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Verizon [17]
2 years ago
6

A cylinder's volume can be calculated by the formula V=Bh, where V stands for volume, B stands for base area, and h stands for h

eight. A certain cylinder's volume can be modeled by 6πx7−6πx4−20πx2 cubic units. If its base area is 2πx2 square units, find the cylinder's height.
Mathematics
2 answers:
Anastaziya [24]2 years ago
6 0

Answer:

h = 3x^5-3x^2-10\text{ units}

Step-by-step explanation:

We are given the following in the question:

Volume of cylinder =

V=Bh

where B is the area of base and h is the height of cylinder.

Volume of cylinder =

V = 6\pi x^7-6\pi x^4-20\pi x^2

Base area =

B = 2\pi x^2

We have to find height of cylinder.

h = \dfrac{V}{B}\\\\h = \dfrac{6\pi x^7-6\pi x^4-20\pi x^2}{2\pi x^2}\\\\h = 3x^5-3x^2-10\text{ units}

Thus, the height of cylinder is 3x^5-3x^2-10 units.

Snowcat [4.5K]2 years ago
6 0

Answer:

h=3x^5-3x^2-10

Step-by-step explanation:

We have been given that volume of a certain cylinder is 6\pi x^7-6\pi x^4-20\pi x^2 and base area is 2\pi x^2. We are asked to find the height of the cylinder.

We know that a cylinder's volume can be calculated by the formula V=Bh, where V stands for volume, B stands for base area, and h stands for height.

Let us solve for h.

h=\frac{V}{B}

Upon substituting our given values, we will get:

h=\frac{6\pi x^7-6\pi x^4-20\pi x^2}{2\pi x^2}

Let us factor out 2\pi x^2 from numerator.

h=\frac{2\pi x^2(3x^5-3x^2-10)}{2\pi x^2}

Upon cancelling out same terms, we will get:

h=3x^5-3x^2-10

Therefore, the height of the cylinder would be 3x^5-3x^2-10 units.

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\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
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x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
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(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
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There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
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\boxed{\boxed{x=1}}
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3241004551 [841]

Hi ;-)

x^2+6x-10=30 \ \ /-30\\\\x^2+6x-40=0\\\\a=1, \ b=6, \ c=-40\\\\\Delta=b^2-4ac=6^2-4\cdot1\cdot(-40)=36+160=196\\\\\sqrt{\Delta}=\sqrt{196}=14\\\\x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-14}{2}=\boxed{-10}\\\\x_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+14}{2}=\boxed4

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