Answer:
A' = 6(3t +1)
Step-by-step explanation:
The area is the square of the side length, so ...
A = (3t +1)²
The derivative with respect to t is then
A' = 2(3t +1)·(3) = 6(3t +1)
If you like, you can multiply out the area expression first, then differentiate.
A = 9t² +6t +1
A' = 18t +6
This can be factored, if you like, to ...
A' = 6(3t +1)
1. Find the equation of the line AB. For reference, the answer is y=(-2/3)x+2.
2. Derive a formula for the area of the shaded rectange. It is A=xy (where x is the length and y is the height).
3. Replace "y" in A=xy with the formula for y: y= (-2/3)x+2:
A=x[(-2/3)x+2] This is a formula for Area A in terms of x only.
4. Since we want to maximize the shaded area, we take the derivative with respect to x of A=x[(-2/3)x+2] , or, equivalently, A=(-2/3)x^2 + 2x.
This results in (dA/dx) = (-4/3)x + 2.
5. Set this result = to 0 and solve for the critical value:
(dA/dx) = (-4/3)x + 2=0, or (4/3)x=2 This results in x=(3/4)(2)=3/2
6. Verify that this critical value x=3/2 does indeed maximize the area function.
7. Determine the area of the shaded rectangle for x=3/2, using the previously-derived formula A=(-2/3)x^2 + 2x.
The result is the max. area of the shaded rectangle.
Answer:
none of the above
Step-by-step explanation:
because 5p + 3p = 8p and positive integers × negative integers = negative integers.
Answer:
the answer is A
Step-by-step explanation:
casue i said so
