exists and is bounded for all . We're told that . Consider the interval [0, 3]. The mean value theorem says that there is some such that
Since , we have
so 19 is the largest possible value.
The midpoint is exactly in the middle
meaning
VW = WX
3x+7 = x + 25
solve from x
3x + 7 = x + 25
subtract 7 from both sides
3x = x + 18
subtract x from both sides
2x = 18
divide both sides by 2
x = 9
Now we have x = 9 but we aren't done.
plug what we found into the expressions
3x + 7
3(9) + 7
27 + 7
34
x + 25
9 + 25
34
34 + 34
68
The length of VX is 68.
Or option D.) 68
Hope this helps
Answer:
f'(x) > 0 on and f'(x)<0 on
Step-by-step explanation:
1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:
To find its decreasing interval :
2) Then let's find the critical point of this function:
2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x''≈0.37 for e≈2.72
3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.
14 = 2•7 and 35 = 5•7, so the GCF of 14k and 35 is 7.
14k + 35 = 2•7k + 5•7 = 7 (2k + 5)
(Remember that the distributive property says a (b + c) = ab + ac.)
Answer:
how many cans?
Step-by-step explanation: