Answer:
length = 2w - 3
l = length
w = width
2(length) + 2(width) = perimeter
2(2w - 3) + 2(w) = 60
6w -6 =60
w = 11
len = 2(11) - 3
Step-by-step explanation:
Answer:
Richard will have $60,000 in his account in 20 years.
Step-by-step explanation:
(1) Multiply $250 x 12
(2) Multiply the answer of $250 x 12 which is 3000 by 20
(3) Final answer would be $60,000
IdentityProperty of Addition OR Zero Property
Answer:
We would have

where " l " is length, " w" is width and "h" is height.
Step-by-step explanation:
Step 1
Remember that
Surface area for a box with no top = 
where " l " is length, " w" is width and "h" is height.
Step 2.
Remember as well that
Volume of the box = 
Step 3
We can now use lagrange multipliers. Lets say,

and

By the lagrange multipliers method we know that

Step 4
Remember that

and

So basically you will have the system of equations

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

Then you would get

You can get rid of
from these equations and you would get

And from those equations you would get

Now remember the original equation

If we plug in what we just got, we would have

Answer:
look up in goog le it will tell u if not i can help what grade is this??
Step-by-step explanation: