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Posted by Deachan Chuskit 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
The graphical representation of 2x+ y= 6 is given in the figure below.
This line divides the xy-plane in two half planes, I and II.
Select a point (not on the line), which lies in one of the half planes, to determine whether the point satisfies the given inequality or not.
We select the point as (0, 0).
It is observed that,
2(0) + 0 ≥ 6 or 0 ≥ 6, which is false.
Therefore, half plane I is not the solution region of the given inequality.
Also, it is evident that any point on the line satisfies the given inequality.
Thus, the solution region of the given inequality is the shaded half plane II including the points on the line.
This can be represented as follows.
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⠀ 5 years, 3 months ago
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Gaurav Seth 5 years, 3 months ago
Let A = (5,– 1,1)
B = (7,– 4, 7)
C = (1, – 6, 10)
D = (– 1, – 3, 4)
Then the position vectors of A,B,C,D are
∴ ABCD forms a Rhombus.
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Yogita Ingle 5 years, 3 months ago
5×2+0×5+8×0+9+8
= 10+0×5+8×0+9+8
= 10 +8×0+9 + 8
= 10 + 9 + 8
= 19 + 8
= 27
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Gaurav Seth 5 years, 3 months ago
7, 77, 777, 7777 ….to n terms.
Sn=7+77+777+7777+……+to n terms
introducing 9
= (7/9) [9+99+999+……+ to n terms]
= (7/9) [(10-1) + (100-1) + (1000-1) +……...+ to n term]
= (7/9) [10+100+1000+……... n terms – (1+1+1+1+…… n terms)
= (7/9) [10+100+1000+………n terms –n]
Here a=10, r=10
= (7/9) [10(10n-1)/ (9) - n]
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Har Har Mahadev 🙏 5 years, 3 months ago
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