This is the concept of application of quadratic expressions. Given that the height of the ball is modeled by the equation;
h=-7.3t^2+8.25t+2.1+5
The time taken for the ball to hit the ground will be given as falls;
-7.3t^2+8.25t+7.1=0
to solve for t we use the quadratic formula;
t=[-b+/-sqrt(b^2-4ac)]/(2a)
a=-7.3, b=8.25, c=2.1
t=[-8.25+/-sqrt[8.25^2+4*7.3*7.1]/(-2*7.3)
t= -0.572
or
t=1.702
since there is not negative time we take the time taken for the ball to hit the ground will be: t=1.702 sec
<span>19 and 43/100................</span>
If both triangles are congruent (that's what the sign means), then AD=CB, and CD=AB. Substituting what we are given into CD=AB,
6x-5=3x+10
Solving the equation:
3x=15
x=5
So we know that 2x+4 (the value of AD) becomes 2*5 +4, and you get 14. Since we know that AD=CB, then CB is also 14. No need to find y!
Answer: CB=14
