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Ask QuestionPosted by Kartik Singh 6 years, 5 months ago
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Posted by Aarti Sharma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
A chemical process in which a substance reacts with oxygen to give off heat and light is called combustion.
Posted by Aarti Sharma 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
- A parallelogram is a quadrilateral with 2 pairs of opposite, equal and parallel sides.
- A rectangle is a quadrilateral with 2 pairs of opposite, equal and parallel sides but also forms right angles between adjacent sides.
Iron Man 6 years, 5 months ago
Posted by Kishan Prajapati 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
You can check solutions here : https://mycbseguide.com/ncert-solutions.html
Posted by Kishan Prajapati 6 years, 5 months ago
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Ꭾꮢꮛꮛꮦ Ꭾꭷꮢꮗꮧꮭ 6 years, 5 months ago
Posted by Jyoti Gupta 6 years, 5 months ago
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Rahul Ranjan Singh 6 years, 5 months ago
Arsh Tutorials 6 years, 5 months ago
Let x be the length and y be the breadth.
{tex}\therefore{/tex} x = y + 4 {tex}\implies{/tex}x - y = 4 --- (1)
P = 2 (x + y) {tex}\implies{/tex}x + y = 36 --- (2)
(1) + (2) {tex}\implies{/tex}2x = 40 {tex}\implies{/tex}x = 20 m {tex}\implies{/tex} y = 16 m
Posted by Raj Sharma 6 years, 5 months ago
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Posted by Misba Patel 6 years, 5 months ago
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?Niharika Sharma ? 6 years, 5 months ago
Posted by Abhay Singh 6 years, 5 months ago
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Posted by Pragya Narayan 6 years, 5 months ago
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Neha Panchal 6 years, 5 months ago
Posted by Riya Sinha 6 years, 5 months ago
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?Niharika Sharma ? 6 years, 5 months ago
Posted by Rama Lakshmi 6 years, 5 months ago
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Posted by Pushpa Keerthi 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let (-1, 6) divides line segment joining the points (-3, 10) and(6, -8) in k:1.
Using Section formula, we get
{tex} - 1 = \frac{{( - 3) \times 1 + 6 \times k}}{{k + 1}}{/tex} {tex}4 \Rightarrow - k - 1 = ( - 3 + 6k){/tex}
⇒ −7k = −2 ⇒ k= {tex}\frac{2}{7}{/tex}
Therefore, the ratio is {tex}\frac{2}{7}:1{/tex} which is equivalent to 2:7.
Therefore, (-1, 6) divides line segment joining the points (-3, 10) and (6, -8) in 2:7.
Posted by Pushpa Keerthi 6 years, 5 months ago
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Posted by Vaibhav Kundekar 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
We have 2x2 - 7x + 3 = 0
{tex}\implies2( x^2 - {7 \over 2}x + {3\over 2}) = 0{/tex}
{tex}\implies x^2 - {7 \over 2}x + {49 \over 16} = {-3 \over 2} +{ 49 \over 16}{/tex} (Adding 49/16 to both sides)
{tex}\implies x^2 -2 \times x \times {7 \over 4} + ({7 \over 4})^2 = {-24 +49 \over 16}{/tex}
{tex}\implies (x-{7\over4})^2 = {25 \over 16}{/tex}
{tex}\implies x-{7\over 4}= \pm \sqrt({25 \over 16}){/tex}
{tex}\implies x={7\over 4} \pm {5 \over 4}{/tex}
{tex}\implies x={7\over 4} + {5 \over 4}\, and \,x={7\over 4} - {5 \over 4}{/tex}
{tex}\implies x=3\, and \,{1\over 2}{/tex}
{tex}\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\over 2{/tex}.
Posted by Abotu Abhishake 6 years, 5 months ago
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Posted by Ayush Gaur 6 years, 5 months ago
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Ayush Gaur 6 years, 4 months ago
Posted by Anil Saini 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
Let the total number of students be x.
Cost of food for each member Rs. {tex}\frac{500}{x}{/tex}
If the number of students decreased by 5. Then,
New cost of food for each member Rs. {tex}\frac{500}{x - 5}{/tex}
According to question,
{tex}\frac{500}{x- 5} - \frac{500}{x} = 5{/tex}
{tex}\Rightarrow \frac{ 500x - 500(x - 5)}{x(x-5)} = 5{/tex}
{tex}\Rightarrow \frac{500x - 500x + 2500}{x^2 - 5x } = 5{/tex}
{tex}\Rightarrow \frac{2500}{ x^2 - 5x } = 5{/tex}
{tex}\Rightarrow \frac{ 2500}{ 5} = x^2 - 5x{/tex}
{tex}\Rightarrow{/tex} {tex}500=x^2-5x{/tex}
{tex}\Rightarrow{/tex} {tex}x^2-5x-500=0{/tex}
{tex}\Rightarrow{/tex}{tex}x^2-25x+20x-500=0{/tex}
{tex}\Rightarrow{/tex} x(x - 25) + 20(x - 25) = 0
{tex}\Rightarrow{/tex} (x + 20)(x - 25) = 0
{tex}\Rightarrow{/tex} x=-20,25
{tex}\Rightarrow{/tex} As number of students can't be negative. {tex}\Rightarrow{/tex} x = 25
Hence, the number of students attended the picnic = x - 5
= 25 - 5 = 20.
