Taking
and differentiating both sides with respect to
yields
![\dfrac{\mathrm d}{\mathrm dx}\bigg[3x^2+y^2\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[7\bigg]\implies 6x+2y\dfrac{\mathrm dy}{\mathrm dx}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B3x%5E2%2By%5E2%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B7%5Cbigg%5D%5Cimplies%206x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D0)
Solving for the first derivative, we have
Differentiating again gives
![\dfrac{\mathrm d}{\mathrm dx}\bigg[6x+2y\dfrac{\mathrm dy}{\mathrm dx}\bigg]=\dfrac{\mathrm d}{\mathrm dx}\bigg[0\bigg]\implies 6+2\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2+2y\dfrac{\mathrm d^2y}{\mathrm dx^2}=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B6x%2B2y%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cbigg%5B0%5Cbigg%5D%5Cimplies%206%2B2%5Cleft%28%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%2B2y%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D0)
Solving for the second derivative, we have
Now, when
and
, we have
Answer:
Step-by-step explanation:
It is a linear equation and the gradient equation is given as:
y=mx+c
Where m is the gradient and c is the y intercept
-3x+2y=1
2y=1+3x
Dividing all through by 2
y=1/2 + 3x/2
Therefore the gradient of this linear equation is 3/2 and the y intercept is 1/2
The equation of a circle centred at point (m,n) and radius r is given by
<span>(x-m)² + (y-n)² = r²
</span>-------------------------------------------------------------
Centre = (4,3)
radius = 5
Equation:
(x - 4)² + (y - 3)² = 5²
⇒ x² - 8x + 16 + y² - 6y + 9 = 25
⇒ x² + y² - 8x - 6y + 25 = 25
⇒ x² + y² - 8x - 6y = 0
The equation of the circle is x² + y² - 8x - 6y = 0
Hope it helps!
Answer:
the bottom left one
Step-by-step explanation:
<span>x^2 - 6x = 7
(x - 7)(x + 1)= 0
x = 7, - 1
The answer is: x = 7, -1.</span>
