The total cost of the mortgage would be $264,192 or C.
Answer:
A) 5.12mm³
Step-by-step explanation:
Step 1
Find the volume of the bigger pyramid
This is a triangular based pyramid.
The formula to use is given as Volume of Pyramid =
1/3 × Area of the triangle × Height
Area of the triangle = 1/2 × 3 mm × 4mm = 6mm²
Volume of the large pyramid = 1/3 × 6mm² × 5mm
= 10mm³
We are given the length of the small pyramid as 4mm
We would be using was is known as scale factor to find the volume of the small pyramid
The scale factor k = (Height of the small pyramid)³/ (Height of the Large pyramid)³
k = 4³/5³
k = Volume of small pyramid/ Volume of large pyramid
Volume of small pyramid = X
Volume of large pyramid = 10mm³
Hence,
4³/5³ = X/10
Cross Multiply
= 4³ × 10 = 5³ × X
X = (4³ × 10)/ 5³
X = 5.12mm³
Answer:
a
Step-by-step explanation:
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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