Posted by Bhavuk Kathuriya 6 years, 5 months ago
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Posted by Adarsh Kumar 6 years, 5 months ago
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Posted by Yoga Shree 6 years, 5 months ago
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Posted by Neeraj Shao 6 years, 5 months ago
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Posted by Vansh Gupta 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
The given equations are
{tex}2x - 3y - 7 = 0,{/tex}
{tex}(k +1) x + (1 - 2k) y + (4 - 5k) = 0.{/tex}
These equations are of the form
{tex}a_1x + b_1y+ c_1 = 0,\ a_2x + b_2y +c_2= 0,{/tex}
where {tex}a_1 = 2,\ b_1= -3,\ c_1= -7\ and\ a_2 =(k + 1),\ b_2= (1 - 2k),\ c_2= (4 - 5k){/tex}
Let the given system of equations have infinitely many solutions.
Then, {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { ( k + 1 ) } = \frac { - 3 } { ( 1 - 2 k ) } = \frac { - 7 } { ( 4 - 5 k ) }{/tex}
{tex}\Rightarrow \frac { 2 } { ( k + 1 ) } = \frac { 3 } { ( 2 k - 1 ) } = \frac { 7 } { ( 5 k - 4 ) }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { ( k + 1 ) } = \frac { 3 } { ( 2 k - 1 ) }{/tex} and {tex}\frac { 3 } { ( 2 k - 1 ) } = \frac { 7 } { ( 5 k - 4 ) }{/tex}
{tex}\Rightarrow{/tex} 4k - 2 = 3k + 3 and 15k -12 = 14k - 7
{tex}\Rightarrow{/tex} k = 5 and k = 5.
Hence, k = 5.
Posted by Shivam Badera 6 years, 5 months ago
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Posted by Nasim Khan 6 years, 5 months ago
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Sirsti Sngm 6 years, 5 months ago
Posted by Nazish Parween 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the consecutive multiples of 3 be 3x and 3(x + 1).
Then, we have
{tex}3x\times3(x + 1) = 648{/tex}
{tex}\Rightarrow{/tex} {tex}9x^2 + 9x - 648 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + x - 72 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 + 9x - 8x - 72 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x(x + 9) - 8(x + 9) = 0{/tex}
{tex}\Rightarrow{/tex} x + 9 = 0 or x - 8 = o
{tex}\Rightarrow{/tex} x = -9 or x = 8
Since x is a positive number, x {tex}\neq{/tex} -9
{tex}\Rightarrow{/tex} x = 8
{tex}\Rightarrow{/tex} 3x = 3(8) = 24 and 3(x + 1) = 3(9) = 27
Hence, the required consecutive multiples of 3 are 24 and 27.
Iron Man 6 years, 5 months ago
Posted by Nazish Parween 6 years, 5 months ago
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Yogita Ingle 6 years, 5 months ago
Let two consecutive multiples of 3 be x and (x+3)
As per given condition
x (x+3) = 648
⇒ x² + 3x = 648
⇒ x² + 3x -648 = 0
⇒ x² + 27x - 24x -648 = 0
⇒ x ( x + 27 ) -24 ( x +27)
⇒ ( x - 24) ( x + 27)
⇒ x = 24 and x = -27
so, we take x = 24.
Required multiples of 3
⇒ x = 24
⇒ x +3 = 24+3 = 27.
Posted by Vaibhav Singh 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the point of x-axis be P(x, 0)
Given A(2, -5) and B(-2, 9) are equidistant from P
That is PA = PB
Hence PA2 = PB2 → (1)
Distance between two points is {tex}\sqrt{[(x_2 - x_1)^2 + (y_2 - y_1)^2]}{/tex}
PA = {tex}\sqrt{[(2 - x)^2 + (-5 - 0)^2]}{/tex}
PA2 = 4 - 4x +x2 + 25
= x2 - 4x + 29
Similarly, PB2 = x2 + 4x + 85
Equation (1) becomes
x2 - 4x + 29 = x2 + 4x + 85
- 8x = 56
x = -7
Hence the point on x-axis is (-7, 0)
Posted by Rajesh Kumar 6 years, 5 months ago
- 2 answers
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