Answer:
y=x+3
Step-by-step explanation:
in the x value you add 3 to get the next value (the next x value).
To prove this inequality we need to consider three cases. We need to see that the equation is symmetric and that switching the variables x and y does not change the equation.
Case 1: x >= 1, y >= 1
It is obvious that
x^y >= 1, y^x >= 1
x^y + y^x >= 2 > 1
x^y + y^x > 1
Case 2: x >= 1, 0 < y < 1
Considering the following sub-cases:
- x = 1, x^y = 1
- x > 1,
Let x = 1 + n, where n > 0
x^y = (1 + n)^y = f_n(y)
By Taylor Expansion of f_e(y) around y = 0,
x^y = f_n(0) + f_n'(0)/1!*y + f_n''(0)/2!*y^2 + ...
= 1 + ln(1 + n)/1!*y + ln(1 + n)^2/2!*y^2 + ...
Since ln(1 + n) > 0,
x^y > 1
Thus, we can say that x^y >= 1, and since y^x > 0.
x^y + y^x > 1
By symmetry, 0 < x < 1, y >= 1, also yields the same.
Case 3: 0 < x, y < 1
We can prove this case by fixing one variable at a time and by invoking symmetry to prove the relation.
Fixing the variable y, we can set the expression as a function,
f(x) = x^y + y^x
f'(x) = y*x^(y-1) + y^x*ln y
For all x > 0 and y > 0, it is obvious that
f'(x) > 0.
Hence, the function f(x) is increasing and hence the function f(x) would be at its minimum when x -> 0+ (this means close to zero but always greater than zero).
lim x->0+ f(x) = 0^y + y^0 = 0 + 1 = 1
Thus, this tells us that
f(x) > 1.
Fixing variable y, by symmetry also yields the same result: f(x) > 1.
Hence, when x and y are varying, f(x) > 1 must also hold true.
Thus, x^y + y^x > 1.
We have exhausted all the possible cases and shown that the relation holds true for all cases. Therefore,
<span> x^y + y^x > 1
----------------------------------------------------
I have to give credit to my colleague, Mikhael Glen Lataza for the wonderful solution.
I hope it has come to your help.
</span>
Answer:
<em>The combined average of Altuve's weighted averages is 1 for 2 or 1/2</em>
Step-by-step explanation:
<u>Combined Average</u>
On Friday, Altuve had 3 out of 5 successes
On Saturday, Altuve had 2 out of 4 successes
On Friday, Altuve had 1 out of 3 successes
The total successes are 3+2+1=6
The total tries are 5+4+3=12
The combined average is:
![\displaystyle \bar x=\frac{6}{12}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbar%20x%3D%5Cfrac%7B6%7D%7B12%7D)
Simplifying:
![\displaystyle \bar x=\frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbar%20x%3D%5Cfrac%7B1%7D%7B2%7D)
The combined average of Altuve's weighted averages is 1 for 2 or 1/2
Answer:
![\boxed{ \bold{ \boxed{ \sf{domain = { (- 2 \: , \: 1 \: , \: 3)}}}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%20%5Cboxed%7B%20%5Csf%7Bdomain%20%3D%20%7B%20%28-%202%20%20%5C%3A%20%2C%20%5C%3A%201%20%5C%3A%20%2C%20%5C%3A%20%203%29%7D%7D%7D%7D%7D)
![\boxed{ \bold{ \boxed{ \sf{range = ( \: - 2 \: , \: 0 \: , \: 4)}}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Cbold%7B%20%5Cboxed%7B%20%5Csf%7Brange%20%3D%20%28%20%5C%3A%20%20-%202%20%5C%3A%20%2C%20%5C%3A%200%20%5C%3A%20%2C%20%5C%3A%204%29%7D%7D%7D%7D)
Step-by-step explanation:
Domain is the set of all x values.
Range is the set of all y - values.
{ ( - 2 , 0 ) , ( 1 , -2 ) , ( 3 , 4 )}
Domain = { -2 , 1 , 3 }
{ ( -2 , 0 ) , ( 1 , -2 ) , ( 3 , 4 )}
Range = { - 2 , 0 , 4 }
Hope I helped!
Best regards! :D
Answer:
y=-5x+12
Step-by-step explanation:
Since slope-intercept form is y=mx+b, m is the slope, and b is the y-intercept. We substitute m for -5, since it is the slope, and we substitute 12 for b, since it is the y-intercept. I hope this helps!