Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis
and lying on the parabola.
y=5-x^2
1 answer:
<span>let 2x be the length of rectangw where x is value of x of point on parabola width is represented as y is the length.
Area = 2x*y = 2x (5-x^2) = 10x -2x^3
maximize Area by finding x value where derivative is zero
dA/dx = 10 -6x^2 = 0
--> x = sqrt(5/3)
optimal dimensions: length = 2sqrt(5/3) width = 10/3</span>
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The expression that could best describe the statement would be " 40 n + 20
260.
For letter the answer is 2
for letter b the answer is 1
Answer:
25
Step-by-step explanation:
5:9
x:40
what is the realationship between 9 and 40? -9x5=40
so 5x5=25 so x=25
5:9x5:5=25:40
3.5h+10=5.25h-11
21=1.75h
12=h
3.5(12)+10=5.25(12)-11
52=52
Answer:
2
Step-by-step explanation:
4x+3=x+9
4x-x=9-3
3x=6
x=6/3
x=2