Find the zeros of h(x)=2(x-11)^2 -8

To find the zeros, we must find when 0=2(x-11)²-8. Adding 8 to both sides using the additive property of equality, we have 8=2(x-11)². Similarly, we can divide 2 from both sides to get 4=(x-11)². Finding the square root of both sides, we have 2=|(x-11)| (note that we include absolute values because the square root must be ≥ 0. If x-11=|x-11|, then 2=x-11 and 13=x. If x-11=-|x-11|, we find that 2=-(x-11)=11-x, and 2+x=11, so x=9, making our zeros 13 and 9.

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step-by-step explanation:

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2