We know that
volume of <span>a rectangular prism =B*h------> equation 1
where
B is the area of the base
h is the height
volume of </span><span>a rectangular pyramid=(1/3)*B*h-----> equation 2
where
</span>B is the area of the base
h is the height
<span>
substitute equation 1 in equation 2
</span>volume of a rectangular pyramid=(1/3)*volume of a rectangular prism
<span>
the answer part a) is
</span>volume of a rectangular pyramid=(1/3)*volume of a rectangular prism
<span>
Part b) </span><span>If the pyramid was full of water, how much of the prism would it fill up?
</span>
the answer part b) is
<span>If the pyramid was filled with water, the prism would only fill 1/3 of its volume
Part c) </span><span>Name another pair of three-dimensional objects that have a relationship similar to this
cones and cylinders
</span>volume of a cylinder =B*h------> equation 1
where
B is the area of the base
h is the height <span>
</span>volume of a cone=(1/3)*B*h-----> equation 2
where
B is the area of the base
h is the height
substitute equation 1 in equation 2
volume of a cone=(1/3)*volume of a cylinder
Answer:
V = 23π/6
Step-by-step explanation:
V = 2π ∫ [a to b] (r * h) dx
y = −x² + 23x − 132
y = −(x² − 23x + 132)
y = −(x − 11) (x − 12)
Parabola intersects x-axis (line y = 0) at x = 11 and x = 12 ----> a = 11, b = 12
r = x
h = −x² + 23x − 132
V = 2π ∫ [11 to 12] x (−x² + 23x − 132) dx
V = 23π/6
<span>Most ancient societies had symbols to represent numbers, but they did not have symbols to represent operations or unknown quantities. Thus, the problems and solutions to the problems had to be written in word form.</span>
X^2(x+4) + 5(x + 4)
(x^2 + 5)(x + 4)
