amplify the following expressions applying the laws of exponents.
(i) 7 to the power 1/3 x 7 to the power−1/2 (ii)(1to the power3 + 2 to the power3 + 3to the power 3) whole into power 1/2
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F1(t) is the cubic function;
F2(t) is the exponential function;
F3(t) is the quadratic function.
Here are some observations which lead to these conclusions.
All power functions ktp with p 0 are zero for t = 0. Thus F2 cannot be a power function.
Assuming that F1(t) = ktn with n = 2 or n = 3, we can evaluate F1(1) to obtain k. In this case, k = 2.5.
Since F1(t) = (2.5)tn with n = 2 or n = 3, we can figure out n by evaluating at t = 1.2.
(2.5)(1.2)2 = 3.6 and (2.5)(1.2)3 = 4.32,
so n = 3.
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