Consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?
2 answers:
<span>The derivative of your function is: 1 + (1*dy/dx + y) - 4y(dy/dx) = 0
2 + dy/dx + y - 4y(dy/dx) = 0
(2 + y) + dy/dx (1 - 4y) = 0
(dy/dx) = -(2 + y)/(1 - 4y) and at point (2,1),
dy/dx = -(2 + 1)/(1 - 4*1) = -3 / -3 = 1
Slope of tangent line = 1</span>
The derivative is actually (-y-1)/(4y+x) then you plug in the points (2,1) ==> (-1-1)/(4*1+2) which equals -1/3 then you just plug in your info into a point-slope formula and get: y-1=-1/3(x-2) which can then be reduced
You might be interested in
Answer:
C) 2
Step-by-step explanation:
Answer:
John- 375
Karen- 125
Step-by-step explanation:
Answer 1...
j =3k
2j + k = 875
substituting the first eqn into the 2nd
2(3k) + k = 875
6k+ k =875
7k = 875
k =875/7 =125
thus j = 3(125) =375
Answer:
Step-by-step explanation:
(Kevin) y= k-6 - k minus six (Melanie) y= 2(k-6) - two times k minus six K, only got that far... sorry
Technically, none of the equations. The equation you're looking for would be: x²+(x+20)²=150²