Answer:
6 seconds
Step-by-step explanation:
<u><em>The question in English is</em></u>
An object is thrown from a platform.
Its height (in meters), x seconds after launch, is modeled by:
h(x)=-5x^2+20x+60
How many seconds after the launch does the object reach the ground?
Let
x ----> the time in seconds
h(x) ---> the height of the object
we have

we know that
When the object hit the ground the height is equal to zero
so
For h(x)=0
we have

The formula to solve a quadratic equation of the form
is equal to
in this problem we have

so
substitute in the formula
The solution is x=6 sec
The he object reach the ground at x=6 seconds
If the function is continuous on (a,c), it has a maximum at b. if it is not continuous, nothing can be said.
You would start at 2 and go down 4 and left 1
Suppose the number is x, its reciprocal is 1/x
x+1/x=13/6
(x^2+1)/x=13/6
6(x^2+1)=13x
6x^2-13x+6=0
(3x-2)(2x-3)=0
x=2/3 or x=3/2
Okay first you have to divide
2 goes into 3
3/2
2 fits into 3 once
1 and the left over is 1 so
1 1/2