Respuesta :
Answer:To calculate a Pythagorean triple select any term of this progression and reduce it to an improper fraction. For example, take the term {\displaystyle 3{\tfrac {3}{7}}}3\tfrac{3}{7}. The improper fraction is {\displaystyle {\tfrac {24}{7}}}{\tfrac {24}{7}}. The numbers 7 and 24 are the sides, a and b, of a right triangle, and the hypotenuse is one greater than the largest side. For example:
{\displaystyle 1{\tfrac {1}{3}}{\text{ }}{\xrightarrow {\text{yields}}}{\text{ }}[3,4,5],{\text{ 2}}{\tfrac {2}{5}}{\text{ }}{\xrightarrow {\text{yields}}}{\text{ }}[5,12,13],{\text{ 3}}{\tfrac {3}{7}}{\text{ }}{\xrightarrow {\text{yields}}}{\text{ }}[7,24,25],{\text{ 4}}{\tfrac {4}{9}}{\text{ }}{\xrightarrow {\text{yields}}}{\text{ }}[9,40,41],{\text{ }}\ldots }1{\tfrac {1}{3}}{\text{ }}{\xrightarrow {{\text{yields}}}}{\text{ }}[3,4,5],{\text{ 2}}{\tfrac {2}{5}}{\text{ }}{\xrightarrow {{\text{yields}}}}{\text{ }}[5,12,13],{\text{ 3}}{\tfrac {3}{7}}{\text{ }}{\xrightarrow {{\text{yields}}}}{\text{ }}[7,24,25],{\text{ 4}}{\tfrac {4}{9}}{\text{ }}{\xrightarrow {{\text{yields}}}}{\text{ }}[9,40,41],{\text{ }}\ldots
Jacques Ozanam[5] republished Stifel's sequence in 1694 and added the similar sequence {\displaystyle 1{\tfrac {7}{8}},{\text{ }}2{\tfrac {11}{12}},{\text{ }}3{\tfrac {15}{16}},{\text{ }}4{\tfrac {19}{20}},\ldots }1{\tfrac {7}{8}},{\text{ }}2{\tfrac {11}{12}},{\text{ }}3{\tfrac {15}{16}},{\text{ }}4{\tfrac {19}{20}},\ldots with terms derived from {\displaystyle n+{\tfrac {4n+3}{4n+4}}}n+{\tfrac {4n+3}{4n+4}}. As before, to produce a triple from this sequence, select any term and reduce it to an improper fraction. The numerator and denominator are the sides, a and b, of a right triangle. In this case, the hypotenuse of the triple(s) produced is 2 greater than the larger side. For example:
{\displaystyle 1{\tfrac {7}{8}}{\xrightarrow {\text{yields}}}[15,8,17],2{\tfrac {11}{12}}{\xrightarrow {\text{yields}}}[35,12,37],3{\tfrac {15}{16}}{\xrightarrow {\text{yields}}}[63,16,65],4{\tfrac {19}{20}}{\xrightarrow {\text{yields}}}[99,20,101],\ldots }1{\tfrac {7}{8}}{\xrightarrow {{\text{yields}}}}[15,8,17],2{\tfrac {11}{12}}{\xrightarrow {{\text{yields}}}}[35,12,37],3{\tfrac {15}{16}}{\xrightarrow {{\text{yields}}}}[63,16,65],4{\tfrac {19}{20}}{\xrightarrow {{\text{yields}}}}[99,20,101],\ldots
Together, the Stifel and Ozanam sequences produce all primitive triples of the Plato and Pythagoras families respectively. The Fermat family must be found by other means.
With a the shorter and b the longer leg of the triangle:
{\displaystyle {\text{Plato: }}c-b=2,\quad \quad {\text{Pythagoras: }}c-b=1,\quad \quad {\text{Fermat: }}\left|a-b\right|=1}{\displaystyle {\text{Plato: }}c-b=2,\quad \quad {\text{Pythagoras: }}c-b=1,\quad \quad {\text{Fermat: }}\left|a-b\right|=1}
Dickson's method
Leonard Eugene Dickson (1920)[6] attributes to himself the following method for generating Pythagorean triples. To find integer solutions to {\displaystyle x^{2}+y^{2}=z^{2}}x^{2}+y^{2}=z^{2}, find positive integers r, s, and t such that {\displaystyle r^{2}=2st}r^{2}=2st is a perfect square.
Then:
{\displaystyle x=r+s\,,\,y=r+t\,,\,z=r+s+t.}x=r+s\,,\,y=r+t\,,\,z=r+s+t.
From this we see that {\displaystyle r}r is any even integer and that s and t are factors of {\displaystyle {\tfrac {r^{2}}{2}}}{\tfrac {r^{2}}{2}}. All Pythagorean triples may be found by this method. When s and t are coprime, the triple will be primitive. A simple proof of Dickson's method has been presented by Josef Rukavicka (2013).[7]
Example: Choose r = 6. Then {\displaystyle {\tfrac {r^{2}}{2}}=18}{\tfrac {r^{2}}{2}}=18. The three factor-pairs of 18 are: (1, 18), (2, 9), and (3, 6). All three factor pairs will produce triples using the above equations.
s = 1, t = 18 produces the triple [7, 24, 25] because x = 6 + 1 = 7, y = 6 + 18 = 24, z = 6 + 1 + 18 = 25.
s = 2, t = 9 produces the triple [8, 15, 17] because x = 6 + 2 = 8, y = 6 + 9 = 15, z = 6 + 2 + 9 = 17.
s = 3, t = 6 produces the triple [9, 12, 15] because x = 6 + 3 = 9, y = 6 + 6 = 12, z = 6 + 3 + 6 = 15. (Since s and t are not coprime, this triple is not primitive.)
1.b
2.refrection
Step-by-step explanation: hope this helps
The function that has the least minimum value is function is h and The function that has the greatest minimum value is function is k .
What is quadratic function ?
we define a quadratic function as an function of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax² + bx + c, where a, b, and c are numbers and a cannot be 0.
According to the question
f(x) = x² - 2x - 6
as per graph
its minimum value at (1,-7)
h(x) = 2(x-1)²
h(x) = 2 (x² + 1 -2x )
h(x) = 2x² - 4x + 2
By plotting the graph
we can observe the
minimum value at (1,0)
k(x) = [tex]x^{4} + 2x^{2} + 8x -4[/tex]
By plotting the graph
we can observe the
minimum value at (-1,-9)
For g(x)
Standard form of quadratic equation :
y = ax² + bx + c
or
g(x) = ax² + bx + c
at x = 0
g(x) = -4
-4 = a* 0 + b* 0 + c
c = -4
at x = 1
g(x) = -5
-5 = a + b - 4
a + b = -1 --------------------(1)
at x = -1
g(x) = -11
-11 = a*(-1)²+ b* (-1) + -4
-7 = a - b --------------------(2)
Adding equation (1) and (2)
a = -4
b = 3
g(x) = -4x² + 3x -4
as x² is negative parabola will be downward
therefore it will have max value
i.e
at (0.375,-3.438)
Now, The function that has the least minimum value is function is h with x = 1 and h(x) = 0
The function that has the greatest minimum value is function is k with x = -1 and k(x) = -9
Hence, The function that has the least minimum value is function is h and The function that has the greatest minimum value is function is k .
To know more about quadratic function here:
https://brainly.com/question/27958964
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