subject
Mathematics, 02.08.2021 05:10 AquaTyrant5054

Please help. Which of these could be the graph of F(x) = In x + 3?

A. Graph A
B. Graph B
C. Graph C
D. Graph D


Please helpppp.

Which of these could be the graph of F(x) = In x + 3?
A. Graph A
B. Graph B
C. Gr

ansver
Answers: 3

Another question on Mathematics

question
Mathematics, 21.06.2019 19:30
John checked his watch and said that it is thursday, 7 am. what will the day and time be 2006 hours plzzz i will give you 100 points
Answers: 1
question
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
question
Mathematics, 21.06.2019 22:00
Which statements describe the solutions to the inequality x< -20 check all that apply. there are infinite solutions. each solution is negative. each solution is positive. the solutions are both positive and negative. the solutions contain only integer values. the solutions contain rational number values.
Answers: 1
question
Mathematics, 21.06.2019 23:00
The equation represents the function f, and the graph represents the function g. f(x)=3(5/2)^x determine the relationship between the growth factors of f and g. a. the growth factor of g is twice the growth factor of f. b. the growth factor of f is twice the growth factor of g. c. the growth factor of f is 2.5 times the growth factor of g. d. the growth factor of f is the same as the growth factor of g.
Answers: 3
You know the right answer?
Please help. Which of these could be the graph of F(x) = In x + 3?

A. Graph A
B....
Questions
question
Chemistry, 31.07.2019 11:30