The given equations are:
{tex}\frac{x}{a}{/tex} + {tex}\frac{y}{b}{/tex} {tex}=(a+b){/tex} {tex}\Rightarrow{/tex}{tex}bx + ay = ab(a + b){/tex}
{tex}\Rightarrow{/tex}{tex}bx + ay - ab(a + b) = 0{/tex} ....(i)
and {tex}\frac { x } { a ^ { 2 } } + \frac { y } { b ^ { 2 } } = 2{/tex} {tex}\Rightarrow{/tex} {tex}b^2x + a^2y = 2a^2b^2{/tex} 
{tex}\Rightarrow{/tex}{tex} b^2x + a^2y -2a^2b^2 = 0{/tex}....(ii)
From eq. (i) and (ii), we get

{tex}\Rightarrow{/tex} {tex}\frac { x } { - 2 a ^ { 3 } b ^ { 2 } + a ^ { 3 } b ( a + b ) }{/tex} = {tex}\frac { - y } { - 2 a ^ { 2 } b ^ { 3 } + a b ^ { 3 } ( a + b ) }{/tex} = {tex}\frac { 1 } { a ^ { 2 } b - a b ^ { 2 } }{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { x } { - a ^ { 3 } b ( 2 b - a - b ) }{/tex} = {tex}\frac { - y } { - a b ^ { 3 } ( 2 a - a - b ) }{/tex} = {tex}\frac { 1 } { a b ( a - b ) }{/tex}
{tex}\Rightarrow{/tex}{tex}\frac { x } { - a ^ { 3 } b ( b - a ) }{/tex} = {tex}\frac { y } { a b ^ { 3 } ( a - b ) }{/tex} = {tex}\frac { 1 } { a b ( a - b ) }{/tex}
{tex}\Rightarrow{/tex} {tex}x = \frac { a ^ { 3 } b ( a - b ) } { a b ( a - b ) }{/tex} = a²;  {tex}y = \frac { a b ^ { 3 } ( a - b ) } { a b ( a - b ) } = b ^ { 2 }{/tex}
The solution is {tex}x = a^2, y = b^2{/tex} .

Padhna aata nhi phle padh toh le likha kya

Don't use phone

So you go and practise for test paper

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.

Consistent

Let angle b be= x. So, angle c be 3x . angle a be x/2. A+B+c =180. X/2+x+3x = 180. X/2+4x=180.(do through out multiplication of 2 in eqn). X+8x=360. 9x=360. X =40. Putting the value of x in a and b and c. a= 40÷2=20. b = 40 . c= 40(3) = 120.

After 578160 minutes

I can solve

40=2³×5. 42=2×3×7. 45= 3²×5. Lcm = 2³×3²×5×7 =8×9×35= 2520.

Answer is 2520

Sorrey

The given equations are
{tex}\frac { 1 } { 2 ( x + 2 y ) } + \frac { 5 } { 3 ( 3 x - 2 y ) } = - \frac { 3 } { 2 }{/tex}.....(1)
and {tex}\frac { 5 } { 4 ( x + 2 y ) } - \frac { 3 } { 5 ( 3 x - 2 y ) } = \frac { 61 } { 60 }{/tex}....(2)
Putting {tex}\frac 1{x+2y}{/tex}=u and {tex}\frac 1{3x-2y}{/tex}= v in equation (1) & equation (2) so that we may get two linear equations in the variables u & v as following:- 
{tex}\frac { 1 } { 2 } u + \frac { 5 } { 3 } v = - \frac { 3 } { 2 }{/tex}.....................(1)
{tex}\frac { 5 } { 4 } u - \frac { 3 } { 5 } v = \frac { 61 } { 60 }{/tex}.................(2)

