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Ask QuestionPosted by Aniket Sharma 4 years, 5 months ago
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Posted by Abhay Yadav Ji Yboys 4 years, 5 months ago
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Yogita Ingle 4 years, 5 months ago
In △ADC,S is the mid-point of AD and R is the mid-point of CD
In △ABC,P is the mid-point of AB and Q is the mid-point of BC
Line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of of it.
∴SR ∥ AC and SR= 1/2 AC ....(1)
∴PQ ∥ AC and PQ= 1/2 AC ....(2)
From (1) and (2)
⇒ PQ = SR and PQ ∥ SR
So,In PQRS,
one pair of opposite sides is parallel and equal.
Hence, PQRS is a parallelogram.
PR and SQ are diagonals of parallelogram PQRS
So,OP=OR and OQ=OS since diagonals of a parallelogram bisect each other.
Hence proved
Posted by God Of Death 4 years, 5 months ago
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Ekta Ekta 4 years, 5 months ago
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Meghna Thapar 4 years, 5 months ago
Surds are numbers left in 'square root form' (or 'cube root form' etc). They are therefore irrational numbers. The reason we leave them as surds is because in decimal form they would go on forever and so this is a very clumsy way of writing them. Surds originated from the Latin word surdus which meant "mute". This muted sound is largely thought that it represents irrational numbers whereas rational numbers would be a pure, clear sound.
Posted by Mudassar Latief 4 years, 5 months ago
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Meghna Thapar 4 years, 5 months ago
The definition of coordination is being able to move and use your body effectively and multiple people or things working well together. An example of coordination is when a gymnast walks on a tightrope without falling. An example of coordination is when two people work together to plan or coordinate a party. Coordinate is one of those words that can mean very different things but is rarely misunderstood in context. It's a great way to describe the work of organizing, planning, and strategizing.
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Shaitan Singh Rajpurohit 4 years, 5 months ago
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Rajan Kumar Pasi 4 years, 5 months ago
I think you are not studying properly, examples are given in the book for this kind of question.
Sol : Let x = 0.373737...
Thus, 100 x = 37.373737...
Subtracting both equations, we get
99 x = 37.000000....
{tex}\huge \boxed{x = \frac {37}{99}}{/tex}
------------------The End--------------------
Posted by Wildheart Devil 4 years, 5 months ago
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Rajan Kumar Pasi 4 years, 5 months ago
{tex}\huge {Q. If\ 2^a = 3^b = 6^c,\ Show\ that,\ \frac1c=\frac{1}{\frac{a+1}{b}}?}{/tex}
| {tex}\huge{Sol:\ Let's\ first\ take\ ,\ 2^a = 3^b }\\ \large{\ \text{(Applying log on both the sides)}}\\ \huge{=>\ \log\ 2^a=\log\ 3^b}\\ \huge{=>\ a\log2=b\log3}\\ \huge{=>\ a={b{\log3\over\log2}}}\\ {/tex} | {tex}\huge{Similarly\ ,\ 3^b=6^c }\\ \large{\ \text{(Applying log on both the sides)}}\\ \huge{=>\ \log\ 3^b=\log\ 6^c}\\ \huge{=>\ b\log3=c\log6}\\ \huge{=>\ c={b{\log3\over\log6}}}\\ {/tex} |
| {tex}\huge{LHS\ ,\ \frac1c = \frac{\log6}{b\log3} }\\ \huge{RHS\ ,\ \frac1a+\frac1b= \frac{\log2}{b\log3}+\frac1b }\\ \huge{=>\frac1b\Bigg( \frac{\log2}{\log3}+1\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log2+\log3}{\log3}\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log(2\times 3)}{\log3}\Bigg) }\\ \huge{=>\frac1b\Bigg( \frac{\log6}{\log3}\Bigg) }\\ {/tex} | {tex}\huge{Thus,\ LHS=RHS}\\ \huge{\frac1c =\ \frac1a+\frac1b= \huge{\frac1b\Bigg( \frac{\log6}{\log3}\Bigg) }\\ }\\ {/tex} |
Please use brackets and mathematical signs in the message of your questions.
