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8 Cards in this Set
- Front
- Back
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My friend told me she had discovered something truly exciting. She told me that I could pick any two integers, and that the sum of their squares, the difference of their squares, and twice their product are the sides of a right triangle. Investigate. Does this appear to work in all cases, or only some cases?
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After investigation, students should discover that this trick does not work in all cases. If the two integers are the same or one of the integers is zero, the trick will not work.
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My friend told me she had discovered something truly exciting. She told me that I could pick any two integers, and that the sum of their squares, the difference of their squares, and twice their product are the sides of a right triangle. Since we now know this trick does not work every time, modify the statement so that it is true every time.
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If two positive integers m and n are chosen such that m>n, then the numbers m²+n², m²-n², and 2mn are the lengths of the sides of a right triangle.
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If two positive integers m and n are chosen such that m>n, then the numbers m²+n², m²-n², and 2mn are the lengths of the sides of a right triangle. Use this trick to find an example of a right triangle in which all of the sides have integer lengths, all three sides are longer than 100 units, and the three side lengths do not have any common factors.
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(One possible solution) m=13 n=6 m²+n²=205 m²-n²=133 and 2mn=156. The side lengths are integers, greater than 100 units and do not have any factors in common.
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What are the expansion formulas used to expand binomials based on their exponential form for (a+b)^n where n=0, 1, 2, 3, and 4?
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(a+b)⁰=1
(a+b)¹=a+b (a+b)²=a²+2ab+b² (a+b)³=a²+3a²b+3ab²+b³ (a+b)⁴=a⁴+4a³b+6a²b²+4ab³+b⁴ |
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Expand using the binomial theorem. (4x+y)³
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(4x)³ + 3(4x)²y + 3(4x)(y)² + y³=
64x³+48x²y+12xy²+y³ |
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Expand using the binomial theorem. (2x+4y)²
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(2x)² + 2(2x)(4y) + (4y)² =
4x²+16xy+16y² = x²+4xy+4y² |
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Expand using the binomial theorem. (5x+2y)¹
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5x+2y
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Expand using the binomial theory. (x+2y)⁴
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(x)⁴ + 4(x)³(2y) + 6(x)²(2y)² + 4(x)(2y)³ + (2y)⁴=
x⁴+4x³2y+6x²4y²+4x8y³+16y⁴= x⁴+8x³y+24x²y²+32xy³+16y⁴ |
