By implicit differentiation:
<span>(x(dy/dx) + y)e^(xy) = 0 </span>
<span>Note that when differentiating e^(xy), apply chain rule. When differentiating xy, use product rule. Also: When differentiating y w/respect to x, think of that as if you are differentiating f(x). </span>
<span>Then, substitute (1,ln(2)) and solve for dy/dx. </span>
<span>(1(dy/dx) + ln(2))e^(1ln(2)) = 0 </span>
<span>((dy/dx) + ln(2))e^(ln(2)) = 0 </span>
<span>Note that e^(ln(2)) = 2 since e and ln are inverse of each other. </span>
<span>2((dy/dx) + ln(2)) = 0 </span>
<span>dy/dx + ln(2) = 0 . . . . You get this expression by dividing both sides by 2 </span>
<span>dy/dx = -ln(2) . . . . . . .Subtract both sides by ln(2) </span>
<span>Therefore, dy/dx = -ln(2) </span>
<span>I hope this helps!</span>
Answer:
259.27
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
Slope or gradient of a line is given by dividing change in y axis by the change in x axis. Slope is represented by m hence 
where 
is the change in y axis while 
is the change in x co-ordinates. Substituting with 6 units while 3 units substitute then the slope will be 
Therefore, the gradient is 2
Answer:
Question 1. (2.2, -1.4)
Question 2. (1.33, 1)
Step-by-step explanation:
Equations for the given lines are
-----(1)
It is given that this line passes through two points (0, 2.5) and (2.2, 1.4).
------(2)
This equation passes through (0, -3) and (2.2, -1.4).
Now we have to find a common point through which these lines pass or solution of these equations.
From equations (1) and (2),
x =
x = 2.2
From equation (2),
y = -1.4
Therefore, solution of these equations is (2.2, -1.4).
Question 2.
The given equations are y = 1.5x - 1 and y = 1
From these equations,
1 = 1.5x - 1
1.5x = 2
x =
Therefore, the solution of the system of linear equations is (1.33, 1).
Answer:
The independent variable is s and the dependent variable is r
Step-by-step explanation:
r = 5s+3
The independent variable is s and the dependent variable is r
