The sign of f on the interval -5/3 is negative
<h3>How to determine the sign?</h3>
The function is given as:
f(x)=(2x−1)(3x+5)(x+1)
The zeros of the function are
x= -5/3, x= -1 and x= 1/2
Next, we plot the graph of the function
At x = -5/3, the function approaches negative infinity
This means that the sign of f on the interval -5/3 is negative
Read more about function intervals at:
brainly.com/question/27831985
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the answer is C. because it is the only one that does not contain a common factor :)
Answer:
B
Step-by-step explanation:
Recall that functions are defined only if for each value in the domain produces one and only one value in the range.
If we view the relations in the questions as x-y coordinates, this means that for every x-value, you can only have one y-value
Lets evaluate the options:
A) we can see that for x = -3, this gives 2 possible values for y i.e (-3,4) and (-3,8) (hence this is not a function)
C) we can see that for x = 3, this gives 2 possible values for y i.e (3,-8) and (3,8) (hence this is not a function)
D) we can see that for x = -3, this gives 2 possible values for y i.e (-3,4) and (-3,8) (hence this is not a function)
the only choice where this doesn't occur is choice B
Answer:
f(x) = sin(x) + 4 Let me know if you don't see why.
Step-by-step explanation:
As much as I hate it I think it might be best to check each one.
sin(x) has a point at (0,0) so that's not right.
sin(x)+4 has (0,4) as a point and a minimum at (3*pi/2+2*pi*n,3) where n is some integer. if we have n = 0 then it becomes (3*pi/2, 3) So that looks like our answer.
cos(x) + 3 has a point at (0,4) then minimums at (pi+2*pi*n, 2) the y coordinate is wrong so this won't work
-3sin(x) has a point at (0,0) so that's wrong
4cos(x) has a point at (0,4) then minimums at (pi+2*pi*n, -4) which again has the wrong y value so this is wrong.
Let me know if you don't understand how I got the results I did, I would be happy to explain.
Answer:
20
Step-by-step explanation:
