Answer:
The Value of d is 5.
Step-by-step explanation:
Given:
We need to solve for d
Hence for solving for the term d we get;
Now we will use Addition property for equality and add both side by 8 we get;
Now we will use Division property for equality and Divide both side by 3 we get;
Hence the Value of d is 5.
Answer: The first one is A. 16 + 10 +10 +10 +10
The second one is 40 units^2
Step-by-step explanation:
Answer:
is there a y part of this equation?
Step-by-step explanation:
<span>If interest rates are at a level of 1% and expected inflation is 2%, it would be preferable to spend your money instead of saving it.
</span>Suppose<span> you have $100 and you save it in a savings account
that pays a 1% interest rate. After a year, you will have $101 in your
account.
During this period, if inflation runs 2%, you would have to
have $102 to make up for the impact of higher prices.
Since you will
only have $101 in your account, you have actually lost some purchasing
power.
If your savings don’t grow to reflect this rise in prices over
time, the effect will be as though you are actually losing money.
This means that if you have $100 which you can use to buy a TV set, and you saved the money instead is a savings account that pays 1% interest.
After 1 year, because of inflation of 2%, the TV set now costs $102 whereas the money in your bang account wil be $101.
Thus, you actually need to get an extra $1 from somewhere to fund the TV set you could have been able to buy a year ago.
</span>
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.
</span>
