A great circle is a section of a sphere that passes through its center. If the earth were a sphere, a great circle would be the equator and its axis would be the line connecting the geographic north and south pole. The length of the axis is then equal to the diameter of the sphere. For this problem, the radius of the sphere is 12 inches. A section is formed by slicing through the sphere and all sections of a sphere are circles. Considering the plane to be cut above and parallel with the equator (which is a great circle), the distance of the plane from the center of the sphere would then be the distance between the centers of the sphere and section. It is also given that the radius of the section is 9 inches. A right triangle is formed by connecting the center of the sphere, an edge of the section, and back to the center of the sphere whose hypotenuse is 12 inches (radius of the sphere), one leg is the 9 inches (radius of the section), and another leg is the distance of the plane from the sphere's center. Thus, the distance can be calculated using the Pythagorean theorem, d = sqrt(12^2 - 9^2) = sqrt(144 - 81) = sqrt(63) = 3*sqrt(7).
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Answer:
y=25.5
Step-by-step explanation:
To find y, you need to get it by itself. Ask yourself, what is being done to y, then to undo that, do the opposite. For example, if 3 was added to y, you would need to subtract 3 to cancel out the 3. In this case, what is being done to y? It's being divided by 3. So, you multiply by 3 to cancel that out. Whatever you do to one side of an equation, you also have to do to the other. So,
8.5=
Multiply both sides by 3
3(8.5) = 3(
) multiply on the left side, and on the right side the 3s cancel out
25.5=y
The product of the probllem is 21
Answer:
Step-by-step explanation:
Statements Reasons
1). SP ≅ TP 1). Given
2). PQ bisects ∠SPT 2). Given
3). ∠SPQ ≅ ∠TPQ 3). Definition of angles bisector
4). PQ ≅ PQ 4). Reflexive property of congruence
5). ΔSPQ ≅ ΔTPQ 5). SAS property of congruence
Answer:
y int. of line 1 = 0
y int. of line 2 = 4
Step-by-step explanation:
hope this helps! moo!