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Ashish Sah 5 years, 1 month ago

please reply
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Sia ? 5 years, 1 month ago

Represent root 11 on the number line, Locate root 11 on the number line, {tex}\sqrt {11}{/tex}on the number line.

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Rishav Jha 5 years, 1 month ago

Its simple if thre is no bar over digit .. Just multiply it by 1000 on upon.. Then the point will be removed.. 0.357 357/1000 Here p and q are integers and q is not 0
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Neeraj Kummithi 5 years, 1 month ago

Archemedes was a greek mathematician
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Kirti Kirti 5 years, 1 month ago

_121221
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Komal Pankhudi 5 years, 1 month ago

Right angled triangle

Ayush Raj 5 years, 1 month ago

Right angled triangle ?

Rishav Jha 5 years, 1 month ago

Its a right angle triangle because P+q+r= 180 angle sum property 60+30+r=180 r=180-90 r=90 These means that right angle is formed at r. Hence it is a right angle triangle

Manglam Yadav 5 years, 1 month ago

Acute angle triangle
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Somanath Nayak 5 years, 1 month ago

When 2 angles firm 180 degrees those angles r said to be complementary of each othee
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Sujeet More 5 years, 1 month ago

(X square + 1 + root 2x)(x square + 1 - root 2x)
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Vinod Singh 5 years, 1 month ago

7
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Sujeet More 5 years, 1 month ago

(2x+5)(3x-4)
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Anshul Kumar 5 years, 1 month ago

First minus .5 from right side and then add .5 in left side then repeat this process in all the numbers
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Yogita Ingle 5 years, 1 month ago

Euclid axioms

  1. Things which are equal to same things are equal to one another
  2. If equals are added to equals, the wholes are equal
  3. If equals are subtracted from equals, the remainders are equal
  4. Things which coincide with one another are equal to one another
  5. This seems to say that if two things are identical (that is, they are the same), then they are equal. In other words, everything equals itself. It is the justification of the principle of superposition.
  6. The whole is greater than the part if a quantity P is part of quantity Q , then Q > P. or we can that there exists a third quantity R such that Q=P+R
  7. Things which are double of the same things are equal to one another
  8. Things which are halves of the same things are equal to one another
  9. If first thing is greater than second and second is greater than third, then first is greater than third

Euclid Postulates

Here are the Euclid Postulates

  1. A straight line may be drawn from one point to another point. As per this postulate,there exists a unique line joining two points
  2. 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
  3. 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
  4. 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
  5. 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
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A K47 5 years, 1 month ago

If the two sides are of length 12 then third will be : 12+12+x = 30 x = 30 - 24 = 6 Then area of triangle By heron's formula √s(s-a)(s-b)(s-c) √15(15-12)(15-6)(15-12) √15*3*9*3 = √1215 = 34.85cm^2
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Yogita Ingle 5 years, 1 month ago

The measurement of the total space occupied by a solid is the volume of a three-dimensional figure. Any object that has length, breadth, and thickness is a three-dimensional figure. The difference between the total amount of space left inside the hollow body and the space occupied by the body is the volume of a hollow 3-dimensional figure.

Saransh Gupta 5 years, 1 month ago

Ghsvb
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Sia ? 5 years, 1 month ago

Given: A quadrilateral ABCD, in which
AC and BD intersect each other at point E.
To Prove: {tex}ar{\text{ }}(AED) \times ar{\text{ }}(BEC){/tex}
{tex}= {\text{ }}ar{\text{ }}(ABE) \times ar{\text{ }}(CDE){/tex}
Construction: A, draw AM {tex} \bot {/tex} BD and from C, draw CN {tex}\bot {/tex} BD.
Proof: ar ({tex}\Delta {/tex}ABE) = {tex}\frac{1}{2} \times BE \times AM{/tex} ……….(i)
And ar ({tex}\Delta {/tex}AED) = {tex}\frac{1}{2} \times DE \times AM\;{/tex}……….(ii)
Dividing eq. (ii) by i), we get,
{tex}\frac{{{\text{ar}}\left( {\Delta {\text{AED}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{ABE}}} \right)}} = \frac{{\frac{1}{2} \times {\text{DE}} \times {\text{AM}}}}{{\frac{1}{2} \times {\text{BE}} \times {\text{AM}}}}{/tex}
{tex}\Rightarrow{/tex} ……….(iii)
Similarly  ……….(iv)
From eq. (iii) and (iv), we get
{tex}\frac{{{\text{ar}}\left( {\Delta {\text{AED}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{ABE}}} \right)}}{/tex} = {tex}\frac{{{\text{ar}}\left( {\Delta {\text{CDE}}} \right)}}{{{\text{ar}}\left( {\Delta {\text{BEC}}} \right)}}{/tex}
{tex} \Rightarrow ar{\text{ }}(AED) \times ar{\text{ }}(BEC){/tex}
{tex} = {\text{ }}ar{\text{ }}(ABE) \times ar{\text{ }}(CDE){/tex}

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Vinod Kumar 5 years, 1 month ago

Where is abcissca in graph

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