Answer:
22 rounds down to 20 if that is what you are asking
Step-by-step explanation:
You did not finish your question i dont think. 22 DOES round down to 20 though.
Answer:
x = 114
Step-by-step explanation:
180 - 105 = 75
39 + 75 = x
x = 114
The answer to your question is 215
Answer:
It would take approximately 6.50 second for the cannonball to strike the ground.
Step-by-step explanation:
Consider the provided function.
![h(t)=-4.9t^2+30.5t+8.8](https://tex.z-dn.net/?f=h%28t%29%3D-4.9t%5E2%2B30.5t%2B8.8)
We need to find the time takes for the cannonball to strike the ground.
Substitute h(t) = 0 in above function.
![-4.9t^2+30.5t+8.8=0](https://tex.z-dn.net/?f=-4.9t%5E2%2B30.5t%2B8.8%3D0)
Multiply both sides by 10.
![-49t^2+305t+88=0](https://tex.z-dn.net/?f=-49t%5E2%2B305t%2B88%3D0)
For a quadratic equation of the form
the solutions are: ![x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
Substitute a = -49, b = 305 and c=88
![t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B-305%2B%5Csqrt%7B305%5E2-4%5Cleft%28-49%5Cright%2988%7D%7D%7B2%5Cleft%28-49%5Cright%29%7D%3D-%5Cfrac%7B-305%2B%5Csqrt%7B110273%7D%7D%7B98%7D%5C%5Ct%20%3D%20%5Cfrac%7B-305-%5Csqrt%7B305%5E2-4%5Cleft%28-49%5Cright%2988%7D%7D%7B2%5Cleft%28-49%5Cright%29%7D%3D%20%5Cfrac%7B305%2B%5Csqrt%7B110273%7D%7D%7B98%7D)
Ignore the negative value of t as time can't be a negative number.
Thus,
![t=\frac{305+\sqrt{110273}}{98}\approx6.50](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B305%2B%5Csqrt%7B110273%7D%7D%7B98%7D%5Capprox6.50)
Hence, it would take approximately 6.50 second for the cannonball to strike the ground.