" 34."7x+(3)/(x)=35(3)/(5)

Solve by factorization: 7x+(3)/(x)=35(3)/(5)

If 2x+3y=34 and (x+y)/(y)=(13)/(8), then find the value of 5y+7x

Solve: (3x)/(7)-(4)/(35)=(2x)/(5)

If x= (3)/(4.8) + (3.5)/(4.8.12)+ (3.5.7)/(4.812.16) + …., then 2x^2+5x=

Distance between the lines 5x+3y-7=0\ a n d\ 15 x+9y+14=0 is (35)/(sqrt(34)) b. 1/(3sqrt(34)) c. (35)/(3sqrt(34)) d. 35//2sqrt(34)

Distance between the lines 5x+3y-7=0\ a n d\ 15 x+9y+14=0 is (35)/(sqrt(34)) b. 1/(3sqrt(34)) c. (35)/(3sqrt(34)) d. 35//2sqrt(34)

If A = [[x, -2],[3, 7]] and A^(-1) = [[(7)/(34), (1)/(17)],[(-3)/(34), (2)/(17)]] , then the value of x is

3((7x+1)/(5x-3))-4((5x-3)/(7x+1))=11;x!=(3)/(5),-(1)/(7)

Assertion (A) : (1)/(5)+(1)/(3.5^(3))+(1)/(5.5^(5))+(1)/(7.5^(7))+…(1)/(2)log((3)/(2)) Reason (R ) : If |x| lt 1 then log_(e )((1+x)/(1-x))=2(x+(x^(3))/(3)+(x^(5))/(5)+…)

Solve the quadratic equation by factorisation: 3((7x+1)/(5x-3))-4((5x-3)/(7x+1))=11 x != 3//5, x != 1//7