" 10."(x-3)/(x+4)>0,x in R

Solve: (x-3)/(x+4)gt0,x in R

Column I ............................................................. Column II y=(x^2-2x+4)/(x^2+2x+4),x in R , then y can't be , p. 1 y=(x^2-3x-2)/(2x-3),x in R , then y can't be , q. 4 y=(2x^2-2x+4)/(x^2-4x+3),x in R , then y can't be , r. -3 x^2-(a-3)x+2<0,AA,x in (-2,3) , then y can't be , s. -10

If f(x)=((3)/(5))^(x)+((4)/(5))^(x)-1,x in R, then the equation f(x)=0 has :

Write the following as intervals : (i) {x : x in R, -4 lt x le 6} (ii) {x : x in R, -12 lt x lt -10} (iii) {x : x in R, 0 le x lt 7} (iv) {x : x in R, 3 le x le 4} .

Write the following as intervals : (i) {x : x in R, -4 lt x le 6} (ii) {x : x in R, -12 lt x lt -10} (iii) {x : x in R, 0 le x lt 7} (iv) {x : x in R, 3 le x le 4} .

Write the following as intervals : (i) {x" ":" "x in R ," "" "4" "<" "xlt= 6} (ii) {x" ":" "x in R ," "" "12" "<" "x" "<" "" "10} (iii) {x" ":" "x in R ," "0lt= x" "<" "7} (iv) {x" ":" "x in R ," "3lt= x lt=4}

The number real of roots of sqrt(x - 3) (x^(2) + 7x + 10)= 0 , where x in R is

A={x/x in R,x!=0,-4<=x<=4 and f:A rarr R is defined by f(x)=(|x|)/(x) for x in A. Then the range of f is

If the function g(x) {{:((e^(px) + log _(e) (1+4x)+q)/(x^(3))","x ne 0),(r"," x=0):} continuous at x=0 , then find the values of p,q and r.

If the function g(x) {{:((e^(px) + log _(e) (1+4x)+q)/(x^(3))","x ne 0),(r"," x=0):} continuous at x=0 , then find the values of p,q and r.