A bag contains 10 blue, 6 green, and 4 red marbles. You choose a marble with one hand and then a second marble with the other ha
2 answers:
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Remark
All Cavaliere's principle says is that a slanted cylinder (for example) has the same volume as a cylinder with perpendicular parallel lines that the cylinder is sitting on.
I little known or maybe not a well emphasized fact is that the base is the same all the way up the cylinder. The coin picture below shows that. The basic formula for a Cavaliere cylinder is the same as for an ordinary cylinder.
See picture below
Formula
V = B * H
Sub and Solve
V = 1024
r = 8
p=3.14
V = pi*r^2 * h
1024 = 3.14*8^2 * h combine the right side.
1024 = 3.14*64 *h
1024 = 200.96 * h divide by 200.96
1024/200.96 = h
You are only getting 5 inches.
None of your answers work.The conversion to meters is going to give you an even smaller number. If you add a comment, I'll come back and edit this question.
All Cavaliere's principle says is that a slanted cylinder (for example) has the same volume as a cylinder with perpendicular parallel lines that the cylinder is sitting on.
I little known or maybe not a well emphasized fact is that the base is the same all the way up the cylinder. The coin picture below shows that. The basic formula for a Cavaliere cylinder is the same as for an ordinary cylinder.
See picture below
Formula
V = B * H
Sub and Solve
V = 1024
r = 8
p=3.14
V = pi*r^2 * h
1024 = 3.14*8^2 * h combine the right side.
1024 = 3.14*64 *h
1024 = 200.96 * h divide by 200.96
1024/200.96 = h
You are only getting 5 inches.
None of your answers work.The conversion to meters is going to give you an even smaller number. If you add a comment, I'll come back and edit this question.
We are given
Since, we have to solve for F
so, we will isolate F on anyone side
step-1:
Multiply both sides by 60
step-2:
Divide both sides by 11
step-3:
Add both sides by 32
so, we get
................Answer
Let
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>
x-------> the width of the rectangular area
y------> the length of the rectangular area
we know that
y=x+15------> equation 1
perimeter of a rectangle=2*[x+y]
2x+2y <= 150-------> equation 2
substitute 1 in 2
2x+2*[x+15] <=150--------> 2x+2x+30 <=150----> 4x <=150-30
4x <= 120---------> x <= 30
the width of the rectangular area is at most 30 ft
y=x+15
for x=30
y=30+15------> y=45
the length of the rectangular area is at most 45 ft
see the attached figure
the solution is<span> the shaded area</span>