5x-2=10-3x
8x=-8
8x/8=-8/8
X=-1
Area of a trapezoid = ½(a + b)h, where a and b are the lengths of the parallel sides and h is its height.
<span>From your information 20 = ½(2z + 3 + 6z – 1)z = ½(8z + 2)z = z(4z + 1) </span>
<span>Solve 20 = 4z² + z which is 0 = 4z² + z – 20 using the quadratic formula
</span>
Answer:
Step-by-step explanation:
area = ![\pi](https://tex.z-dn.net/?f=%5Cpi)
so it's ![\pi](https://tex.z-dn.net/?f=%5Cpi)
= 153.938... ![cm^{2}](https://tex.z-dn.net/?f=cm%5E%7B2%7D)
circumference is 2
r = 2*
*7 = 43.98229 cm
Note: It the given equation the coefficient of
must be -16 instead of 16.
Given:
Consider the height of the water balloon over time can be modeled by the function
![y=-16x^2+160x+50](https://tex.z-dn.net/?f=y%3D-16x%5E2%2B160x%2B50)
To find:
The maximum height of the water balloon after it was thrown.
Solution:
We have,
![y=-16x^2+160x+50](https://tex.z-dn.net/?f=y%3D-16x%5E2%2B160x%2B50)
Here, leading coefficient is negative. So, it is a downward parabola and vertex of a downward parabola, is the point of maxima.
If a parabola is
, then
![Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)](https://tex.z-dn.net/?f=Vertex%3D%5Cleft%28-%5Cdfrac%7Bb%7D%7B2a%7D%2Cf%28-%5Cdfrac%7Bb%7D%7B2a%7D%29%5Cright%29)
Here,
. So,
![-\dfrac{b}{2a}=-\dfrac{160}{2(-16)}](https://tex.z-dn.net/?f=-%5Cdfrac%7Bb%7D%7B2a%7D%3D-%5Cdfrac%7B160%7D%7B2%28-16%29%7D)
![-\dfrac{b}{2a}=-\dfrac{160}{-32}](https://tex.z-dn.net/?f=-%5Cdfrac%7Bb%7D%7B2a%7D%3D-%5Cdfrac%7B160%7D%7B-32%7D)
![-\dfrac{b}{2a}=5](https://tex.z-dn.net/?f=-%5Cdfrac%7Bb%7D%7B2a%7D%3D5)
Now, put x=5 in the given equation.
![y=-16(5)^2+160(5)+50](https://tex.z-dn.net/?f=y%3D-16%285%29%5E2%2B160%285%29%2B50)
![y=-16(25)+800+50](https://tex.z-dn.net/?f=y%3D-16%2825%29%2B800%2B50)
![y=-400+850](https://tex.z-dn.net/?f=y%3D-400%2B850)
![y=450](https://tex.z-dn.net/?f=y%3D450)
The vertex of the given parabolic equation is (5,450).
Therefore, the maximum height of the balloon is 450 units after 5 units of time.