Answer:
![\frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B-8x%5E3-5y%5E3%7D%7B25y%5E5%7D)
Step-by-step explanation:
Given, ![2x^3+5y^3=7](https://tex.z-dn.net/?f=2x%5E3%2B5y%5E3%3D7)
![6x^2+15y^2\frac{dy}{dx}=0](https://tex.z-dn.net/?f=6x%5E2%2B15y%5E2%5Cfrac%7Bdy%7D%7Bdx%7D%3D0)
![\Rightarrow \frac{dy}{dx}=\frac{-6x^2}{15y^2}=\frac{-2x^2}{5y^2}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B-6x%5E2%7D%7B15y%5E2%7D%3D%5Cfrac%7B-2x%5E2%7D%7B5y%5E2%7D)
Now find the second differentiation w.r.t.
.
![\frac{d^2y}{dx^2}=\frac{15y^2(-12x)-(-6x^2)(30y.\frac{dy}{dx})}{225y^4}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B15y%5E2%28-12x%29-%28-6x%5E2%29%2830y.%5Cfrac%7Bdy%7D%7Bdx%7D%29%7D%7B225y%5E4%7D)
![=\frac{-180xy^2+180x^2y\frac{dy}{dx}}{225y^4}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B-180xy%5E2%2B180x%5E2y%5Cfrac%7Bdy%7D%7Bdx%7D%7D%7B225y%5E4%7D)
![=\frac{180xy(x\frac{dy}{dx}-y)}{225y^4}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B180xy%28x%5Cfrac%7Bdy%7D%7Bdx%7D-y%29%7D%7B225y%5E4%7D)
![=\frac{4(x\frac{dy}{dx}-y)}{5y^3}\quad \quad ...(i)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B4%28x%5Cfrac%7Bdy%7D%7Bdx%7D-y%29%7D%7B5y%5E3%7D%5Cquad%20%5Cquad%20...%28i%29)
put value of
in equation
.
![\frac{d^2y}{dx^2}=\frac{4(x(\frac{-2x^2}{5y^2})-y}{5y^3}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B4%28x%28%5Cfrac%7B-2x%5E2%7D%7B5y%5E2%7D%29-y%7D%7B5y%5E3%7D)
![=\frac{\frac{-8x^3}{5y^2}-y}{5y^3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cfrac%7B-8x%5E3%7D%7B5y%5E2%7D-y%7D%7B5y%5E3%7D)
![\Rightarrow \frac{d^2y}{dx^2}=\frac{-8x^3-5y^3}{25y^5}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%3D%5Cfrac%7B-8x%5E3-5y%5E3%7D%7B25y%5E5%7D)
Hence,
.
Answer:
It takes 1 second for the tape to reach the ground.
Equation to use: ![y(t)=16-\frac{1}{2} g\,t^2[/tex with acceleration due to gravity "g" = 32ft/s^2]Step-by-step explanation:This is an object moving vertically under the action of the acceleration of gravity (32 ft/s^2), with a starting position of 16 feet, and with NO initial velocity (drops from the roof).The equation that describes the"y" position of the object as a function of time (t) will be written as:[tex]y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5B%2Ftex%20with%20acceleration%20due%20to%20gravity%20%22g%22%20%3D%2032ft%2Fs%5E2%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThis%20is%20an%20object%20moving%20vertically%20under%20the%20action%20of%20the%20acceleration%20of%20gravity%20%2832%20ft%2Fs%5E2%29%2C%20with%20a%20starting%20position%20of%2016%20feet%2C%20and%20with%20NO%20initial%20velocity%20%28drops%20from%20the%20roof%29.%3C%2Fp%3E%3Cp%3EThe%20equation%20that%20describes%20the%22y%22%20%20position%20of%20the%20object%20as%20a%20function%20of%20time%20%28t%29%20will%20be%20written%20as%3A%3C%2Fp%3E%3Cp%3E%5Btex%5Dy%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
As explain above, the initial position
s 16 ft, there is no initial velocity, so
, and the acceleration of gravity is 32 ft/s^2, and should be considered negative [as pointing down in the y-direction], so the equation simplifies to:
![y(t) = y_0+ v_0*t-\frac{1}{2} g\,t^2\\y(t)=16-\frac{1}{2} g\,t^2](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_0%2B%20v_0%2At-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5Cy%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2)
In order to find the tima it takes it to hit the ground, we simple solve the equation for t when y(t) = 0 (the tape has reached the ground (zero height in the y-direction):
![y(t)=16-\frac{1}{2} g\,t^2\\0=16-\frac{1}{2} g\,t^2\\-16=-\frac{1}{2} 32\,t^2\\-16=-16\,t^2\\t^2=1\\t=+/- 1 \,second](https://tex.z-dn.net/?f=y%28t%29%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C0%3D16-%5Cfrac%7B1%7D%7B2%7D%20g%5C%2Ct%5E2%5C%5C-16%3D-%5Cfrac%7B1%7D%7B2%7D%2032%5C%2Ct%5E2%5C%5C-16%3D-16%5C%2Ct%5E2%5C%5Ct%5E2%3D1%5C%5Ct%3D%2B%2F-%201%20%5C%2Csecond)
We select the positive time (+1 second) which is what makes physical sense, since a negative value in time would mean time before the tape was dropped.
So the answer is: It takes 1 second for the tape to reach the ground.
Answer:
D
Step-by-step explanation:
The slope intercept form is this: mx +b=y
m=slope
b= y intercept
so all you have to do is input them in their place
15/2 = 7.5
7.5 times 6 = 45
Check work: 45/6 = 7.5