$x^{3}=-1$
x-1=x+1
9-3/(1/3)+1
4(3x+1)=64
(3a²)$^{3}$
(3a²)3
Has a gradient value of 2 and passes through the point (3,10)
$\sqrt{1-x}$
2+2
By making an appropriate substitution, we find that the value of ∫1/5−1/5(5x+2)11dx
is:
Question 1Answer
A.
132,8603
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