The math teachers inspect the homework assignments from a random sample of 50 students at the school. Only 68% of the students c
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One thing to bear in mind with trigonometric expressions is that, the exponent location is next to the function name, instead of the whole expression, though it applies to the whole thing, namely cos²(x), is really [ cos(x) ]², and that matters for using the chain rule.
![\bf y=cos^2(x)-sin^2(x)\implies y=[cos(x)]^2-[sin(x)]^2 \\\\\\ \cfrac{dy}{dx}=\stackrel{chain~rule}{2cos(x)\cdot -sin(x)}~~-~~\stackrel{chain~rule}{2sin(x)\cdot cos(x)} \\\\\\ \cfrac{dy}{dx}=-2cos(x)sin(x)-2cos(x)sin(x) \\\\\\ \cfrac{dy}{dx}=-4cos(x)sin(x)](https://tex.z-dn.net/?f=%5Cbf%20y%3Dcos%5E2%28x%29-sin%5E2%28x%29%5Cimplies%20y%3D%5Bcos%28x%29%5D%5E2-%5Bsin%28x%29%5D%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Cstackrel%7Bchain~rule%7D%7B2cos%28x%29%5Ccdot%20-sin%28x%29%7D~~-~~%5Cstackrel%7Bchain~rule%7D%7B2sin%28x%29%5Ccdot%20cos%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-2cos%28x%29sin%28x%29-2cos%28x%29sin%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D-4cos%28x%29sin%28x%29)
The value of x from the given conditions is 35.
The line TV and RI intersect at E.
The measure of IVE=50° and the measure of VIE=70°.
Since, the points V, I and E makes a triangle.
Therefore, the total angle inside the triangle is 180°.
Hence,
∠IVE + ∠VIE + ∠VEI = 180°
50°+70°+ ∠VEI = 180°
∠VEI = 180°-50°-70°
∠VEI = 60°
Also the angles ∠VEI and the angle ∠RET are congruent.
Therefore, ∠VEI = ∠RET
60° = 2x-10
2x = 70
x = 70/2
x = 35.
The lines TV and RI intersect at E. If the measure of IVE=50, the measure of VIE=70 and the measure of RET=2X-10. Then the value of x = 35.
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