The slope-intercept is y = mx+b. You can use this as a technique for graphing to make the experience easier. The "5" would be the slope or rise/run of the line. The 10 would be where it started on the y-intercept. You could also say it would be (0,10). Since the slope is positive than it would be going in an upward direction. That is why this technique is good for graphing. I hope this helps you and please put me as the brainliest answer. :)
For this case we have the following function that models the problem is:
![y = 10 + 95x-16x ^ 2 ](https://tex.z-dn.net/?f=y%20%3D%2010%20%2B%2095x-16x%20%5E%202%0A)
The time for which the maximum height occurs is obtained by deriving the equation:
![y '= 95-32x ](https://tex.z-dn.net/?f=y%20%27%3D%2095-32x%0A)
We equal zero and clear x:
![95-32x = 0 32x = 95 x = 95/32 x = 2.97 ](https://tex.z-dn.net/?f=95-32x%20%3D%200%0A%0A%2032x%20%3D%2095%0A%0A%20x%20%3D%2095%2F32%0A%0A%20x%20%3D%202.97%20%0A)
The maximum height is obtained by evaluating the time obtained in the given equation:
![y = 10 + 95 * (2.97) -16 * (2.97) ^ 2 y = 151.02](https://tex.z-dn.net/?f=y%20%3D%2010%20%2B%2095%20%2A%20%282.97%29%20-16%20%2A%20%282.97%29%20%5E%202%0Ay%20%3D%20151.02)
Then, by the time it reaches the ground we have:
![10 + 95x-16x ^ 2 = 0 ](https://tex.z-dn.net/?f=10%20%2B%2095x-16x%20%5E%202%20%3D%200%0A)
From here, we clear x solving the quadratic equation:
![x = \frac{-95+/- \sqrt{95^2-4(-16)(10)} }{2(-16)}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B-95%2B%2F-%20%5Csqrt%7B95%5E2-4%28-16%29%2810%29%7D%20%7D%7B2%28-16%29%7D%20)
We discard the negative root, and obtain as a result:
Answer:<span>
2.The pumpkin's maximum height is 151.02 feet and it hits the ground after 6.04 seconds </span>
The answer is option (c) 7xy^2-2xy^2
Answer:
![- 2 {x}^{2}](https://tex.z-dn.net/?f=%20-%202%20%7Bx%7D%5E%7B2%7D%20)
Step-by-step explanation:
on the picture
if it's helpful ❤❤❤
THANK YOU.
The answer is answer choice D.