Radical form for 112 would be 4{7 and then I think you can just search up how to reduce the 4 to 7 i don’t have the symbol btw
Answer:
part 1: white blood cells
part 2: C=3h(6)
Step-by-step explanation:
I hope this is correct pls let me
know in the comments below if i am correct
Answer: Not a function
<u>Step-by-step explanation:</u>
The graph fails the vertical line test. In other words, when you draw a vertical line anywhere through the curve, the line passes through two points.
Bonus: If you do not have a graph, it is not a function if there is one x-value that has two different y-values.
For example, If (1, 2) and (1, -2) lie on the curve, then it is NOT a function because the x-value (1) has two different y-values (2 & -2).
Answer:
An apple, potato, and onion all taste the same if you eat them with your nose plugged
Step-by-step explanation:
E) 2
Remember that the first derivative of a function is the slope of the function at any specified point. We've been told that f(0) = -5 and that f'(x) is always less than or equal to 3. So let's look at the available options and see what the average slope would have to be in order to get the specified value of f(2).
A) -10: (-10 - -5)/(2 - 0) = -5/2 = -2.5
B) -5: (-5 - -5)/(2 - 0) = 0/2 = 0
C) 0: (0 - -5)/(2 - 0) = 5/2 = 2.5
D) 1: (1 - -5)/(2 - 0) = 6/2 = 3
E) 2: (2 - -5)/(2 - 0) = 7/2 = 3.5
Now taking into consideration the mean value theorem, the value of the function f'(x) has to have the value equal to the average slope between the two points at at least one point between the two given values. For options A, B, C, and D it's possible for f'(x) to return values that make that slope possible. However, for option E, the mean value theorem indicates that f'(x) has to have the value of 3.5 for at least 1 point between x=0 and x=2. And since we've been told that f'(x) is less than or equal to 3 for all possible values of x, that is in conflict and f(2) can not have the value of 2.
