The length of a rectangle is 12 inches more than its width. What is the width of the rectangle if the perimeter is 42 inches?
x best represents .
x + 12 best represents
<h3>Answer: C) none of the equations are identities</h3>
If you plugged theta = 0 into the first equation, then you would have
sin(45) + cos(45) = sin(0) + cos(0)
sqrt(2) = 1
which is a false equation. We don't have an identity here.
The same story happens with the second equation. Plug in theta = 0 and it becomes
cos(60) - sin(60) = cos^2(0) + tan(0)
1/2 - sqrt(3)/2 = 1 + 0
-0.37 = 1
which is false.
The base case is the claim that
![\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}](https://tex.z-dn.net/?f=%5Cdfrac11%20%2B%20%5Cdfrac12%20%3E%20%5Cdfrac%7B2%5Ccdot2%7D%7B2%2B1%7D)
which reduces to
![\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86](https://tex.z-dn.net/?f=%5Cdfrac32%20%3E%20%5Cdfrac43%20%5Cimplies%20%5Cdfrac46%20%3E%20%5Cdfrac86)
which is true.
Assume that the inequality holds for <em>n</em> = <em>k </em>; that
![\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}](https://tex.z-dn.net/?f=%5Cdfrac11%20%2B%20%5Cdfrac12%20%2B%20%5Cdfrac13%20%2B%20%5Ccdots%20%2B%20%5Cdfrac1k%20%3E%20%5Cdfrac%7B2k%7D%7Bk%2B1%7D)
We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that
![\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}](https://tex.z-dn.net/?f=%5Cdfrac11%20%2B%20%5Cdfrac12%20%2B%20%5Cdfrac13%20%2B%20%5Ccdots%20%2B%20%5Cdfrac1k%20%2B%20%5Cdfrac1%7Bk%2B1%7D%20%3E%20%5Cdfrac%7B2%28k%2B1%29%7D%7Bk%2B2%7D)
By the induction hypothesis,
![\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}](https://tex.z-dn.net/?f=%5Cdfrac11%20%2B%20%5Cdfrac12%20%2B%20%5Cdfrac13%20%2B%20%5Ccdots%20%2B%20%5Cdfrac1k%20%2B%20%5Cdfrac1%7Bk%2B1%7D%20%3E%20%5Cdfrac%7B2k%7D%7Bk%2B1%7D%20%2B%20%5Cdfrac1%7Bk%2B1%7D%20%3D%20%5Cdfrac%7B2k%2B1%7D%7Bk%2B1%7D)
Now compare this to the upper bound we seek:
![\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}](https://tex.z-dn.net/?f=%5Cdfrac%7B2k%2B1%7D%7Bk%2B1%7D%20%20%3E%20%5Cdfrac%7B2k%2B2%7D%7Bk%2B2%7D)
because
![(2k+1)(k+2) > (2k+2)(k+1)](https://tex.z-dn.net/?f=%282k%2B1%29%28k%2B2%29%20%3E%20%282k%2B2%29%28k%2B1%29)
in turn because
![2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0](https://tex.z-dn.net/?f=2k%5E2%20%2B%205k%20%2B%202%20%3E%202k%5E2%20%2B%204k%20%2B%202%20%5Ciff%20k%20%3E%200)
Answer:
34
Step-by-step explanation: