Answer:
Slope intercept form would be, 3y = 2x + 6
Simplified it would be, y = 2/3x + 2
The tangent to through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces and at that point.
Let . Then is the level curve . Recall that the gradient vector is perpendicular to level curves; we have
so that the gradient of at (1, 1, 1) is
For the surface , we have
so that . We can obtain a vector normal to by taking the cross product of the partial derivatives of , and evaluating that product for :
Now take the cross product of the two normal vectors to and :
The direction of vector (24, 8, -8) is the direction of the tangent line to at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by . Then adding (1, 1, 1) shifts this line to the point of tangency on . So the tangent line has equation
The answer is H) 0.75
all you do is divide 4.5 by 6 and you'll get 0.75
Multiply the first equation my 2 and than subtract the two equations to eliminate x
6x+16y=60-
6x+7y=6
9y=54
Y=6
Then plug in y to one of the equations
6x+7(6)=6
6x+42=6
6x=-36
X=-6
X=-6 Y=6
It’s a black picture can’t see