Answer:
x = -2
Step-by-step explanation:
3(5x + 2) = 2(3x - 6)
<em>Expand the brackets. </em>
15x + 6 = 6x - 12
<em>Isolate the unknown variable x. </em>
15x - 6x = -12 - 6
<em>Evaluate. </em>
9x = -18
<em>Find x. </em>
x = -18 ÷ 9
x = -2
<h3>
Answer: C. If two lines are parallel, then the alternate interior angles formed are congruent.</h3>
This is through the alternate interior angles theorem. Angles Q and T pair up as one alternate interior set of angles that are the same measure. The same thing applies to angles X and R.
The identical arrow markers on segments XQ and TR show that those segments are parallel. Segment TQ is one transversal cut (forming alternate interior angles Q and T). Segment XR is the other transversal cut (forming alternate interior angles X and R).
We could say "angle XRT" or "angle TRX" instead of "angle R", though its ideal to use shortcuts whenever possible. The same applies for the other angles as well.
You could use perturbation method to calculate this sum. Let's start from:
On the other hand, we have:
So from (1) and (2) we have:
Now, let's try to calculate sum
, but this time we use perturbation method.
but:
![S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\= \sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\= \sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\ \boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}](https://tex.z-dn.net/?f=S_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%5Ccdot0%21%2B%5Csum%5Climits_%7Bk%3D1%7D%5E%7Bn%2B1%7Dk%5Ccdot%20k%21%3D0%2B%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%2B1%29%28k%2B1%29k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B2k%2B1%29k%21%3D%5C%5C%5C%5C%5C%5C%3D%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%5Cleft%5B%28k%5E2%2B1%29k%21%2B2k%5Ccdot%20k%21%5Cright%5D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B%5Csum%5Climits_%7Bk%3D0%7D%5En2k%5Ccdot%20k%21%3D%5C%5C%5C%5C%5C%5C%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%7D)
When we join both equation there will be:
![\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\ S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\ \sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\= (n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\= n(n+1)!+1](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7DS_%7Bn%2B1%7D%3DS_n%2B%28n%2B1%29%28n%2B1%29%21%5C%5C%5C%5CS_%7Bn%2B1%7D%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%0AS_n%2B%28n%2B1%29%28n%2B1%29%21%3D%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%2B2S_n%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Csum%5Climits_%7Bk%3D0%7D%5E%7Bn%7D%28k%5E2%2B1%29k%21%3DS_n-2S_n%2B%28n%2B1%29%28n%2B1%29%21%3D%28n%2B1%29%28n%2B1%29%21-S_n%3D%5C%5C%5C%5C%5C%5C%3D%0A%28n%2B1%29%28n%2B1%29%21-%5Csum%5Climits_%7Bk%3D0%7D%5Enk%5Ccdot%20k%21%5Cstackrel%7B%28%5Cstar%29%7D%7B%3D%7D%28n%2B1%29%28n%2B1%29%21-%5B%28n%2B1%29%21-1%5D%3D%5C%5C%5C%5C%5C%5C%3D%28n%2B1%29%28n%2B1%29%21-%28n%2B1%29%21%2B1%3D%28n%2B1%29%21%5Ccdot%5Bn%2B1-1%5D%2B1%3D%5C%5C%5C%5C%5C%5C%3D%0An%28n%2B1%29%21%2B1)
So the answer is:
Sorry for my bad english, but i hope it won't be a big problem :)
Answer:
Step-by-step explanation:
The temperature at a given altitude is
y = 36 - 3x
The temperature on the surface of the planet is the point (0,t) where t is the temperature for the given height.
y = 36 - 3*0
y = 36
So at the surface of the planet is 36 degrees C.
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Effectively it is the slope of the equation which is - 3
So ever km going up will mean a loss of 3 degrees. I think they want you to write -3
