Explanation Formula used: The standard equation of the circle with center at ( h , k ) and radius r is ( x − h ) 2 + ( y − k ) 2 = r 2 . The area of the semicircle is π r 2 2 where, r is radius of the circle. The area of the rectangle is l w where, l is length of the rectangle and w is width of the rectangle. Calculation: Consider the given circle x 2 + y 2 − 3.00 y + 1.25 = 0 . The given equation of the circle is not in the standard form rewrite the equation into standard form as follows, x 2 + y 2 − 3.00 y + 1.25 = 0 x 2 + y 2 − 3.00 y = − 1.25 In order to complete the square of the y terms add ( 1.500 ) 2 on both sides of the equation as follows, x 2 + y 2 − 3.00 y + ( 1.500 ) 2 = − 1.25 + ( 1.500 ) 2 x 2 + ( y 2 − 3.00 y + ( 1.500 ) 2 ) = − 1

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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Author: Allyn Washington, Richard Evans

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Chapter 21.3, Problem 68E

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To find: The area of the window when an architect designs a Norman windows, which has the form of a semicircle surmounted on a rectangle and the circular part is on the circle x2+y2−3.00y+1.25=0.

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Chapter 21.3, Problem 67E

Chapter 21.4, Problem 1PE

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(a) (b) (c) (d) de unique? Answer the following questions related to the linear system x + y + z = 2 x-y+z=0 2x + y 2 3 rewrite the linear system into the matrix-vector form A = 5 Fuse elementary row operation to solve this linear system. Is the solution use elementary row operation to find the inverse of A and then solve the linear system. Verify the solution is the same as (b). give the null space of matrix A and find the dimension of null space. give the column space of matrix A and find the dimension of the column space of A (Hint: use Rank-Nullity Theorem).

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Chapter 21 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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The area of the window when an architect designs a Norman windows, which has the form of a semicircle surmounted on a rectangle and the circular part is on the circle x 2 + y 2 − 3.00 y + 1.25 = 0 .

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To find: The area of the window when an architect designs a Norman windows, which has the form of a semicircle surmounted on a rectangle and the circular part is on the circle x2+y2−3.00y+1.25=0.

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Chapter 21 Solutions