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Ask QuestionPosted by Nandini Shukla 6 years, 2 months ago
- 14 answers
Aseem Mahajan 6 years, 2 months ago
Aseem Mahajan 6 years, 2 months ago
Aseem Mahajan 6 years, 2 months ago
Aseem Mahajan 6 years, 2 months ago
Posted by Meena Madhu 6 years, 2 months ago
- 1 answers
Posted by Janhwi Rai 6 years, 2 months ago
- 2 answers
Posted by Aayush Kumar Chandrawanshi 6 years, 2 months ago
- 16 answers
Aadya Singh ? 6 years, 2 months ago
(3x+1-9)/3 =0
3x-8 =0×3
3x-8=0 ...dekho...aise 3 aayega????
Aseem Mahajan 6 years, 2 months ago
Aseem Mahajan 6 years, 2 months ago
Aadya Singh ? 6 years, 2 months ago
=> 3x-8 = 0
D= b^2 - 4ac
3^2+ 4×0×8
=> 9
Posted by Aayush Kumar Chandrawanshi 6 years, 2 months ago
- 1 answers
Yogita Ingle 6 years, 2 months ago
Considering AZ = ZB, that means {tex}A Z^{2}=Z B^{2}{/tex}
{tex}(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2}{/tex}
Squaring both the sides we get the factors and then simplifying them we get:
{tex}x^{2}+25-10 x+y^{2}+1-2 y=x^{2}+1+2 x+y^{2}+25-10 y{/tex}
12x = 8y
3x = 2y
Hence proved
Posted by Aayush Jain 6 years, 2 months ago
- 7 answers
Mani Kumar 6 years, 2 months ago
Diksha? 6 years, 2 months ago
Posted by Vipul Kumar Gupta 6 years, 2 months ago
- 2 answers
Yogita Ingle 6 years, 2 months ago
{tex}Distance\: from\\ \:the \:point \:P(3,2)\: and\\ \:the \:origin\\=5{/tex}
Step-by-step explanation:
{tex}\boxed{Distance\: from\\ \:the \:point \:(x,y)\: and\\ \:the \:origin\: is\: \sqrt{x^{2}+y^{2}}}{/tex}
{tex}Distance\: from\\ \:the \:point \:P(3,2)=(x,y)\: and\\ \:the \:origin\\= \sqrt{3^{2}+2^{2}}\\=\sqrt{9+4}\\=\sqrt{13}\\{/tex}
Therefore,
{tex}Distance\: from\\ \:the \:point \:P(3,2)\: and\\ \:the \:origin\\=\sqrt{13}{/tex}
Manisha Sharma 6 years, 2 months ago
Posted by Varun Chadha 6 years, 2 months ago
- 1 answers
Posted by Ansh Pathak 6 years, 2 months ago
- 0 answers
Posted by ? Roman ? 6 years, 2 months ago
- 3 answers
Yogita Ingle 6 years, 2 months ago
It is given that, difference of two natural numbers is 5
Let the x, (x+5) are two natural numbers.
Reciprocals of the numbers are
x and 1/(x+5)
According to the problem given,
{tex}\frac{1}{x}-\frac{1}{x+5}=\frac{1}{10}{/tex}
{tex}\implies \frac{x+5-x}{x(x+5)}=\frac{1}{10}{/tex}
{tex}\implies \frac{x}{x^{2}+5x}=\frac{1}{5}{/tex}
Do the cross multiplication, we get
{tex}\implies 50=x^{2}+5x{/tex}
=> x²+5x-50=0
Splitting the middle term, we get
=> x²+10x-5x-50=0
=> x(x+10)-5(x+10)=0
=>(x+10)(x-5)=0
=> x+10 = 0 Or x-5=0
=> x = -10 Or x = 5
But the two numbers are natural numbers.
x = 5
Therefore,
Required two natural numbers are,
x = 5
and
x+5 = 5+5 = 10
Posted by Dinesh Chettri 6 years, 2 months ago
- 1 answers
Aadya Singh ? 6 years, 2 months ago
Posted by Utkarsh Vatsa 6 years, 2 months ago
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Posted by Khushboo Meena 6 years, 2 months ago
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Posted by Aniket Anand 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
A polynomial is defined as an expression which contains two or more algebraic terms. It is made up of two terms namely Poly (meaning “many”) and Nominal (meaning “terms.”). Polynomials are composed of:
- Constants such as 1, 2, 3, etc.
- Variables such as g, h, x, y, etc.
- Exponents such as 5 in x5 etc.
Posted by Aniket Anand 6 years, 2 months ago
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Diksha? 6 years, 2 months ago
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Ankesh ...... 6 years, 2 months ago
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Nandini Shukla 6 years, 2 months ago
1Thank You