H(x) = 0.15x - 0.19
p(x) = 0.29x - 0.16
(p · h)(-8) = (0.29(-8) - 0.16)(0.15(-8) - 0.19)
(p · h)(-8) = (-2.32 - 0.16)(-1.2 - 0.19)
(p · h)(-8) = (-2.48)(-1.39)
(p · h)(-8) = 3.4472
It is approximately equal to 3.
Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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The formula of the midpoint of HE:
We have H(0; 0) and E(2a; 2a). Substitute:
Answer:
scale is 4...... ........
