" Hivstration "7.37" Prove that "sum_(a+p=y=10)(10!)/(a^(2)beta^(1)y!)=3

Prove that sum_(alpha+beta+gamma=10)(10!)/(alpha!beta!gamma!)=3^(10)

Prove that sum_(alpha+beta+gamma = 10) (10 !)/(alpha!beta!gamma!)=3^(10)dot

Prove that sum_(alpha+beta+gamma = 10) (10 !)/(alpha!beta!gamma!)=3^(10)dot

Prove that sum_(alpha+beta+gamma = 10) (10 !)/(alpha!beta!gamma!)=3^(10)dot

3-2x=7 2y+1=10-2(1)/(2)y

Subtract: (-2y^(2)+(1)/(2)y-3) om 7y^(2)-2y+10

If the normal at (x_(i) y_(i)) i=1,2,3,4 on xy=c^2 meet at the point (alpha, beta) show that sum x_(i)=alpha, sum y_(i)=beta, sum x_(i)^(2)=alpha^(2), sum y^(2)=beta^(2), x_(1)x_(2)x_(3)x_(4)=y_(1)y_(2)y_(3)y_(4)=-c^(4)

The coordinates of the foot of the perpendicular from the point (2,3) on the line -y+3x+4=0 are given by (a)((37)/(10),-(1)/(10))(b)(-(1)/(10),(37)/(10))(c)((10)/(37),-10)(d)((2)/(3),-(1)/(3))

The equation to the locus of a point P for which the distance from P to (0,5) is double the distance from P to y -axis is 1) 3x^(2)+y^(2)+10y-25=0 2) 3x^(2)-y^(2)+10y+25=0 3) 3x^(2)-y^(2)+10y-25=0 4) 3x^(2)+y^(2)-10y-25=0

If (3.7)^x=(.37)^y=1000 ,then prove that 1/x-1/y=1/3