Giannisdagoatidn
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The volume of a prism is the product of its height and the area of its base, [tex]\( V = Bh \)[/tex].

A rectangular prism has a volume of [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex] cubic units. Which could be the base area and height of the prism?

A. Base area: [tex]\( 4y \)[/tex] square units, Height: [tex]\( 4y^2 + 4y + 12 \)[/tex] units

B. Base area: [tex]\( 8y^2 \)[/tex] square units, Height: [tex]\( y^2 + 2y + 4 \)[/tex] units

C. Base area: [tex]\( 12y \)[/tex] square units, Height: [tex]\( 4y^2 + 4y + 36 \)[/tex] units

D. Base area: [tex]\( 16y^2 \)[/tex] square units, Height: [tex]\( y^2 + y + 3 \)[/tex] units

Sagot :

GinnyAnsweridn
To determine which of the given pairs of base area and height correctly represent the volume of the rectangular prism, we'll evaluate each of the provided options step-by-step. We know the volume [tex]\( V \)[/tex] of the prism is given as [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex] cubic units. The volume formula for a prism is [tex]\( V = B \cdot h \)[/tex], where [tex]\( B \)[/tex] is the base area and [tex]\( h \)[/tex] is the height.

### Option 1:
Base Area: [tex]\( 4y \)[/tex] square units
Height: [tex]\( 4y^2 + 4y + 12 \)[/tex] units

Calculate the volume:
[tex]\[ V_1 = (4y) \cdot (4y^2 + 4y + 12) \][/tex]
[tex]\[ V_1 = 4y \cdot 4y^2 + 4y \cdot 4y + 4y \cdot 12 \][/tex]
[tex]\[ V_1 = 16y^3 + 16y^2 + 48y \][/tex]

This volume is:
[tex]\[ 16y^3 + 16y^2 + 48y \][/tex]

which does not match the given volume [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex].

### Option 2:
Base Area: [tex]\( 8y^2 \)[/tex] square units
Height: [tex]\( y^2 + 2y + 4 \)[/tex] units

Calculate the volume:
[tex]\[ V_2 = (8y^2) \cdot (y^2 + 2y + 4) \][/tex]
[tex]\[ V_2 = 8y^2 \cdot y^2 + 8y^2 \cdot 2y + 8y^2 \cdot 4 \][/tex]
[tex]\[ V_2 = 8y^4 + 16y^3 + 32y^2 \][/tex]

This volume is:
[tex]\[ 8y^4 + 16y^3 + 32y^2 \][/tex]

which does not match the given volume [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex].

### Option 3:
Base Area: [tex]\( 12y \)[/tex] square units
Height: [tex]\( 4y^2 + 4y + 36 \)[/tex] units

Calculate the volume:
[tex]\[ V_3 = (12y) \cdot (4y^2 + 4y + 36) \][/tex]
[tex]\[ V_3 = 12y \cdot 4y^2 + 12y \cdot 4y + 12y \cdot 36 \][/tex]
[tex]\[ V_3 = 48y^3 + 48y^2 + 432y \][/tex]

This volume is:
[tex]\[ 48y^3 + 48y^2 + 432y \][/tex]

which does not match the given volume [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex].

### Option 4:
Base Area: [tex]\( 16y^2 \)[/tex] square units
Height: [tex]\( y^2 + y + 3 \)[/tex] units

Calculate the volume:
[tex]\[ V_4 = (16y^2) \cdot (y^2 + y + 3) \][/tex]
[tex]\[ V_4 = 16y^2 \cdot y^2 + 16y^2 \cdot y + 16y^2 \cdot 3 \][/tex]
[tex]\[ V_4 = 16y^4 + 16y^3 + 48y^2 \][/tex]

This volume is:
[tex]\[ 16y^4 + 16y^3 + 48y^2 \][/tex]

which does match the given volume [tex]\( 16y^4 + 16y^3 + 48y^2 \)[/tex].

So, the correct answer is:
a base area of [tex]\( 16y^2 \)[/tex] square units and height of [tex]\( y^2 + y + 3 \)[/tex] units.
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