A Pythagorean triple is a set of thre integer numbers, a, b and c that meet the Pythgorean theorem a^2 + b^2 = c^2
Use Euclide's formula for generating Pythagorean triples.
This formula states that given two arbitrary different integers, x and y, both greater than zero, then the following numbers a, b, c form a Pythagorean triple:
a = x^2 - y^2
b = 2xy
c = x^2 + y^2.
From a = x^2 - y^2, you need that x > y, then you can discard options A and D.
Now you have to probe the other options.
Start with option B, x = 4, y = 3
a = x^2 - y^2 = 4^2 - 3^2 = 16 -9 = 7
b = 2xy = 2(4)(3) = 24
c = x^2 9 y^2 = 4^2 + 3^2 = 16 + 9 = 25
Then we could generate the Pythagorean triple (7, 24, 25) with x = 4 and y =3.
If you want, you can check that a^2 + b^2 = c^2; i.e. 7^2 + 24^2 = 25^2
The answer is the option B. x = 4, y = 3
Undefined slope
slope=rise/run
if run=0, we get an undefined slope
means that it does go left or right, only up and down
the equaiton will be x=something
x intercept is -7 so the equation is x=-7
Answer:
92%
Step-by-step explanation:
This is because all you have to do to turn a decimal into percentage is to move the decimal over 2 spots, and drop the Zero.
The value of x is 1/13 (4th choice)
Step-by-step explanation:
![{a}^{ \frac{m}{n} } = \sqrt[n]{ {a}^{m} }](https://tex.z-dn.net/?f=%20%7Ba%7D%5E%7B%20%5Cfrac%7Bm%7D%7Bn%7D%20%7D%20%20%3D%20%20%5Csqrt%5Bn%5D%7B%20%7Ba%7D%5E%7Bm%7D%20%7D%20)
If 14^x = ¹³√14, what value of <em>x </em>makes this equation true?
![\sqrt[13]{14} = {14}^{x} \\ \sqrt[13]{14} = {14}^{ \frac{1}{13} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B13%5D%7B14%7D%20%20%3D%20%20%7B14%7D%5E%7Bx%7D%20%20%5C%5C%20%20%5Csqrt%5B13%5D%7B14%7D%20%20%3D%20%20%7B14%7D%5E%7B%20%5Cfrac%7B1%7D%7B13%7D%20%7D%20)
So, the value of <em>x</em> is 1/13 (4th choice)
<em>Hope</em><em> </em><em>it </em><em>helpful </em><em>and </em><em>useful </em><em>:</em><em>)</em>
