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Mathematics, 21.06.2019 16:00
Let the closed interval [a , b] be the domain of function f. the domain of f(x - 3) is given by (a) the open interval (a , b) (b) the closed interval [a , b] (c) the closed interval [a - 3 , b - 3] (d) the closed interval [a + 3 , b + 3]
Answers: 2
Mathematics, 21.06.2019 20:30
Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x^2 + 3x + 4, [−1, 1] no, f is continuous on [−1, 1] but not differentiable on (−1, 1). no, f is not continuous on [−1, 1]. yes, f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous and differentiable on . there is not enough information to verify if this function satisfies the mean value theorem. yes, it does not matter if f is continuous or differentiable; every function satisfies the mean value theorem.
Answers: 1
Mathematics, 21.06.2019 21:30
Zack notices that segment nm and segment pq are congruent in the image below: which step could him determine if δnmo ≅δpqr by sas? (5 points) segment mo ≅ segment qr segment on ≅ segment qp ∠ n ≅ ∠ r ∠ o ≅ ∠ q
Answers: 3
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