Problem 5)
The bases are the same, so we add the exponents when multiplying something like this out. We have -5+7 = 2 as the final exponent meaning that 5^2 = 25
<h3>Answer: Choice B) 25</h3>
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Problem 6)
We can multiply the exponents (t^8)^2 = t^(8*2) = t^16
Or we can expand out (t^8)^2 to get (t^8)*(t^8), then add the exponents from here to get t^(8+8) = t^16. Either way we get the same answer
<h3>Answer: Choice C) t^16</h3>
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Problem 7)
The negative exponent indicates we will apply the reciprocal to get that exponent to a positive number. The rule is x^(-k) = 1/(x^k)
So 5^(-4) = 1/(5^4) = 1/625 and (t^3)^(-4) = t^(3*(-4)) = t^(-12) = 1/(t^12)
Put together, we end up with 1/(625t^12)
<h3>Answer: Choice C) 1 over 625t^12</h3>
Answer:
sin(x)−1/3(sin3(x))+C
Step-by-step explanation:
use u substitution method to integrate
The answer would be C because they say if 3 new students arrived each year.
