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We need to use Newtons law F=ma going down the slope. We can see that the only forces acting in this direction are the component of the weight and friction, so we have that: F = Wsin30 - μR = 3a
( sec2(x))/((sec(x)+1)(sec(x)-1))Then, by the rule of 'difference of two squares', we know that this equals= (sec2(x))/(sec2(x)-1)= (sec2x/tan2x)s...
You cannot work with this equation in the current form so you must use identities to find an equivalent form that you can work with. It is known that tan(2A) = 2tan(A) / 1-tan2(A) so set this e...
We must first use the identity cosecθ = 1/sinθ. Now the equation becomes (1/sinθ)(3 cos 2θ+7)+11=0. Since we know that the question is asking for the answer in the form of asin2θ + bsinθ + c = ...
So let's start with taking the natural log on both sides of y=ax, giving us ln(y) = ln(ax). Using the laws of logarithms we can write this as ln(y) = xln(a).Next, we differentiate bo...
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