Answer:
the distance between points M(2,6) and N(3,1)
MN =
=![\sqrt{26}](https://tex.z-dn.net/?f=%5Csqrt%7B26%7D)
Answer:
436000=to=$tandard form is = 4.36×10^5
Step-by-step explanation:
hope it helps
You said that 51 + (the other number) = 136 .
This is an equation.
I can do anything I want to one side of it,
as long as I do exactly the same thing to
the other side.
Let's subtract 51 from each side: (the other number) = 136 - 51
= 85 .
The factors of 51 are 1, 3, 17, and 51 .
The factors of 85 are 1, 5, 17, and 85 .
The common factors are 1 and 17 .
8.8 meters. In order to round to the nearest tenth you have to look at the hundredth place, 1-4 and you keep the tenth the same 5-9 you increase the tenth by 1
Step-by-step explanation:
![here \: is \: your \: solution \\ \\ base \: = 4 \: cm \\ \\ height \: = 6 \: cm \\ \\ area \: of \: traingle = (1 \div 2) \times base \times height \\ \\ area \: = (4 \times 6) \div 2 \\ \\ area = 12 \: cm.sq \\ \\ hope \: it \: helps](https://tex.z-dn.net/?f=here%20%5C%3A%20is%20%5C%3A%20your%20%5C%3A%20solution%20%5C%5C%20%20%5C%5C%20base%20%5C%3A%20%20%3D%204%20%5C%3A%20cm%20%5C%5C%20%20%5C%5C%20height%20%5C%3A%20%20%3D%206%20%5C%3A%20cm%20%5C%5C%20%20%5C%5C%20area%20%5C%3A%20of%20%5C%3A%20traingle%20%3D%20%281%20%5Cdiv%202%29%20%5Ctimes%20base%20%5Ctimes%20height%20%5C%5C%20%20%5C%5C%20area%20%5C%3A%20%20%3D%20%284%20%5Ctimes%206%29%20%5Cdiv%202%20%5C%5C%20%20%5C%5C%20area%20%3D%2012%20%5C%3A%20cm.sq%20%5C%5C%20%20%5C%5C%20hope%20%5C%3A%20it%20%5C%3A%20helps)