![\bf 2sec^2(\theta )-tan(\theta )-3=0 \\\\\\ 2[1+tan^2(\theta )]-tan(\theta )-3=0\implies 2+2tan^2(\theta )-tan(\theta )-3=0 \\\\\\ 2tan^2(\theta )-tan(\theta )-1=0\implies [2tan(\theta )+1][tan(\theta )-1]=0\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%202sec%5E2%28%5Ctheta%20%29-tan%28%5Ctheta%20%29-3%3D0%0A%5C%5C%5C%5C%5C%5C%0A2%5B1%2Btan%5E2%28%5Ctheta%20%29%5D-tan%28%5Ctheta%20%29-3%3D0%5Cimplies%202%2B2tan%5E2%28%5Ctheta%20%29-tan%28%5Ctheta%20%29-3%3D0%0A%5C%5C%5C%5C%5C%5C%0A2tan%5E2%28%5Ctheta%20%29-tan%28%5Ctheta%20%29-1%3D0%5Cimplies%20%5B2tan%28%5Ctheta%20%29%2B1%5D%5Btan%28%5Ctheta%20%29-1%5D%3D0%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
bear in mind that tangent is sine/cosine or y/x
for the tangent to be negative, the signs of "y" and "x" must differ, and that happens only on the II and IV quadrants
and for the tangent to be positive, the signs must be same, and that's only on I and III quadrants.
This isnt 30 points this is 15
<span><span>2.5−<span>4<span>(<span>1.5−10</span>)</span></span></span>+3</span><span>=39.5</span>
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Answer:
635
Step-by-step explanation:
so basically you need to make an equation to find out what number, when put through all of those operations, will still equal the same number in the end. the way you do that is with an equation. so let's say our starting number is <em>x</em>, since we don't know its value yet.
the final equation is (((((((x/5)+6)/7)-8)9)+114)3)-4 = x
basically it's just x, all those different operations in that order, which is why you need parentheses, equals to x again because it's the same in the end. when you solve that you get 635
