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Yogita Ingle 6 years, 6 months ago
Postulate – I :A straight line segment can be formed by joining any two points in space.
Postulate – II: Any straight line can be extended indefinitely on both sides.
Postulate – III: A circle can be drawn with any centre and any radius. For any line segment, a circle can be drawn with its centre at one endpoint and the radius of the circle as the length of the line segment.
Postulate – IV: All right angles are congruent or equal to one another.
Postulate – V: Two lines are parallel to each other if they intersect the third line and the interior angle between them is 180 degrees.
Posted by Geeta Bohra 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let p(x) = 3x3 + x2 - 22x + 9 and q(x) = 3x2 + 7x - 6
We know if p(x) is divided by q(x) which is quadratic polynomial therefore
if p(x) is not exactly divisible by q(x) then the remainder be r(x) and degree of r(x) is less than q(x) (or Divisor)
∴By long division method
Let we added ax + b (linear polynomial) is p(x), so that p(x) + ax + b is exactly divisible by 3x2 + 7x - 6.
Hence p(X) + ax + b = s(x) = 3x3 + x2 - 22x + 9
ax + b = 3x3 + x2 - x(22 - a) + (9 + b)
Hence, x(a – 2 + b – 3 = 0 . x + 0)
⇒ a - 2 = 0 & b - 3 = 0 ⇒ a = 2 or b = 3
Hence, if we add ax + b or 2x + 3 in p(x) then it is exactly divisible by 3x2 + 7x - 6.
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Yogita Ingle 6 years, 6 months ago
Linear equations in two variables are equations which can be expressed as ax + by + c = 0, where a, b and c are real numbers and both a, and b are not zero. The solution of such equations is a pair of values for x and y which makes both sides of the equation equal.
Alishka 02 6 years, 6 months ago
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Yogita Ingle 6 years, 6 months ago
(x + a) (x + b) = x2 + (a+b)x + ab
(X+4)(x+10) = x2 + (10+4)x + 10(4)
= x2 + 14x + 40
Mihir Kumar 6 years, 6 months ago
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Smita Jaiswal 6 years, 6 months ago
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