You could use simple factoring to solve this. You would factor out a 7 and then have
7(3+4m) = A
l x w = A
w= 3 +4m
Answer:
b
Step-by-step explanation:
the step by step explanation is that because the word increase is the clave word just belive in me
Set the height to h, and the width to w.
We know that wh=190 and h=2w-1.
Substituting 2w-1 for h, we have:
w(2w-1)=190
So:
2w^2-w-190=0
Factoring this equation, we get (2w+19)(w-10)=0. The solutions to this equation are -9.5 and 10, but clearly the width must be positive, Substituting 10 for the width, we get 10*2-1=19 for the height.
I am going to do number 2 so when you subtract decimals you have to add two zeros like this 10.00 or 0.25 so now when you subtract you get $9.75
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.