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Answered on 16 Apr Learn Pair of linear equations in two variable
Sadika
To solve the equations graphically, let's first rewrite them in slope-intercept form:&space;\(y&space;=&space;-2x&space;+&space;2\)&space;2.&space;\(2y&space;-&space;x&space;=&space;4\)&space;\(y&space;=&space;\frac{1}{2}x&space;+&space;2\)&space;Now,&space;let's&space;plot&space;these&space;lines&space;on&space;a&space;graph:&space;&space;Now,&space;to&space;find&space;the&space;area&space;of&space;the&space;triangle&space;formed&space;by&space;the&space;two&space;lines&space;and&space;the&space;x-axis&space;(y=0),&space;we&space;need&space;to&space;find&space;the&space;x-coordinates&space;of&space;the&space;points&space;where&space;the&space;lines&space;intersect&space;with&space;the&space;x-axis.&space;For&space;the&space;first&space;line:&space;\[y&space;=&space;-2x&space;+&space;2\]&space;\[0&space;=&space;-2x&space;+&space;2\]&space;\[2x&space;=&space;2\]&space;\[x&space;=&space;1\]&space;For&space;the&space;second&space;line:&space;\[y&space;=&space;\frac{1}{2}x&space;+&space;2\]&space;\[0&space;=&space;\frac{1}{2}x&space;+&space;2\]&space;\[\frac{1}{2}x&space;=&space;-2\]&space;\[x&space;=&space;-4\] "1. \(2x + y=2\) \(y = -2x + 2\) 2. \(2y - x=4\) \(y = \frac{1}{2}x + 2\) Now, let's plot these lines on a graph:  Now, to find the area of the triangle formed by the two lines and the x-axis (y=0), we need to find the x-coordinates of the points where the lines intersect with the x-axis. For the first line: \[y = -2x + 2\] \[0 = -2x + 2\] \[2x = 2\] \[x = 1\] For the second line: \[y = \frac{1}{2}x + 2\] \[0 = \frac{1}{2}x + 2\] \[\frac{1}{2}x = -2\] \[x = -4\]")
Answered on 16 Apr Learn Pair of linear equations in two variable
Sadika
To find the value of kk for which the pair of equations has no solution, we need the equations to represent parallel lines.
Answered on 16 Apr Learn Arithmetic Progression
Sadika
The difference between any two terms in an arithmetic progression (AP) is constant and equal to the common difference (d).
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 16 Apr Learn Arithmetic Progression
Sadika
In this arithmetic progression, the common difference is
Answered on 16 Apr Learn Arithmetic Progression
Sadika
Let's denote the sum of the first nn terms of the first arithmetic progression
Answered on 16 Apr Learn Polynomials
Sadika
We are given that : :
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 16 Apr Learn Polynomials
Sadika
To find the quadratic polynomial whose zeroes are
Answered on 16 Apr Learn Polynomials
Sadika
To find the zeroes of the quadratic polynomial
Answered on 16 Apr Learn Quadratic Equations
Sadika
To solve the quadratic equation
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 16 Apr Learn Quadratic Equations
Sadika
To solve the quadratic equation
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