Answer:
The statement is false.
Step-by-step explanation:
Let a and b represent two odd integers.
The average (A) is defined by:
A=
In order to prove the statement, you have to show that it's true for all odd integers. But, to disprove it, you just have to find a counterexample where the statement is false.
Notice that it is easier to try to find a counterexample.
For example, a=3 and b=5
A= 
The result of the average is even, therefore the statement is false.
The previous calculation is a counterexample.
The greatest common factor of 19x7 AND 3x5 is: X5
hope this helps
Answer:
Fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
Step-by-step explanation:
Given the figure with dimensions. we have to find the area of given figure.
Area of figure=ar(1)+ar(2)+ar(3)
Area of region 1 = ar(ANGI)+ar(AIB)
![=L\times B+\frac{1}{2}\times base\times height\\\\=[1500\times (5000-2000-1500)]+\frac{1}{2}\times (3000-1500)\times (5000-2000-1500)\\\\=3375000m^2=337.5ha](https://tex.z-dn.net/?f=%3DL%5Ctimes%20B%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height%5C%5C%5C%5C%3D%5B1500%5Ctimes%20%285000-2000-1500%29%5D%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%283000-1500%29%5Ctimes%20%285000-2000-1500%29%5C%5C%5C%5C%3D3375000m%5E2%3D337.5ha)
Area of region 2 = ar(DHBC)
Area of region 3 = ar(GFEH)
Hence, Area of figure=ar(1)+ar(2)+ar(3)=337.5ha+300ha+350ha
=987.5 ha
Now, we have to do straight-line fencing such that area become half and cost of fencing is minimum.
Let the fencing be done through x m downward from B which divides the two into equal area.
⇒ Area of upper part above fencing=Area of lower part below fencing
⇒
Hence, fencing is done along KL which is (1500+520.8=2020.8 m) from the top left corner and divides the property into half.
The answer will be 250 because you divide 2 by 500
The missing number is 4/6
the number line goes 1/6 2/6 3/6 4/6 5/6 6/6
