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Ask QuestionPosted by Sanjit Ram 5 years, 5 months ago
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Posted by Sumit Thakur 5 years, 5 months ago
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Sia ? 5 years, 5 months ago
We have, x2 + 4 ix - 4 = 0 ...(i)
On comparing Eq. (i) with ax2 + bx + c = 0, we get
a = 1, b = 4 i and c = - 4
{tex}\because{/tex} x = {tex}\frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }{/tex}
{tex}\therefore{/tex} x = {tex}\frac { - 4 i \pm \sqrt { ( 4 i ) ^ { 2 } - 4 \times 1 \times ( - 4 ) } } { 2 \times 1 }{/tex}
= {tex}\frac { - 4 i \pm \sqrt { - 16 + 16 } } { 2 }{/tex}
= {tex}\frac { - 4 i } { 2 }{/tex} = - 2 i
Hence, the roots of the given equation are - 2 i and - 2 i.
Posted by Charna Jak 5 years, 5 months ago
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Posted by Jai Phadtare 5 years, 5 months ago
- 1 answers
Gaurav Seth 5 years, 5 months ago
( √3 + √2 ) / ( √3 - √2 )
= [(√3 + √2)(√3 + √2)]/[(√3-√2)(√3+√2)]
= ( √3 + √2 )² / [ (√3 )² - ( √2 )² ]
= [ (√3)² +2×(√3 )(√2) + (√2)² ]/(3-2)
= 3 + 2√6 + 2
= 5 + 2√6
Posted by Mayank Sinha 5 years, 5 months ago
- 3 answers
Posted by Mdmoheen Fda 5 years, 5 months ago
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Purva Dhammi 5 years, 5 months ago
(4,0)
As it lies on x-axis so coordinate of y would be zero.
And right to the origin implies x to be positive
Posted by Atharv Patange 5 years, 5 months ago
- 4 answers
Aditya Mishra Smhs Chintal, X-A Hyd 5 years, 5 months ago
Posted by Sanjay Kumar 5 years, 5 months ago
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Posted by Champa Devi 5 years, 5 months ago
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Posted by Badal Sahu 5 years, 5 months ago
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Posted by Devraj Singh 5 years, 5 months ago
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Posted by Parmod Kumar 5 years, 5 months ago
- 2 answers
Samridhi ?? 5 years, 5 months ago
Posted by Atulya Mishra 5 years, 5 months ago
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Posted by Srishti Nagar 5 years, 5 months ago
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Posted by Asha Mahesh 5 years, 5 months ago
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Satyam Lodhi Rajput 5 years, 5 months ago
Posted by G Kr 5 years, 5 months ago
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Posted by Shivaraj Dumbali 5 years, 5 months ago
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Posted by Ashmeet Sandhu 5 years, 5 months ago
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Posted by Lal Prabhakar Singh 5 years, 5 months ago
- 2 answers
Yogita Ingle 5 years, 5 months ago
(a + b + c)² = a² + b² + c² +2(ab + bc + ca)
=250 + 2 × 3
= 256
a + b + c = √256
= 16
Dishita Tuteja 5 years, 5 months ago
Posted by Suneel Kumar 5 years, 5 months ago
- 2 answers
Yogita Ingle 5 years, 5 months ago
Given Equation is 2a^3 + 16b^3 - 5a - 10b.
This can be written as
2(a^3 + 8b^3) - 5a - 10b
We know that a 3 + b 3 = (a+b)(a2-ab+b2)
= > 2(a + 2b)(a2 - 2ab + 4b2) - 5a - 10b
= > 2(a + 2b)(a2 - 2ab + 4b2) - 5(a + 2b)
= > (a + 2b)(2(a 2 -2ab + 4b2) - 5)
= > (a + 2b)(2a2 - 4ab + 8b2 - 5)
Therefore the equation can be factorized as (a + 2b)(2a2 - 4ab + 8b2 - 5)
Posted by Adarsh Jain 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
=√50-√98+√162
=5√2-7√2+9√2
Taking root 2 common,
=√2(5-7+9)
=7√2
Posted by Tanisha Rawat 5 years, 5 months ago
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Posted by Meenakshi Kansal 5 years, 5 months ago
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Posted by Divya Gupta 4 years, 7 months ago
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Sia ? 5 years, 5 months ago
According to question, x : y = 11 : 19
Therefore, x = 11a and y = 19 a
Now, AD is parallel to BE
So
{tex}\angle{/tex}BDC = {tex}\angle{/tex}ABD
y = 32
19 a = 32
So a = {tex}\frac{32}{19}{/tex}
Thus x = {tex}11 \times \frac{32}{19}{/tex}
x = 18.52
Since {tex}\angle{/tex}DCE ={tex}\angle{/tex}ADC
So. {tex}\angle{/tex}DCE = x + 32 = 18.52 + 32 = 50.32
Posted by Nikhil Saxena 5 years, 5 months ago
- 2 answers
Posted by Meenakshi Kansal 5 years, 5 months ago
- 1 answers
Sia ? 5 years, 5 months ago
49x2 - b = {tex}\left( {7x + \frac{1}{2}} \right)\left( {7x - \frac{1}{2}} \right){/tex}
49x2 - b = (7x)2 - {tex}{\left( {\frac{1}{2}} \right)^2}{/tex}
49x2 - b = 49x2 - {tex}\frac 14{/tex}
b = {tex}\frac 14{/tex}
Posted by Meenakshi Kansal 5 years, 5 months ago
- 0 answers
Posted by Meenakshi Kansal 5 years, 5 months ago
- 1 answers
Yogita Ingle 5 years, 5 months ago
If f(x) is a polynomial such that f (-1/3) =0
So, f(x) = polynomial with variable x
If f(-1/3) = 0 (it's the value of the polynomial at that point)
So, one factor of f(x) will be (x + 1/3)
Because, if the zero of the polynomial is a negative number, the factor will be a positive number.
Posted by Meenakshi Kansal 5 years, 5 months ago
- 1 answers
Sia ? 5 years, 5 months ago
We know, √3 = 1.732 and √5 = 2.23
Hence, irrational numbers between them are 1.737337333..... or 2.101001000.....
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Samridhi ?? 5 years, 5 months ago
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