25 degrees and your welcome
Answer:
8 sides for this one
Step-by-step explanation:
hope this helps
Given:
x and y are both differentiable functions of t.
![4x^2+3y^3=28](https://tex.z-dn.net/?f=4x%5E2%2B3y%5E3%3D28)
![x=-1\text{ and }\dfrac{dy}{dt}=8](https://tex.z-dn.net/?f=x%3D-1%5Ctext%7B%20and%20%7D%5Cdfrac%7Bdy%7D%7Bdt%7D%3D8)
To find:
The value of
.
Solution:
We have,
...(i)
At x=-1,
![4(-1)^2+3y^3=28](https://tex.z-dn.net/?f=4%28-1%29%5E2%2B3y%5E3%3D28)
![4+3y^3=28](https://tex.z-dn.net/?f=4%2B3y%5E3%3D28)
![3y^3=28-4](https://tex.z-dn.net/?f=3y%5E3%3D28-4)
![3y^3=24](https://tex.z-dn.net/?f=3y%5E3%3D24)
Divide both sides by 3.
![y^3=8](https://tex.z-dn.net/?f=y%5E3%3D8)
Taking cube root on both sides.
![y=2](https://tex.z-dn.net/?f=y%3D2)
So, y=2 at x=-1.
Differentiate (i) with respect to t.
![4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0](https://tex.z-dn.net/?f=4%282x%5Cdfrac%7Bdx%7D%7Bdt%7D%29%2B3%283y%5E2%5Cdfrac%7Bdy%7D%7Bdt%7D%29%3D0)
Putting x=-1, y=2 and
, we get
![4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0](https://tex.z-dn.net/?f=4%282%28-1%29%5Cdfrac%7Bdx%7D%7Bdt%7D%29%2B3%283%282%29%5E28%29%3D0)
![-8\dfrac{dx}{dt}+9(4)(8)=0](https://tex.z-dn.net/?f=-8%5Cdfrac%7Bdx%7D%7Bdt%7D%2B9%284%29%288%29%3D0)
![-8(\dfrac{dx}{dt}-9(4))=0](https://tex.z-dn.net/?f=-8%28%5Cdfrac%7Bdx%7D%7Bdt%7D-9%284%29%29%3D0)
Divide both sides by -8.
![\dfrac{dx}{dt}-36=0](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D-36%3D0)
![\dfrac{dx}{dt}=36](https://tex.z-dn.net/?f=%5Cdfrac%7Bdx%7D%7Bdt%7D%3D36)
Therefore, the value of
is 36.
![\huge{\mathfrak{\purple{Answer}}}](https://tex.z-dn.net/?f=%5Chuge%7B%5Cmathfrak%7B%5Cpurple%7BAnswer%7D%7D%7D)
Equation: x² - 15x = 16
Factors: x = -16 and x = 1
![\huge{\mathfrak{\purple{Explanation}}}](https://tex.z-dn.net/?f=%5Chuge%7B%5Cmathfrak%7B%5Cpurple%7BExplanation%7D%7D%7D)
Given
A number squared minus fifteen times the number is equal to 16.
<u>Equation</u>: x² - 15x = 16
Factors:
x² - 15x - 16 = 0
So,
(x - 16) (x + 1)
=> x - 16 = 0, x + 1 = 0
= x = -16 and x = 1
#HopeItHelps