35^2=30(0.7)d\implies \cfrac{35^2}{30(0.7)}=d\implies 58~ft\approx d$