I’m pretty sure it’s C. because half of 12 is 6 so if they’re buying a pound and a half it would be 12+6, which is 18
Answer:
C is the answer.
Step-by-step explanation:
Answer: well your answer is 18360
Step-by-step explanation: so let's say 51,000 is 100% and x is what where looking for, we know x is 36% so write x=36% the you have 51,000=100% now you have two simple equations.
Answer:
the equation is
y = 0.75x
And, the constant of proportionality is 0.75
Step-by-step explanation:
The computation is shown below:
Given that
4y = 3x
And,
12y = 9x
Here the equation that represent proportional relationship is
4y = 3x
Divide both sides by 4
y = 3x ÷ 4
Now
12y = 9x
divides both sides by 12
y = 9 ÷ 12x
y = 3 ÷ 4x
y = 0.75x
so the equation is
y = 0.75x
And, the constant of proportionality is 0.75
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.