If xy then 6 plus 4(x-y)?

If xy then 6 plus 4(x-y)?
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How xy plus x plus y plus 1 will solve?

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If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

Description : If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

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If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

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If `x+y=1` then find the maximum value of the function `xy^2`

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If `x=1+log_(a) bc, y=1+log_(b) ca, z=1+log_(c) ab`, then `(xyz)/(xy+yz+zx)` is equal to

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If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

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Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

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X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

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Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

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Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

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Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

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X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

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If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

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Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

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