Because they're all the same distance from the x axis on a coordinate plane. Also, remember that in quadrant I, all trig values are positive. In Q II, only sine and cosecant are positive. In Q III, only tangent and cotangent are positive. In Q IV, only cosine and secant are positive. Think of it as <u>A</u>ll <u>S</u>tudents <u>T</u>ake <u>C</u>alculus.
The slope of the line is 2
V(1)=65t+280 answer 65 gallons per minute
Now, the cosecant of θ is -6, or namely -6/1.
however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.
we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now
![\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}](https://tex.z-dn.net/?f=%5Cbf%20csc%28%5Ctheta%29%3D-6%5Cimplies%20csc%28%5Ctheta%29%3D%5Ccfrac%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%7B%5Cstackrel%7Bopposite%7D%7B-1%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%0A%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%5Csqrt%7B6%5E2-%28-1%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B35%7D%3Da%5Cimplies%20%5Cstackrel%7BIV~quadrant%7D%7B%2B%5Csqrt%7B35%7D%3Da%7D)
recall that
![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad\qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \\\\\\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \\\\\\ % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%5Cqquad%20%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cotangent%0Acot%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bopposite%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20cosecant%0Acsc%28%5Ctheta%29%3D%5Ccfrac%7Bhypotenuse%7D%7Bopposite%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20secant%0Asec%28%5Ctheta%29%3D%5Ccfrac%7Bhypotenuse%7D%7Badjacent%7D)
therefore, let's just plug that on the remaining ones,
![\bf sin(\theta)=\cfrac{-1}{6} \qquad\qquad cos(\theta)=\cfrac{\sqrt{35}}{6} \\\\\\ % tangent tan(\theta)=\cfrac{-1}{\sqrt{35}} \qquad \qquad % cotangent cot(\theta)=\cfrac{\sqrt{35}}{1} \\\\\\ sec(\theta)=\cfrac{6}{\sqrt{35}}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7B-1%7D%7B6%7D%0A%5Cqquad%5Cqquad%20%0Acos%28%5Ctheta%29%3D%5Ccfrac%7B%5Csqrt%7B35%7D%7D%7B6%7D%0A%5C%5C%5C%5C%5C%5C%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7B-1%7D%7B%5Csqrt%7B35%7D%7D%0A%5Cqquad%20%5Cqquad%20%0A%25%20cotangent%0Acot%28%5Ctheta%29%3D%5Ccfrac%7B%5Csqrt%7B35%7D%7D%7B1%7D%0A%5C%5C%5C%5C%5C%5C%0Asec%28%5Ctheta%29%3D%5Ccfrac%7B6%7D%7B%5Csqrt%7B35%7D%7D)
now, let's rationalize the denominator on tangent and secant,
Answer:
672
Step-by-step explanation:
Okay so, the surface area of a figure is just all the areas of the faces added together. The front face ( 9x16 ) would be 144. The back face would be 12x9 or 109. The base would be 20x9 or 180. Now, this is where it might get tricky. The triangles will be 12x20x0.5 or 1/2. The answer for that would be 120. Now since there are two, multiply the 120 by two. ( Technically, since there are two of the same triangles you would not need to multiply by 0.5, but just to be safe i'll do it here. ) Finally, add all the areas together. The final answer is 672. Hope this helps!