n > -6 is the inequality for 16 plus 5 times a number is more than the number minus 8
<em><u>Solution:</u></em>
Given statement is:
16 plus 5 times a number is more than the number minus 8
We have to write a inequality
From given statement,
16 plus 5 times a number is more than the number minus 8
Let the number be "n"
Therefore,
16 plus 5 times "n" is more than "n" minus 8
Here, "plus" means addition and "minus" means subtraction and "times" means multiplication
Thus the statement is translated into inequality as:
Simplify the inequality
Thus the soultion is n > -6
Answer:
1. x = independent variable: Age
y = dependent variable: Accidents
2. Age scale and Accidents scale
The scale for age ranges from 15 - 30 with the value of 5 difference between each number.
While the accidents scale ranges from 0 -1 with no difference in between.
3. In my opinion, the scatter plot looks as though it decreases in accidents once the age rises. As seen from the data shown.
Step-by-step explanation:
1) x is considered as the independent variable, while y is considered the dependent value.
2) There are two sets of scales, one for the x and the other for y. These scales are the values that are interpreted in order to show data of the graph, due to their number and various size(s).
3) If you were to draw a line through the graph, you can somewhat create the image that the line would go down into the right corner! Meaning that it is decreasing over time.
Answer:
surface area of a cuboid=2(L*W)+ 2(L*H)+ 2(W*H)
2(9*5)+2(9*4)+2(5*4)
Answer:
3.370459909
Step-by-step explanation:
Figure out the arithmetic mean of the numbers (22.8)
Subtract the mean from each number and square the difference
Add up the squared differences and divide the number by 5
Square root answer
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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