Posted by Nitya Singh 4 years, 5 months ago
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Posted by Vikash Kumar 4 years, 5 months ago
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Rajan Kumar Pasi 4 years, 5 months ago
{tex}\huge{Q.\ Simplify,}\\ \huge { A) \sqrt{45}+\sqrt{80}-3\sqrt{20}}\\ \huge {B)7\sqrt{6}-\sqrt{252}-\sqrt{294}+6\sqrt{7}}\\ \huge{ C)4\sqrt{28} +3\sqrt{7} }\\{/tex}
Sol:
| {tex}\large{A)\ \sqrt{45}+\sqrt{80}-3\sqrt{20}}\\ \large{=>\ \sqrt{9\times5}+\sqrt{16\times5}-3\sqrt{4\times5}}\\ \large{=>\ 3\sqrt{5}+4\sqrt{5}-3\times2\sqrt{5}}\\ \large{=>\ 3\sqrt{5}+4\sqrt{5}-6\sqrt{5}}\\ \large{=>\ (3+4-6)\sqrt{5}}\\ \large{=>\ (1)\sqrt{5}=>\boxed{\sqrt5}}\\{/tex} | {tex}\large{B)\ 7\sqrt{6}-\sqrt{252}-\sqrt{294}+6\sqrt{7}}\\ \large{=>\ 7\sqrt{6}-\sqrt{7\times6\times6}-\sqrt{6\times7\times7}+6\sqrt{7}}\\ \large{=>\ 7\sqrt{6}-6\sqrt{7}-7\sqrt{6}+6\sqrt{7}}\\ \large{=>\boxed{0}}\\{/tex} |
| {tex}\large{C)\ 4\sqrt{28}+3\sqrt{7}}\\ \large{=>\ 4\sqrt{7\times4}+3\sqrt{7}}\\ \large{=>\ 4\times2\sqrt{7}+3\sqrt{7}}\\ \large{=>\ 8\sqrt{7}+3\sqrt{7}}\\ \large{=>\ (8+3)\sqrt{7}}\\ \large{=>\boxed{11\sqrt7}}\\{/tex} | Do something by yourself too, not every question is hard. |
Posted by Vikash Kumar 4 years, 5 months ago
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Rajan Kumar Pasi 4 years, 5 months ago
{tex}Q.\text{ If x = 2 + }\sqrt5\text{, prove that: }{x^2+\frac1{x^2}=18.}{/tex}
| {tex}Sol.\text{ If x = 2 + }\sqrt5,\\ {=>\ x^2=4+4\sqrt5+5= 9+4\sqrt5}\\ So,{x^2+\frac1{x^2}={(9+4\sqrt5)+\frac1{(9+4\sqrt5)}}}\\ (Rationalising,)\\ =>\ {(9+4\sqrt5)+\frac1{(9+4\sqrt5)}\times\frac{(9-4\sqrt5)}{(9-4\sqrt5)}}\\ =>\ {(9+4\sqrt5)+\frac{(9-4\sqrt5)}{(81-80)}}\\ =>\ {(9+4\sqrt5)+\frac{(9-4\sqrt5)}{(1)}}\\ =>\ {(9+4\sqrt5)+{(9-4\sqrt5)}}\\ =>18\\ =>LHS=RHS\\ Hence Proved...{/tex} | {tex}\huge{Identities\ used:}\\ \huge{1.\ (a+b)^2=a^2+2ab+b^2}\\ \huge{2.\ (a+b)(a-b)=a^2-b^2}\\{/tex} |
Posted by Vikash Kumar 4 years, 5 months ago
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Rajan Kumar Pasi 4 years, 5 months ago
I have answered another question like this one, try to understand it and solve this by yourself.
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Bangtan Army Forever 4 years, 5 months ago
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