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Insha Naaz 4 years, 10 months ago
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Posted by Jasvirkaur Jasvirkaur 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
s = (a + b + c)/2
Area of a triangle = √ s (s -a) (s - b) (s -c)
Posted by Avinash S 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
The circle drawn with S (circumcenter) as center and passing through all the three vertices of the triangle is called the circumcircle.
The point of concurrency of the perpendicular bisectors of the sides of a triangle is called the circumcenter and is usually denoted by S.
Posted by Emi Lima 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
3,6,9,12,15,18 are first three multiples of 3.
we know that ,
{tex}mean =\frac{sum \: of \: the \: observations }{number \: of \: observations }{/tex}
{tex}mean =\frac{3+6+9+12+15+18}{6}\\=\frac{63}{6}\\=\frac{21}{2}\\=10.5{/tex}
Therefore,
mean = 10.5
Posted by Nidhi Oberoi 4 years, 10 months ago
- 3 answers
Yogita Ingle 4 years, 10 months ago
Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'.
Hence OA = OB = 5 cm
➾ O'A = O'B
➾ 3 cm
OO' is the perpendicular bisector of chord AB.
Therefore, AC = BC
Given, OO' = 4 cm
Let OC = x
Hence O'C = 4 − x
In right angled ΔOAC, by Pythagoras theorem OA²
➾ OC² + AC²
➾ 5² = x² + AC²
➾ AC² = 25 − x²à
(1) In right angled ΔO'AC, by Pythagoras theorem O'A²
➾ AC² + O'C²
➾ 3² = AC² + (4 – x)²
➾ 9 = AC² + 16 + x² − 8x
➾ AC² = 8x − x² − 7 à
(2) From (1) and (2),
we get 25 − x² = 8x − x² − 7 8x = 32
Therefore, x = 4
Hence, the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.
AC² = 25 − x²
➾ 25 − 4²
➾ 25 − 16 = 9
Therefore, AC = 3 m Length of the common chord,
➾ AB = 2AC
➾ 6 cm
Posted by Sachin Hota 4 years, 10 months ago
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Insha Naaz 4 years, 10 months ago
Posted by Abhishek Patil 4 years, 10 months ago
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Mohak Shah 4 years, 10 months ago
Mr. X 4 years, 10 months ago
Posted by Abhishek Patil 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
- A straight line may be drawn from one point to another point. As per this postulate,there exists a unique line joining two points
- A terminated line can be produced indefinitely. A terminated line or line segment is the segment of line joined by two separate point. As per this postulate, a line segment can be extended in in both direction to make a line
- A circle can be drawn with any centre and any radius. As per this postulate, we can draw circles with any centre and any radius. For example,we can draw a circle on line segment with centre at its end point and radius as the length of the line segments
- All right angles are equal to one another. A right angle is angle measuring 900. As per this postulate ,all the right angles are equal and if we place one right angle over top of another right angle, they will coincide
- If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the straight lines if produced indefinitely meet on that side on which the angles are less than the two right angles
Posted by Abhishek Patil 4 years, 10 months ago
- 1 answers
Yogita Ingle 4 years, 10 months ago
X = 2 , y = 14
1. x + y = 16
2. 2 x + y = 18
There are infinite no: of lines passing through (2 , 14) because an equation in two lines have infinite no: of solutions.
Posted by Ishan Kohli 4 years, 10 months ago
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Posted by Gurjent Singh 4 years, 10 months ago
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Posted by Chiku Kalsi 4 years, 10 months ago
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Posted by Vijay Laxmi 4 years, 10 months ago
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Yogita Ingle 4 years, 10 months ago
p(x) = 3x3 - 4x2 + 7x - 5
Remainder when x = 3 divides p(x) is
p(3) = 3 
33 - 4 32 + 7 3 - 5
= 3 27 - 4 9 + 21 - 5 = 61
Remainder when x = - 3 divides p(x) is
p(-3) = 3 (-3)3 - 4(-3)2 + 7 (-3) - 5
= 3 (-27) - 4 9 - 21 - 5
= -143
Therefore, the required remainders are 61 and -143.
Posted by Manish Yadav 4 years, 10 months ago
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Posted by Manish Yadav 4 years, 10 months ago
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