I know my math and your answer is going to be A.
Answer: Choice D) 
The steps to finding the inverse will have us swap x and y. Afterward, we solve for y

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Extra info:
The inverse relation is not a function because of the plus minus. For instance, plugging x = 4 into
leads to y = -3 and y = 3 simultaneously. You would have to apply a domain restriction on
to make it a one-to-one function, to make the inverse a function. One possible domain restriction is
which would lead to the inverse function 
Answer: x=3
Step-by-step explanation:
4(2x-7)-2x=-10
Multiply 4 with 2x and -7
8x-28-2x=-10
Add like terms which are 8x and -2x
6x-28=-10
Add 28 to both sides so the -28 cancels out
6x=18
Divide both sides by 6 to isolate the x
Answer= x=3
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.