Answer:
![y=20(2)^x](https://tex.z-dn.net/?f=y%3D20%282%29%5Ex)
Step-by-step explanation:
We want to write an exponential function that goes through the points (0, 20) and (6, 1280).
The standard exponential function is given by:
![y=ab^x](https://tex.z-dn.net/?f=y%3Dab%5Ex)
The point (0, 20) tells us that <em>y</em> = 20 when <em>x</em> = 0. Hence:
![20=a(b)^0](https://tex.z-dn.net/?f=20%3Da%28b%29%5E0)
Simplify:
![20=a(1)\Rightarrow a=20](https://tex.z-dn.net/?f=20%3Da%281%29%5CRightarrow%20a%3D20)
So, our exponential function is now:
![y=20(b)^x](https://tex.z-dn.net/?f=y%3D20%28b%29%5Ex)
Next, the point (6, 1280) tells us that <em>y</em> = 1280 when <em>x</em> = 6. Thus:
![1280=20(b)^6](https://tex.z-dn.net/?f=1280%3D20%28b%29%5E6)
Solve for <em>b</em>. Divide both sides by 20:
![64=b^6](https://tex.z-dn.net/?f=64%3Db%5E6)
Therefore:
![b=\sqrt[6]{64}=2](https://tex.z-dn.net/?f=b%3D%5Csqrt%5B6%5D%7B64%7D%3D2)
Hence, our function is:
![y=20(2)^x](https://tex.z-dn.net/?f=y%3D20%282%29%5Ex)
It hits the ground at h=0
0=-16t^2-16t+96
undistribute -16
0=-16(t^2+t-6)
factor
0=-16(t-2)(t+3)
set to zero
0=t-2
2=t
t+3=0
t=-3, nope, we can't have negative time
at t=2, or 2 seconds
Answer:
x + y = ±15
Step-by-step explanation:
Step 1: Write out systems of equations
x - y = 1
xy = 56
Step 2: Rearrange 1st equation
x = y + 1
Step 3: Substitution
(y + 1)y = 56
Step 4: Distribute
y² + y = 56
Step 5: Solve for <em>y</em>
y² + y - 56 = 0
(y - 7)(y + 8) = 0
y = -8, 7
Step 6: Plug in <em>y </em>to find <em>x</em>
x - 7 = 1
x = 8
x - (-8) = 1
x + 8 = 1
x = -7
Step 7: Find answer
x + y = ?
8 + 7 = 15
-7 + -8 = -15
So both 15 and -15 could be the answer.
You’d have to try to get r on one side of the equation.
You can do that by subtracting 12 from both sides which would make it so
R = -9