The ball will hit the ground at 2 seconds.
Step-by-step explanation:
Given that,
The path of the ball = h(x)=−2x2+1x+6
Here,
x is the time while h is the height of ball.
When the ball will hit the ground, the height will become zero. Therefore,
h(x)=−2x2+1x+6
0 =−2x2+1x+6
or
2x2 -1x - 6 = 0
This is a quadratic equation, hence by applying quadratic equation formula:

here,
a = 2
b = -1
c = -6
Putting these values in formula, we get




x = 2, -3/2
As the time cannot be negative. Therefore, the ball will hit the ground at 2 seconds.
The other two vertices are (-3,6) and (-6,3)
Answer:
The correct option is the last option d.) y = negative 3 plus-or-minus StartRoot 5 x + seven-halves EndRoot
Step-by-step explanation:
the given equation is 
Therefore we can write 
To find the inverse of the above function let us replace x with y and y with x.
Therefore we get

Now we write the above equation with only y on the Left hand side and we will obtain the inverse of the given function

Therefore the correct option is the last option d.) y = negative 3 plus-or-minus StartRoot 5 x + seven-halves EndRoot , 
Let x be the no. of seconds it takes for the balloons to be at the same height.
7+3x = 12+2x
Deduct both sides by 2x.
7+x = 12
Deduct both sides by 7.
x = 5
Ans: It will take 5 seconds for the balloons to be at the same height.
<em>2 solutions</em>
<em>X= 16</em>
<em>X=49</em>
Step-by-step explanation:
<em>original equation</em>
<em>x-11√x+28 = 0</em>
<em> Isolate</em>
<em> -11√x = -x-28+0</em>
<em> Tidy up</em>
<em> 11√x = x+28</em>
<em> Raise both sides to the second power</em>
<em> (11√x)2 = (x+28)2</em>
<em>After squaring</em>
<em> 121x = x2+56x+784</em>
<em> Plug in 49 for x </em>
<em> 11√(49) = (49)+28</em>
<em>Simplify</em>
<em> 11√49 = 77</em>
<em> Solution checks !!</em>
<em> Solution is:</em>
x = 49
<em>Plug in 16 for x </em>
<em> 11√(16) = (16)+28</em>
<em>Simplify</em>
<em> 11√16 = 44</em>
<em> Solution checks !!</em>
<em> Solution is:</em>
x = 16