Answer:
w= 6
Step-by-step explanation:
I just started by making an educated guess using the values already given. Then I inserted that into the problem to see if it worked.
L= 2w - 5
I used 6 as a random, educated guess for the value of w.
L = 2(6) - 5
L = 12-5
L = 7
Then, multiply L by 2 to account for both side lengths of the rectangle.
7(2)= 14
Subtract that value from the total perimeter to find what the width must equal.
26 - 14 = 12
Divide that answer by 2 since there are two sides for width.
12/2 = 6
I know this was kind of long, but I hope it helps! :)
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em><em>:</em><em>)</em>
Hi there,
Since Eric ran 200 m/min for 4 minutes, then 4×200 = 800m
and he walked 80 m/min for 8 minutes, then 8×80 = 640m
and in total he ran 800 + 640 = 1440m
so he still have 1600 - 1440 = 160m left to reach 1600m
Hope this helped :P
Answer: 
; 5
Step-by-step explanation:
Given series : ![[\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cdfrac%7B5%7D%7B1%5Ccdot2%7D%5D%2B%5B%5Cdfrac%7B5%7D%7B2%5Ccdot3%7D%5D%2B%5B%5Cdfrac%7B5%7D%7B3%5Ccdot4%7D%5D%2B....%2B%5B%5Cdfrac%7B5%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D)
Sum of series = ![S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]](https://tex.z-dn.net/?f=S_n%3D%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5C%20%5B%5Cdfrac%7B5%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D%3D5%5B%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5Cdfrac%7B1%7D%7Bn%5Ccdot%28n%2B1%29%7D%5D)
Consider 
⇒ ![S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]](https://tex.z-dn.net/?f=S_n%3D5%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5Cdfrac%7B1%7D%7Bn%5Ccdot%28n%2B1%29%7D%3D5%5Csum%5E%7B%5Cinfty%7D_%7B1%7D%5B%5Cdfrac%7B1%7D%7Bn%7D-%5Cdfrac%7B1%7D%7Bn%2B1%7D%5D)
Put values of n= 1,2,3,4,5,.....n
⇒ 
All terms get cancel but First and last terms left behind.
⇒
Formula for the nth partial sum of the series :
Also, 
