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Kaylis [27]
3 years ago
11

What is the volume of the cube below?

Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0

Answer:

D. 9h

Step-by-step explanation:

Volume of a cube = Length × Width× Height

= 3×3×h

= 9h

jeka57 [31]3 years ago
5 0

Answer:

D. 9h

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Geometry</u>

Volume of a Cuboid Formula: V = a²h

  • <em>a</em> is the sides of the cube
  • <em>h</em> is the height

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

<em>a</em> = 3

<em>h</em> = <em>h</em>

<em />

<u>Step 2: Find Volume</u>

  1. Substitute in variables [Volume of a Cuboid Formula]:                                 V = 3²h
  2. Evaluate exponents:                                                                                         V = 9h
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The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

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If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

here:

  • (x₁, y₁) and (x₂, y₂) are two points on the straight line

Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

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Now, our region bounded by the three lines are:

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Similarly, the change in polar coordinates is:

  • x = rcosθ,
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where;

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  • rsinθ = 4(rcosθ)  ⇒ tan θ = 4;  θ = tan⁻¹ (4)

  • ⇒ r = 0   to   r = 2/cosθ
  •    θ = 0  to    θ = tan⁻¹ (4)

Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

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