Answer:
True
Step-by-step explanation:
anything over 9 that is not 9 is a repeating decimal
Answer:
A. 10 + 2
Step-by-step explanation:
288 ÷ 24 is shown below. 240 ÷ 24 = 10 288 – 240 = 48 48 ÷ 24 = 2
After obtaining the quotient of 288 ÷ 24 = 10,
Multiplying the divisor 24 by 10 = 240 and subtracting 240 from 288 to obtain the remainder = 48, the divisor is again used in 48 and the quotient obtained is added to the intula quotient value of 10 ; (10 + 2) and the process repeated
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.
Answer:
y= -6
Step-by-step explanation:
-9y+6=-10y
+10y. +10y
y+6=0
-6. -6
y=-6