Which is True?????/ Value of x???
1 answer:
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See the attached figure to better understand the problem
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
let
L-----> length side of the cuboid
W----> width side of the cuboid
H----> height of the cuboid
we know that
One edge of the cuboid has length 2 cm-----> <span>I'll assume it's L
so
L=2 cm
[volume of a cuboid]=L*W*H-----> 2*W*H
40=2*W*H------> 20=W*H-------> H=20/W------> equation 1
[surface area of a cuboid]=2*[L*W+L*H+W*H]----->2*[2*W+2*H+W*H]
100=</span>2*[2*W+2*H+W*H]---> 50=2*W+2*H+W*H-----> equation 2
substitute 1 in 2
50=2*W+2*[20/W]+W*[20/W]----> 50=2w+(40/W)+20
multiply by W all expresion
50W=2W²+40+20W------> 2W²-30W+40=0
using a graph tool------> to resolve the second order equation
see the attached figure
the solutions are
13.52 cm x 1.48 cm
so the dimensions of the cuboid are
2 cm x 13.52 cm x 1.48 cm
or
2 cm x 1.48 cm x 13.52 cm
<span>Find the length of a diagonal of the cuboid
</span>diagonal=√[(W²+L²+H²)]------> √[(1.48²+2²+13.52²)]-----> 13.75 cm
the answer is
the length of a diagonal of the cuboid is 13.75 cm
At the start, the tank contains
(0.25 lb/gal) * (100 gal) = 25 lb
of sugar. Let be the amount of sugar in the tank at time
. Then
.
Sugar is added to the tank at a rate of <em>P</em> lb/min, and removed at a rate of
and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,
Separate variables, integrate, and solve for <em>S</em>.
Use the initial value to solve for <em>C</em> :
The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time
has passed. At this time, we want the tank to contain
(0.5 lb/gal) * (5 gal) = 2.5 lb
of sugar, so we pick <em>P</em> such that