Answer:
D
Step-by-step explanation:
Hello! And thank you for your question!
First use the rule: A/b x c/d = ac/bd
3 x 2 over 5 x 3
Simplify 3 x 2:
6 over 5 x 3
Simplify 5 x 3:
6/15
Simplify further:
2/5
Final Answer:
2/5
Answer:
Answer is D.
Step-by-step explanation:
Which is not true?
So, lets assess the answers given to the question.
A. It is the equation of a horizontal line.
Yes, it is. Since there is only one y - value which is y = -4, we can confirm this is the y intercept, so therefore the x value is zero. But it's not the answer we're looking for.
B. It's slope is zero, yes the slope is zero. It only lists the y-intercept which confirms that the slope is 0.
C. Yes, I mean i can't really explain how. Just use desmos calculator, the graph is exactly perpendicular.
D. The line actually does cross the x-axis, -4 is the y intercept and the x value is 0, and zero is still a point. So D is the one which isn't correct making it the correct answer.
8/2 pound, simplified 4/1 pound
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
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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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