0=40+3x
0=120x
Calculate the product
0=120x
Swap sides of the equation
120x=0
Divide both sides
X=0
When the first equation is written in slope-intercept form it becomes
y = 3x + 3
This shows it to describe a line with the same slope but different intercept as the line described by the second equation.
The lines are parallel.
Answer:
![x=4y^2-24y+30](https://tex.z-dn.net/?f=x%3D4y%5E2-24y%2B30)
![(x-4.5)^2+(y-2.5)^2=23](https://tex.z-dn.net/?f=%28x-4.5%29%5E2%2B%28y-2.5%29%5E2%3D23)
Step-by-step explanation:
Standard form of a sideways parabola: ![x=ay^2+by+c](https://tex.z-dn.net/?f=x%3Day%5E2%2Bby%2Bc)
Given equation:
![4y^2-x-24y+30=0](https://tex.z-dn.net/?f=4y%5E2-x-24y%2B30%3D0)
Add x to both sides:
![4y^2-24y+30=x](https://tex.z-dn.net/?f=4y%5E2-24y%2B30%3Dx)
![\implies x=4y^2-24y+30](https://tex.z-dn.net/?f=%5Cimplies%20x%3D4y%5E2-24y%2B30)
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<u>Standard form of circle equation</u>
![(x-a)^2+(y-b)^2=r^2](https://tex.z-dn.net/?f=%28x-a%29%5E2%2B%28y-b%29%5E2%3Dr%5E2)
(where (a,b) is the center and r is the radius)
Given equation:
![2x^2+2y^2-18x-10y+7=0](https://tex.z-dn.net/?f=2x%5E2%2B2y%5E2-18x-10y%2B7%3D0)
Group like terms:
![2x^2-18x+2y^2-10y+7=0](https://tex.z-dn.net/?f=2x%5E2-18x%2B2y%5E2-10y%2B7%3D0)
Divide by 2:
![x^2-9x+y^2-5y+3.5=0](https://tex.z-dn.net/?f=x%5E2-9x%2By%5E2-5y%2B3.5%3D0)
Factor by completing the square for each variable:
![(x-4.5)^2-20.25+(y-2.5)^2-6.25+3.5=0](https://tex.z-dn.net/?f=%28x-4.5%29%5E2-20.25%2B%28y-2.5%29%5E2-6.25%2B3.5%3D0)
Rearrange into standard form:
![(x-4.5)^2+(y-2.5)^2=23](https://tex.z-dn.net/?f=%28x-4.5%29%5E2%2B%28y-2.5%29%5E2%3D23)
Therefore, the circle has a center at (4.5, 2.5) and a radius of √23
We want to know the time, <em>t</em>, it takes the ball to reach a height (<em>y</em>) of 0.
![0=-16t^2+24t](https://tex.z-dn.net/?f=0%3D-16t%5E2%2B24t)
We can factor out the GCF first. The largest number that will divide evenly into 16 and 24 is 8. Also, both terms have a <em>t</em>, so we can factor that out as well:
![0=8t(-2t+3)](https://tex.z-dn.net/?f=0%3D8t%28-2t%2B3%29)
(-16/8 = -2 and 24/8 = 3)
Using the zero product property, we know that either 8t=0 or -2t+3=0. Solving the first equation, we would divide both sides by 8:
8t/8=0/8
t=0
This is at 0 seconds, before the ball is in the air at all.
Solving the second equation, we start by subtracting 3 from both sides:
-2t+3-3=0-3
-2t=-3
Now we divide both sides by -2
-2t/-2=-3/-2
t=1.5
After 1.5 seconds, the ball will hit the ground again.