Question

4. Find the area of a triangle whose vertices are (3, 0), (7, 0) and (8, 4). 1. Find the roots of the quadratic equation 3x 2 +2√6x+2=0 . 2. Find the 4th term from the end of the AP -11, -8, -5, .........., 49. 3. In fig. 1, if length of AB is 9 cm, then, find the length of CP.

Answer 1

1)area of the Triangle=1/2x|3(0-4)+7(4-0)+8(0-0)|=1/2x|-12+28+0|=1/2*16=8sq.unit 2)3x^2+√6x+√6x+2=0 =>√3x(√3x+√2)+√2(√3x+√2)=0=>(√3x+√2)(√3x+√2)=0 so x=-√2/√3..both the roots are equal...3)a=-11..d=3..l=49..so..49=-11+(n-1)*3...60=3n-3=>n=21..so...4th term from last...a=49..n=4 ,d=-3...so..49+3*(-3)=>49-9=>40(ans)...plzz clear ur last question!!..

Answer 2

Answer:

Step-by-step explanation:

i dont understand plz check its incorrect