Here are the answers to your questions:
Question 1
A) The equation of the circle is given by the formula \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
- Center: (-8, 3)
- Radius: 4
Thus, the equation becomes: \[ (x + 8)^2 + (y - 3)^2 = 16 \]
Question 2
A) Hemisphere - A hemisphere can be rotated around its diameter to form a sphere, which is a common 3D shape.
Question 3
A) Option 4: the formula for the volume of a cylinder - A stack of pennies resembles a cylinder and can be used to represent the volume of a cylinder.
Question 4
A) The formula for the volume of a cylinder is given by \(V = \pi r^2 h\). Let's calculate the radius:
- The candle's diameter = 2 × distance of the wick to the edge = 2 × 334 = 668 inches.
- Radius (r) = 668 / 2 = 334 inches.
Now, substituting in the volume formula with volume = 175 in³: \[ 175 = \pi(334^2)h \] Now we solve for height \(h\): \[ h = \frac{175}{\pi(334^2)} \] Calculating \(h\) yields approximately: \[ h \approx 0.00158 \text{ inches,} \] Which rounds to approximately 0 inches.
Question 5
A) The volume of a cylinder is \(V_{cylinder} = V_{cone} \times 3\), so if the volume of the cone is \(11\pi\), \[ V_{cylinder} = 11\pi \cdot 3 = 33\pi \text{ cubic units.} \]
Question 6
A) A two-dimensional vertical cross-section of a pyramid has 3 sides.
Question 7
A) The volume of a pyramid is given by the formula \(\frac{1}{3} \times base \times height\). Since the cube has a volume of 27 cubic inches:
- The side length of the cube (s) = \(3 \text{ inches}\) (since \(s^3 = 27\))
- Base area = \(3 \times 3 = 9 \text{ square inches}\) (base area of the pyramid).
Thus, the pyramid's volume: \[ V = \frac{1}{3} \times 9 \times 3 = 9 \text{ cubic inches.} \]
Question 8
A) A) The volume of the shape (which is a cube) with height 12 m and base edge 6 m is given by: \[ V = \text{base area} \times h = 6 \times 6 \times 12 = 432 \text{ m}^3. \]
B) Given volume = 48 units³ and length of one side of the base = 4 units: \[ V = \text{base area} \times h = 4 \times 4 \times h \] Thus, \[ 48 = 16h \Rightarrow h = \frac{48}{16} = 3 \text{ units.} \]
C) A cube has a volume of 81 cubic units, thus each side is: \[ s = 81^{1/3} = 4.326 \text{ units.} \] Since it has the same height and base area as the pyramid: \[ V_{pyramid} = \frac{1}{3} \times base \times height = \frac{1}{3} \times 4.326^2 \times 4.326 = 25.836. \]
Question 9
A) The volume of the shape with a radius of 14 is given by: \[ V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(14)^3 = \frac{4}{3}\pi(2744) \approx 11438.6. \]
B) Given volume = 50965 units³ find diameter: Using volume formula \(V = \frac{1}{3}\pi r^2 h\), rearranging gives: \[ 50965 = \frac{1}{3} \pi (r^2) h. \] Then, solving for diameter (d = 2r) using assumed height can be cumbersome without further data.
C) The volume of a soccer ball with a diameter of 22 cm: \[ V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(11)^3 \approx 5575 \text{ cm}³. \]
Question 10
A) Volume of a cylinder: \[ V = \pi r^2 h = \pi(9^2)(17) = 1365\pi. \]
B) Given volume is \(225\pi\) and radius is 5, find height: \[ 225\pi = \pi(5^2)h \Rightarrow 225 = 25h \Rightarrow h = 9 \text{ units.} \]
Question 11
A) Depending on the shape defined, we might need more information to calculate the complete volume.
Question 12
A) The volume of the prism: \[ V = length \times width \times height = 6 \times 6.5 \times 8 = 312 \text{ in}³. \]
B) Volume of the hemisphere: \[ V = \frac{2}{3}\pi r^3 \text{ (r = } \frac{2.5}{2} = 1.25) \Rightarrow \frac{2}{3}\pi(1.25)^3 = \frac{2}{3}\pi(1.953125) \approx 4.08 \text{ in}³. \]
C) The volume of the entire solid: \[ Total ; volume = \text{Volume of the prism} - \text{Volume of the hemisphere} = 312 - 4.08 = 307.92 \text{ in}³. \]
These calculations provide the necessary answers and can be completed or adjusted based on further inputs or specific numeric methods as needed.