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The condition that the equation $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ can take the form $a x squared plus 2 h x y plus b y squared equals 0$ when shifting the origin is

$a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$2 f g h - b g^{2} - c h^{2} = 0$
$2 f g h - a f^{2} - c h^{2} = 0$
$2 f g h - a f^{2} - b g^{2} = 0$

hint Hint:

solve for the quadratic equation in x and then apply D=0 to find the equal roots. Then solve for the resultant quadratic equation in y and solve for the equal and real roots.

The correct answer is: $a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent equals 0$

abc + 2fgh -af²-bg²-ch²=0

In the first equation, if we apply the formula to find the roots, we get
X=(-2(hy+g)±√(4(hy+g)²-4a(by²+2fy+c))/2
For a linear relationship between x and y, D=0
Hence,
4(hy+g²)-4a(by²+2fy+c)=0
This is a quadratic in y, whose D=0 since the roots are equal
abc + 2fgh -af²-bg²-ch²=0

When the roots are equal and real, then the discriminant value of the the quadratic equation becomes 0.

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MolecularorbitalelectronicconfigurationofBe2, willbe-
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b)KK,σ2s²,σ2s²
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Which of the following species will have the minimum bond energy:

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The transformed equation of $14 x to the power of 2 end exponent minus 4 x y plus 11 y to the power of 2 end exponent minus 36 x plus 48 y plus 41 equals 0$ when the axes are translated to (1, –2) is

shifting of origin is done when the aes are shifted parallely to a different location on the cartesian plane.
this results in change of coordinates as: x=x'+h and y=y'+k
where h,k are the new origin.

The transformed equation of $14 x to the power of 2 end exponent minus 4 x y plus 11 y to the power of 2 end exponent minus 36 x plus 48 y plus 41 equals 0$ when the axes are translated to (1, –2) is

Maths-General

shifting of origin is done when the aes are shifted parallely to a different location on the cartesian plane.
this results in change of coordinates as: x=x'+h and y=y'+k
where h,k are the new origin.

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When an aqueous solution of copper sulphate is electrolysed using copper electrodes the reaction at the anode is represented by

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The condition that the equation can take the form when shifting the origin is

solve for the quadratic equation in x and then apply D=0 to find the equal roots. Then solve for the resultant quadratic equation in y and solve for the equal and real roots.

The correct answer is:

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The condition that the equation $a x to the power of 2 end exponent plus 2 h x y plus b y to the power of 2 end exponent plus 2 g x plus 2 f y plus c equals 0$ can take the form $a x squared plus 2 h x y plus b y squared equals 0$ when shifting the origin is

The correct answer is: $a b c plus 2 f g h minus a f to the power of 2 end exponent minus b g to the power of 2 end exponent minus c h to the power of 2 end exponent equals 0$

MolecularorbitalelectronicconfigurationofBe2, willbe-
a)KK, σ2s²
b)KK,σ2s²,σ2s²
c)σ1s²,σ1s²,σ2s²,σ2s²
d)σ1s²,σ1s²,σ2s²,2p²Correctansweris-

MolecularorbitalelectronicconfigurationofBe2, willbe-
a)KK, σ2s²
b)KK,σ2s²,σ2s²
c)σ1s²,σ1s²,σ2s²,σ2s²
d)σ1s²,σ1s²,σ2s²,2p²Correctansweris-

The transformed equation of $14 x to the power of 2 end exponent minus 4 x y plus 11 y to the power of 2 end exponent minus 36 x plus 48 y plus 41 equals 0$ when the axes are translated to (1, –2) is

The transformed equation of $14 x to the power of 2 end exponent minus 4 x y plus 11 y to the power of 2 end exponent minus 36 x plus 48 y plus 41 equals 0$ when the axes are translated to (1, –2) is

N₂and O₂are converted into mono an ions, N₂and O₂respectively. Which of the following statements is wrong?

N₂and O₂are converted into mono an ions, N₂and O₂respectively. Which of the following statements is wrong?

Whichofthefollowingspeciesarepolar:
1)CH
2)XeF
3)SO₂
4)SF₄
5)SF₆correctansweris:

Whichofthefollowingspeciesarepolar:
1)CH
2)XeF
3)SO₂
4)SF₄
5)SF₆correctansweris:

$open square brackets fraction numerator blank P e r m e a b i l i t y blank over denominator blank P e r m i t t i v i t y blank end fraction close square brackets$ will have the dimensional formula of :

$open square brackets fraction numerator blank P e r m e a b i l i t y blank over denominator blank P e r m i t t i v i t y blank end fraction close square brackets$ will have the dimensional formula of :