I’m eye billing it I would say $300 bekuz the rest are to low loll
Answer:
D) 12,656
Step-by-step explanation:
First of all, as they have given us 3 quantities of plants in 1/2 an acre, we can work out the average amount of plants in 1 whole acre.
In the first 1/2 acre there were:
In the second 1/2 acre there were:
In the third 1/2 acre there were:
to work out the average, we add all these together then divide them by 3 (as there are 3 examples)
291 + 327 + 286 = 904
904 ÷ 3 = 301.3 (this is the average amount of plants in every 1/2 acre.) If we multiply this answer by 2 we will get the average amount of plants in every acre.
301.3 x 2 = 602.67 (average plants in every 1 acre) multiply by total acres (21) for your answer
602.67 x 21 = 12656 average plants in total
I hope this was helpful :-)
Answer:
39
Step-by-step explanation:
Start with
![\dfrac{dy}{dx}=\dfrac{\sin(x)}{y}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B%5Csin%28x%29%7D%7By%7D)
Separate the variables:
![y\;dy = \sin(x)\;dx](https://tex.z-dn.net/?f=y%5C%3Bdy%20%3D%20%5Csin%28x%29%5C%3Bdx)
Integrate both parts:
![\displaystyle \int y\;dy = \int\sin(x)\;dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20y%5C%3Bdy%20%3D%20%5Cint%5Csin%28x%29%5C%3Bdx)
Which implies
![\dfrac{y^2}{2} = -\cos(x)+c](https://tex.z-dn.net/?f=%5Cdfrac%7By%5E2%7D%7B2%7D%20%3D%20-%5Ccos%28x%29%2Bc)
Solving for y:
![y = \sqrt{-2\cos(x)+c}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%7B-2%5Ccos%28x%29%2Bc%7D)
Fix the additive constant imposing the condition:
![y(0) = \sqrt{-2+c}=2 \iff -2+c=4 \iff c=6](https://tex.z-dn.net/?f=y%280%29%20%3D%20%5Csqrt%7B-2%2Bc%7D%3D2%20%5Ciff%20-2%2Bc%3D4%20%5Ciff%20c%3D6)
So, the solution is
![y(x) = \sqrt{-2\cos(x)+6}](https://tex.z-dn.net/?f=y%28x%29%20%3D%20%5Csqrt%7B-2%5Ccos%28x%29%2B6%7D)
Answer:
It will take 0.72s for the football to hits the ground.
Step-by-step explanation:
We have that the equaation for the height of the football is
![H(t) = -16t^{2} + 6t + 4](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-16t%5E%7B2%7D%20%2B%206t%20%2B%204)
The football will hit the ground when
.
![H(t) = -16t^{2} + 6t + 4](https://tex.z-dn.net/?f=H%28t%29%20%3D%20-16t%5E%7B2%7D%20%2B%206t%20%2B%204)
![-16t^{2} + 6t + 4 = 0](https://tex.z-dn.net/?f=-16t%5E%7B2%7D%20%2B%206t%20%2B%204%20%3D%200)
Multiplying by -1
![16t^{2} - 6t - 4 = 0](https://tex.z-dn.net/?f=16t%5E%7B2%7D%20-%206t%20-%204%20%3D%200)
To solve this equation, we need the bhaskara formula:
Given a second order polynomial expressed by the following equation:
.
This polynomial has roots
such that
, given by the following formulas:
![x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%20%5Cfrac%7B-b%20%2B%20%5Csqrt%7B%5Cbigtriangleup%7D%7D%7B2%2Aa%7D)
![x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}](https://tex.z-dn.net/?f=x_%7B2%7D%20%3D%20%5Cfrac%7B-b%20-%20%5Csqrt%7B%5Cbigtriangleup%7D%7D%7B2%2Aa%7D)
![\bigtriangleup = b^{2} - 4a](https://tex.z-dn.net/?f=%5Cbigtriangleup%20%3D%20b%5E%7B2%7D%20-%204a)
In this problem, we have that:
![16t^{2} - 6t - 4 = 0](https://tex.z-dn.net/?f=16t%5E%7B2%7D%20-%206t%20-%204%20%3D%200)
So
![a = 16, b = -6, c = -4](https://tex.z-dn.net/?f=a%20%3D%2016%2C%20b%20%3D%20-6%2C%20c%20%3D%20-4)
![\bigtriangleup = (-6)^{2} - 4*16*(-4) = 292](https://tex.z-dn.net/?f=%5Cbigtriangleup%20%3D%20%28-6%29%5E%7B2%7D%20-%204%2A16%2A%28-4%29%20%3D%20292)
![t_{1} = \frac{-(-6) + \sqrt{292}}{2*16} = 0.72](https://tex.z-dn.net/?f=t_%7B1%7D%20%3D%20%5Cfrac%7B-%28-6%29%20%2B%20%5Csqrt%7B292%7D%7D%7B2%2A16%7D%20%3D%200.72)
![t_{2} = \frac{-(-6) - \sqrt{292}}{2*16} = -0.35](https://tex.z-dn.net/?f=t_%7B2%7D%20%3D%20%5Cfrac%7B-%28-6%29%20-%20%5Csqrt%7B292%7D%7D%7B2%2A16%7D%20%3D%20-0.35)
It cannot take negative seconds for the ball to hit the ground.
So it will take 0.72s for the football to hits the ground.