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Ask QuestionPosted by Krishnaansh Viz 6 years, 6 months ago
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Posted by Vansh Gupta 6 years, 6 months ago
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Tripti Rawat 6 years, 6 months ago
Using identity ---- a^3 - b^3= (a-b)(a^2+a×b+b^2),
= x {(1)^3 - (2×y)^3}
= x {1 - 2y} { (1)^2 + 1×2y + (2y)^2 }
= x {1 - 2y} { 1 + 2y + 4(y)^2 }
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Yogita Ingle 6 years, 6 months ago
Let us draw the graphs of both the equations. The tables for these equations are:
| x | 0 | 6 |
| y = (6 – x)/3 | 2 | 0 |
| x | 0 | 3 |
| y = (2x – 12)/3 | – 4 | – 2 |
Now, take a graph paper and plot the following points:
- A (0, 2)
- B (6, 0)
- P (0, – 4)
- Q (3, – 2)
Next, draw the lines AB and PQ as shown below.
From the figure above, you can see that the two lines intersect at the point B (6, 0). Therefore, x = 6 and y = 0 is the solution of this pair of equations in two variables. Hence, it is Consistent.
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45516111? 46 6 years, 6 months ago
Yogita Ingle 6 years, 6 months ago
Irrational numbers: Irrational numbers are those which can’t be expressed in fractional form, i.e., in pq form. They neither terminate nor do they repeat. They are also known as non- terminating non-repeating numbers.
Shambhavi Srivastav 6 years, 6 months ago
Posted by Brije Mohan Singh 6 years, 6 months ago
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Veera Sharma 6 years, 6 months ago
Yogita Ingle 6 years, 6 months ago
2x2 - 7x - 15
= 2x2 - 10x + 3x - 15
= 2x( x - 5) + 3 (x - 5)
= (2x + 3) ( x-5 )
Posted by Technical 6G World 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
{tex}{\mathrm T}_\mathrm n\;=\;{\mathrm S}_\mathrm n\;-\;{\mathrm S}_{\mathrm n-1}{/tex}
{tex}{\mathrm T}_{\mathrm n-1}\;=\;{\mathrm S}_{\mathrm n-1}\;-\;{\mathrm S}_{\mathrm n-2}{/tex}
{tex}{\mathrm S}_\mathrm n\;-\;2{\mathrm S}_{\mathrm n-1}\;+\;{\mathrm S}_{\mathrm n-2}\;=\;{\mathrm S}_\mathrm n\;-\;{\mathrm S}_{\mathrm n-1}\;-\;{\mathrm S}_{\mathrm n-1}\;+\;{\mathrm S}_{\mathrm n-2}{/tex}
= {tex}({\mathrm S}_\mathrm n\;-\;{\mathrm S}_{\mathrm n-1})\;-\;({\mathrm S}_{\mathrm n-1}\;-\;{\mathrm S}_{\mathrm n-2}){/tex}
= {tex}({\mathrm T}_\mathrm n\;-\;{\mathrm T}_{\mathrm n-1}){/tex}
= d
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Sia ? 6 years, 6 months ago
Consider the following pair of linear equations in two variables,
The solution of this pair would be a pair (x, y). Let’s find the solution, geometrically. The tables for these equations are:
Now, take a graph paper and plot the following points:
Next, draw the lines AB and PQ as shown below.
From the figure above, you can see that the two lines intersect at the point Q (4, 2). Therefore, point Q lies on the lines represented by both the equations, x – 2y = 0 and 3x + 4y = 20. Hence, (4, 2) is the solution of this pair of equations in two variables. Let’s verify it algebraically:
1Thank You