F(x+1) = f(x) + 2.5 with f(x) is the current number.
Answer:
x-5+ 
Step-by-step explanation:
If x+2 is a factor, -2 is a zero. So we can put -2 in the left-hand corner of your synthetic division, bring the coefficients down so you'll have a set up like this:
-2 | 1 -3 -7
____________
1. bring down the 1:
-2 | 1 -3 -7
____________
1
2. multiply that by -2 and enter the result underneath 3:
-2 | 1 -3 -7
-2
___________
1
a. then, add through:
-2 | 1 -3 -7
-2
___________
1 -5
3. Do the same for -5 (multiply by the zero -2), enter that underneath -7 and add through:
-2 | 1 -3 -7
-2 10
___________
1 -5 3
Now, reassign the coefficients to their x. Because we already factored out an x with the zero, the 1 is assigned to 'x' rather than back to '
'. The '-5' will not have an x attached. '3' is your remainder and cannot be divided out any further, so it will be written as . Finally, just put them back together to get x-5+.
Hope this helped!
X=5
90°-50°=40°
5x+15=40°
5x=25°
x=5
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.
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