Algebra 1 Common Core Answers Chapter 10 Radical Expressions and Equations Exercise 10.1
Algebra 1 Common Core Solutions
Chapter 10 Radical Expressions and Equations Exercise 10.1 1LC




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Hence the lengths 12 cm. 35 cm and 37 cm form right triangIe
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Consider the conditional “If you study math, then you are student? The converse of the conditional is “If you are a student, then you study math”
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By the converse of the Pythagorean Theorem, the lengths 15 ft. 36 ft and 39 ft could be the sides of the right triangle
Chapter 10 Radical Expressions and Equations Exercise 10.1 24E

By the converse of the Pythagorean Theorem, the lengths 12 m,60 m 61 m are not the sides of the right triangle
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By the converse of the Pythagorean Theorem, the lengths 13 in. 35 in and 38 in are not the sides of the right triangle
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By the converse of the Pythagorean Theorem, the lengths 16 cm, 63 cm and 65 cm could be the sides of the right triangle.
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By the converse of the Pythagorean Theorem, the lengths 14 in. 48 in and 50 in could be the sides of the right triangle.
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By the converse of the Pythagorean Theorem, the lengths 16 yd. 30 yd and 34 yd could be the sides of the right triangle
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Hence the distance between the swimmers head and the life guards head is 10ft
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By the converse of tile Pythagorean Theorem, the numbers 11. 60 and 61 are Pythagorean Triple.
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By the converse of the Pythagorean Theorem, the numbers 13. 84 and 85 are Pythagorean Triple.
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By the converse of the Pythagorean Theorem, the numbers 40. 41 and 58 are not Pythagorean Triple.
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By the converse of the Pythagorean Theorem, the numbers 50. 120 and 130 are Pythagorean Triple
Chapter 10 Radical Expressions and Equations Exercise 10.1 34E

By the converse of the Pythagorean Theorem, the numbers 32. 126 and 130 are Pythagorean Triple
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By the converse of the Pythagorean Theorem, the numbers 28, 45 and 53 are Pythagorean Triple.
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Hence the length of the third side is 6 in
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Hence the two forces 50 lb and 120 lb are pulling at right angles and the resultant force is 130
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Hence the diameter of the smallest circular opening is about 5.7 cm.
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Hence the area of the larger square is equal to sum of the areas of 4 triangles and area of the smaller square.
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Hence the positive solution of the equation is t = 5.4.
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