A rectangle is 18 inches long and 9 inches wide. Find it’s area.

Mathematics

1 answer:

galina1969 [7]3 years ago

4 0

Answer:

162 square inch

Step-by-step explanation:

Area = Length * Width

Area = $18*9=162$

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Find the area of the largest rectangle (with sides parallel to the coordinate axes) that can be inscribed in the region enclosed

garik1379 [7]

F(x) = 18-x^2 is a parabola having vertex at (0, 18) and opening downwards.
g(x) = 2x^2-9 is a parabola having vertex at (0, -9) and opening upwards.
By symmetry, let the x-coordinates of the vertices of rectangle be x and -x => its width is 2x.
Height of the rectangle is y1 + y2, where y1 is the y-coordinate of the vertex on the parabola f and y2 is that of g.
=> Area, A
= 2x (y1 - y2)
= 2x (18 - x^2 - 2x^2 + 9)
= 2x (27 - 3x^2)
= 54x - 6x^3
For area to be maximum, dA/dx = 0 and d²A/dx² < 0
=> 54 - 18x^2 = 0
=> x = √3 (note: x = - √3 gives the x-coordinate of vertex in second and third quadrants)

d²A/dx² = - 36x < 0 for x = √3
=> maximum area
= 54(√3) - 6(√3)^3
= 54√3 - 18√3
= 36√3.

4 0

2 years ago

In another baseball division, one team had a winning percentage of 0.444 repeating What fraction of their games did this team wi

UkoKoshka [18]

Answer:

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