At the library, Newton borrows 8 books and Descartes borrows 4 books. 7 of their books are nonfiction. The rest are fiction. How
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Answer:
The radius of a sphere is 2 millimeters
The surface area of a sphere is 50.24 square millimeters.
The circumference of the great circle of a sphere is 12.56 millimeters.
Step-by-step explanation:
<u><em>Verify each statement</em></u>
case A) The radius of a sphere is 8 millimeters
The statement is false
we know that
The volume of the sphere is equal to
we have
substitute and solve for r
case B) The radius of a sphere is 2 millimeters
The statement is True
(see the case A)
case C) The circumference of the great circle of a sphere is 9.42 square millimeters
The statement is false
The units of the circumference is millimeters not square millimeters
The circumference is equal to
we have
substitute
case D) The surface area of a sphere is 50.24 square millimeters.
The statement is True
Because
The surface area of the sphere is equal to
we have
substitute
case E) The circumference of the great circle of a sphere is 12.56 millimeters.
The statement is true
see the case C
case F) The surface area of a sphere is 25.12 square millimeters
The statement is false
because the surface area of the sphere is
see the case D
An equation is formed of two equal expressions. The correct options are C, D, and E.
<h3>What is an equation?</h3>An equation is formed when two equal expressions are equated together with the help of an equal sign '='.
The value of x in the given equation is,
Substitute the value of x in other option to check if the equations are equal are not. Therefore, the options can be written as,
A.) 18x - 15 = 72
18(4.5) - 15 = 66
Thus, this will not return the same value.
B.) 50x - 25 = 72
50(4.5) - 25 = 200
Thus, this will not return the same value.
C.) 18x - 9 = 72
18(4.5) - 9 = 72
Thus, this will return the same value.
D.) 3(6x - 3) = 72
3[6(4.5) - 3] = 72
Thus, this will return the same value.
E.) x = 4.5
Thus, this will return the same value.
Hence, the correct options are C, D, and E.
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