Multiplying (1) by 36 and (2) by 100, we get
{tex}{/tex}
{tex}{/tex}
{tex}18u + 60v = -54{/tex}...............(3)
{tex}125 u - 60 v = \frac { 305 } { 3 }{/tex}.............(4)
Adding (3) and (4),we get
{tex}143 u = \frac { 305 } { 3 } - 54 = \frac { 305 - 162 } { 2 } = \frac { 143 } { 3 }{/tex}
{tex}\therefore \quad u = \frac { 1 } { 3 } = \frac { 1 } { x + 2 y }{/tex}
{tex}\therefore{/tex}{tex} x + 2y = 3{/tex}...........(5)
Putting value of u in (3), we get
{tex}1 + 10v = -9{/tex} (after dividing by 3)
{tex}\therefore 10 \mathrm { v } = - 10{/tex} or {tex}\mathrm { v } = - 1 {/tex}
{tex}\Rightarrow - 1 = \frac { 1 } { 3 x - 2 y }{/tex}
{tex}\Rightarrow 3 x - 2 y = - 1{/tex}......(6)
Adding (5) and (6), we get
{tex}4 x = 2 \quad{/tex}
{tex} \therefore x = \frac { 1 } { 2 }{/tex}
Putting value of x in (5),
{tex}\frac { 1 } { 2 } + 2 y = 3{/tex}
{tex} \text { or } 2 y = 3 - \frac { 1 } { 2 } = \frac { 5 } { 2 }{/tex}
{tex}\therefore \quad y = \frac { 5 } { 4 }{/tex}
The required solution is {tex}\mathrm { x } = \frac { 1 } { 2 } , \mathrm { y } = \frac { 5 } { 4 }{/tex}

Point-slope form, standard form, and slope-intercept form.

Yes

Ok

No

Find LCM of 48,60&72 

Which would be 720km.after covering 720km each they will meet,till then they all have covered two times the field

Please specify questions...are you talking about NCERT?

Yup

Consider equal sideof isoscelous right angle triangle be x Then hypotenous will be √2x Then apply the formula to find area of both triangle and compare it?????

Nth term=Sn-S(n-1)

Since the factors of 13 and 15 are 1 hence the fhcf is 1

Hcf is highest common factor and there is no common factor other than 1 there So hcf is 1

Sorry lcm is 195 But hcf is 1?????

1

HCF of 13 and 15 is 1.

195

Please give solution

Am i right

45 year

Where are the quadratic equation sis@?

For unique solution, -1÷3kis not equal to 1÷2. Hence k is not equal to -2÷3

let root 3 is rational number. root 3=p/q [where,p and q are integers,q is not equal to 0 and p and q are co-primes]. root 3q=p squaring both sides, 3q2 =p2...............[1] 3 divides p2 3 divides p.............[2] p2=3q2 p2=3m Put the value of p2 in eqn [1]. p2=3q2 [3m]2=3q2 9m2=3q2 3m2=q2 q2=3m2 3 divides q2 3 divides q [3] By eqn [2] and [3].... 3 is the common factor of both p and q . This is contradiction that and q are co-primes. Our assumption is wrong. Hence , root 3 is irrational number.

 let us assume that {tex}\sqrt 3{/tex} be a rational number.

{tex}\sqrt { 3 } = \frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \neq{/tex}0
Squaring both sides, we have
{tex}\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}
or, {tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)
a² is divisible by 3.
Hence a is divisible by 3..........(ii)
Let a = 3c ( where c is any integer)

squaring on both sides we get
(3c)² = 3b²
9c² = 3b²
b² = 3c²
so b² is divisible by 3
hence, b is divisible by 3..........(iii)
From equation(ii) and (iii), we have
3 is a factor of a and b which is contradicting the fact that a and b are co-primes.
Thus, our assumption that {tex}\sqrt 3{/tex} is rational number is wrong.
Hence, {tex}\sqrt 3{/tex} is an irrational number.

When y = x

x	1	2
y	1	2

When 3y = x,

x	6	3
y	2	1

when x + y = 8 or y = 8 - x

x	4	5
y	4	3

 

There is no competition in our school . Project is for holiday homework

What type of math model is there any type of competition in your school? ?????