what is your question then?
2/5 is equivalent to 4/10 so at the end of February 4/10 was used up
And at the end of January, 3/10 was used
so 4/10 + 3/10 = 7/10
At the end of February, 7/10 of the wood was used
Answer:
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The volume of the open-topped box is equal to

where

substitute

Convert to expanded form

using a graphing tool
Graph the cubic equation
Remember that
The domain for x is the interval -----> (0,1)
Because
If x>1
then
the width is negative (W=2-2x)
so
The maximum is the point (0.46,3.02)
see the attached figure
therefore
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Answer:
-15e+14f-30
Step-by-step explanation:
First using distributive property to expand the first term:
-21e+14f-28+6e-2
Combining like terms:
-15e+14f-30