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Posted by Tisha Nainani 6 years, 4 months ago
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Posted by Tisha Nainani 6 years, 4 months ago
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Posted by Tisha Nainani 6 years, 4 months ago
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Posted by Jeyanthi Sermaraja 6 years, 4 months ago
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Akshay Krishnan 6 years, 4 months ago
Santosh Kesarwani 6 years, 4 months ago
Ritik Raj 6 years, 4 months ago
Posted by Tisha Nainani 6 years, 4 months ago
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Raj Singh 6 years, 4 months ago
Posted by Mrinal Yash 6 years, 4 months ago
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Posted by Renuka Bhardwaj 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
x + y = 5 ...(1)
2x + 2y = 10 ...(2)
Here, a1 = 1, b1 = 1, c1 = -5
a2 = 2, b2 = 2, c2 = -10
We see that {tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}{/tex}
Hence, the lines represented by the equations (1) and (2) are coincident.
Therefore, equations (1) and (2) have infinitely many common solutions, i.e., the given pair of linear equations is consistent.
Graphical Representation, we draw the graphs of the equations (1) and (2) by finding two solutions for each if the equations. These two solutions of the equations (1) and (2) are given below in table 1 and table 2 respectively.
For equation (1) x + y = 5 {tex}\Rightarrow{/tex} y = 5 - x
Table 1 of solutions
| x | 0 | 5 |
| y | 5 | 0 |
For equations (2) x + 2y = 10
{tex}\Rightarrow{/tex} 2y = 10 - 2x
{tex}\Rightarrow y = \frac{{10 - 2x}}{2} \Rightarrow{/tex} y = 5 - x
Table 2 of solutions
| x | 1 | 2 |
| y | 4 | 3 |
We plot the points A(0, 5) and B(5, 0) on a graph paper and join these points to form the line AB representing the equation (1) as shown in the figure, Also, we plot the points C(1, 4) and D (2, 3) on the same graph paper and join these points to form the line CD representing the equation (2) as shown in the same figure.
In the figure we observe that the two lines AB and CD coincide.
Posted by Alisha Vir 6 years, 4 months ago
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Harsh Mishra 6 years, 4 months ago
If you mean, to prove (tan²θ - sin²θ) = tan²θ × sin²θ , then the answer will Be.....
L.H.S = tan²θ - sin²θ
= sin²θ/cos²θ - sin²θ
= sin²θ [ 1/cos²θ - 1]
[ we know, sec x = 1/cos x so, 1/cos²θ = sec²θ]
= sin²θ [ sec²θ - 1]
we also know that, sec² x - tan² x=1
so,sec² θ - tan² θ=1
or,sec² θ - 1 = tan² θ
then, sin² θ [sec²θ - 1] = sin² θ × tan² θ = RHS
Hence, (tan² θ - sin² θ) = tan² θ × sin² θ.
Posted by Karan Sahu 6 years, 4 months ago
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Posted by Sukhdev Singh 6 years, 4 months ago
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Posted by Aditya Bhashkar 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
Let a be the positive integer.
And, b = 4 .
Then by Euclid's division lemma,
We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .
°•° Then, possible values of r is 0, 1, 2, 3 .
Taking r = 0 .
a = 4q .
Taking r = 1 .
a = 4q + 1 .
Taking r = 2
a = 4q + 2 .
Taking r = 3 .
a = 4q + 3 .
But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .
•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .
Hence , it is solved
Posted by Trishna Kumari 6 years, 4 months ago
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Iron Man 6 years, 4 months ago
Posted by Prabhat Singh Baghel 6 years, 4 months ago
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Posted by Pranjal Rai 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).
{tex}\Rightarrow (x+1)^2=(x-5)^2{/tex}
⇒ x2 + 2x + 1 = x2 - 10x + 25
⇒ 2x + 1 = -10x + 25
⇒ 2x + 10x = 25 - 1
⇒ 12x = 24
⇒ x = 24/12
{tex}\Rightarrow x= 2{/tex}
Therefore , Required point is (2,0).
Posted by Khevna Modi 6 years, 4 months ago
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Posted by Parm Basra 6 years, 4 months ago
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Pranjal J. 6 years, 4 months ago
Posted by Neha Verma 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
α,β are zeros of ax2+bx+c
{tex}\mathrm{Then}\;\mathrm\alpha+\mathrm\beta=-\frac{\mathrm b}{\mathrm a}\;\mathrm{and}\;\mathrm{αβ}=\frac{\mathrm c}{\mathrm a}{/tex}
α4 + β4= (α2 + β2)2 - 2α2β2
=((α + β)2 - 2αβ)2 - (2αβ)2
{tex}={\left[ {\left( { - \frac{b}{a}} \right) - 2\left( {\frac{c}{a}} \right)} \right]^2} - \left[ {2{{\left( {\frac{c}{a}} \right)}^2}} \right]{/tex}
{tex} = {\left[ {\frac{{{b^2} - 2ac}}{{{a^2}}}} \right]^2} - \frac{{2{c^2}}}{{{a^2}}}{/tex}
{tex} = \frac{{{{\left( {{b^2} - 2ac} \right)}^2} - 2{a^2}{c^2}}}{{{a^4}}}{/tex}.
Posted by Priyanshi Jain 6 years, 4 months ago
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Ã.P.W.B.Đ ⚡ 6 years, 4 months ago
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Krishna Sharma 6 years, 4 months ago
Posted by Rohit Prasad 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given, an = 7 - 3n
Put n = 1, a1 = 7 - 3 {tex}\times{/tex} 1 = 7 - 3 = 4
Put n = 2, a2 = 7 - 3 {tex}\times{/tex} 2 = 7 - 6 = 1
Common difference(d) = 1 - 4 = -3
Sn = {tex}\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}
S25 = {tex} \frac { 25 } { 2 } [ 2 \times 4 + ( 25 - 1 ) ( - 3 ) ]{/tex}
{tex}= \frac { 25 } { 2 } [ 8 - 72 ]{/tex}
{tex}= \frac { 25 } { 2 } \times - 64{/tex}
= -800
Posted by Priyanshi P. 6 years, 4 months ago
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Posted by Aadil Aadhi 6 years, 4 months ago
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Nishant Bansal 6 years, 4 months ago
3Thank You