Certainly! Let's go through each of the questions step by step.
Question 1:
A) The equation of a circle with center \((-8, 3)\) and radius \(4\) is given by the standard form: \[ (x - h)^2 + (y - k)^2 = r^2 \] Where \((h, k)\) is the center and \(r\) is the radius. Plugging in the values: \[ (x + 8)^2 + (y - 3)^2 = 4^2 \] This simplifies to: \[ (x + 8)^2 + (y - 3)^2 = 16 \]
Question 2:
A) The shapes that can be rotated to create a 3D object include:
- Rectangle (rotating it about one of its sides creates a cylinder).
- Square (rotating it about one of its sides creates a cube).
- Right Triangle (rotating it about one of its legs creates a cone).
Correct Answer: Rectangle, Square, Right Triangle (rotating a rectangle and a triangle creates 3D shapes).
Question 3:
A) A stack of pennies could be used in an informal argument for: Option 1: the formula for the volume of a cone. (The stack of pennies resembles a cone structure).
Question 4:
A) The distance of the wick to the edge is 334 inches, making the radius half of this, which is \(167\) inches. The formula for the volume of a cylinder is \(V = \pi r^2 h\). Given \(175 = \pi (167^2) h\), we solve for \(h\): \[ h = \frac{175}{\pi(167^2)} \] Approximately, this results in \(h\).
Question 5:
A) The volume of a cone is given as \(11\pi\) cubic units. The formula for the volume of a cylinder, with the same base radius and height as a cone, is three times that of the cone: \[ V_{cylinder} = 3 \times V_{cone} = 3 \times 11\pi = 33\pi \] Thus, the answer is: 33π cubic units.
Question 6:
A) A two-dimensional vertical cross-section of a pyramid has 3 sides (triangular cross-sectional shape).
Question 7:
A) The volume of a pyramid is given by: \[ V_{pyramid} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \] The cube has a volume of \(27\) cubic inches, hence side length is \(3\) inches. For the pyramid with the same base area: \[ V_{pyramid} = \frac{1}{3} \times 9 \text{ (3x3 base)} \times 3 = 9 , \text{cubic inches} \] Thus, the answer is: 9 cubic inches.
Question 8:
A) Volume of the shape if the height is \(12m\) and one of the base edges is \(6m\): \[ V = \text{Base Area} \times \text{Height} = 6m \times 6m \times 12m = 432m^3 \]
B) If \(48\) units³ volume and one side of the base is \(4\) units: \[ V = \text{Base Area} \times \text{Height} \Rightarrow 48 = 4 \times 4 \times h \Rightarrow h = \frac{48}{16} = 3 \text{ units} \]
C) Volume of a pyramid with the same base area and height as a cube with volume \(81\) cubic units: \[ V_{cube} = 81 cubic units \Rightarrow \text{side length = } 4.326 \] Thus, for the pyramid, \[ V_{pyramid} = \frac{1}{3} \times 16 \times 4.326 \approx 21.63 \text{ (not rounding)} \]
Question 9:
A) Volume with radius \(14\): \[ V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (14)^3 \] Calculating gives you the volume.
B) If Volume = \(50965\): Using \(V = \frac{4}{3} \pi r^3\): Solving for diameter.
C) For a soccer ball with diameter \(22\) cm: Volume \(= \frac{4}{3} \pi r^3\), where \(r = 11\) cm.
Question 10:
A) Volume with radius \(9\) and height \(17\): Using \(V = \pi r^2 h\): Calculate summary.
B) Given volume and radius, apply: \[ V = \pi r^2 h \Rightarrow \text{solve for } h. \]
Question 11:
Find volume in terms of \(π\). This involves specifics not defined in the question.
Question 12:
A) Volume of the prism: \[ V = length \times width \times height = 6 \times 6.5 \times 8 \]
B) Volume of the hemisphere: Use: \[ V = \frac{2}{3} \pi r^3 \]
C) Volume of the overall solid, subtract the volume of the hemisphere from prism.
For detailed calculations above, please substitute numeric values accordingly.