Answer:

Step-by-step explanation:
GIVEN: A farmer has
of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is
.
TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.
SOLUTION:
Let the length of rectangle be
and
perimeter of rectangular pen 


area of rectangular pen 

putting value of 


to maximize 



but the dimensions must be lesser or equal to than that of barn.
therefore maximum length rectangular pen 
width of rectangular pen 
Maximum area of rectangular pen 
Hence maximum area of rectangular pen is
and dimensions are 
SORRY IM LATE
TRY 5814
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please mark brainliest! <3
1. Number that adds up to -12 and multiplied by the same # that gives u -13. Set it up like this because both are negative so, u need 1 negative and 1 positive.
(x-13)(x+1)
2. Same rule applies. GCF
8y(4y+1)
3. (x+3)(x-3)
im sorry but i cant seem to open your image?
3/2 and 9/10. Look at the denominators. 2 times what number equals 10. That number would be 5. So if you multiply the first fractions denominator by 5 you get 10. Do the same to the top. you get a new fracrion which is 15/10. Add normally. 15/10 + 9/10 = 24/10. In lowest terms it is 2 2/5 (2 wholes and 2 out of 5)