First write out the multiples of 6 and 26. Then find the common multiple they share that is immediately less than 85. (I do 26 first since it will have less multiple and it will give me an idea what to look for within the multiples for 6)
26: 26, 52, 78 (anything more will be over 85 and we need a number less than 85)
6: 6, 12, 18, 24, 30, 36, 42, 48, 56, 72, 78 (I stopped at 78 since any number after this will be larger than 85)
I now look to find the multiple that both 26 & 6 shares that is also less than 85 and it land me on the number 78!
The Answer: 78
Hope this helps!
so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%5Cqquad%20%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B%5B8-5%5D%5E2%2B%5B-3-%28-4%29%5D%5E2%7D%5Cimplies%20P1P2%3D%5Csqrt%7B%288-5%29%5E2%2B%28-3%2B4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B3%5E2%2B1%5E2%7D%5Cimplies%20%5Cboxed%7BP1P2%3D%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%5B7-8%5D%5E2%2B%5B-10-%28-3%29%5D%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B%287-8%29%5E2%2B%28-10%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%28-1%29%5E2%2B%28-7%29%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B50%7D%5Cimplies%20%5Cboxed%7BP2P3%3D5%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%5Cqquad%20%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%5B5-7%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B%285-7%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%28-2%29%5E2%2B6%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B40%7D%5Cimplies%20%5Cboxed%7BP3P1%3D2%5Csqrt%7B10%7D%7D%20)
Imagine, at the first day you have only one penny. Then tomorrow you have 2 pennies, next day you have 4 (2x2), next day you have 8 (4x2), next day you have 16 (8x2).... and so forth.
It looks like geometric sequence isn't it? (the ratio between the number of pennies that you have from the 2nd day and the 1st day is 2)
So, by using geometric sequence theorem we can total those pennies until day-27
S (total pennies at day-27) = (1)(2^27-1) / 2-1 = 2^26 pennies
So, you have 2^26 pennies.. a big number of pennies huh?=))
{The formula is: S = a( r^n-1) / r-1
Where:
a= the number of pennies that you've got at the 1st day
n= number of days you spent to collect those pennies)
r= the ratio of the number of pennies}
Answer:
Length =5cm
Step-by-step explanation:
When there's a problem ask
3 1/4 cups of flour for one batch
he will need 1/2 of 3 1/4cups for the 1/2 batch
1/2 x 13/4=13/8= 1 5/8 more cups