<span>x^2 • y^–3 • x^4
= x^6 / y^3</span>
Step-by-step explanation:
<h2>-20 is the right answer.</h2>
Y = xe^x
dy/dx(e^x x)=>use the product rule, d/dx(u v) = v*(du)/(dx)+u*(dv)/(dx), where u = e^x and v = x:
= e^x (d/dx(x))+x (d/dx(e^x))
y' = e^x x+ e^x
y'(0) = 1 => slope of the tangent
slope of the normal = -1
y - 0 = -1(x - 0)
y = -x => normal at origin
Answer:
B, E
Step-by-step explanation:
The equation of that line in slope intercept form is y = -2/5x + 2
The slope of a line parallel to that line will be the same as its slope, so -2/5.
To find the y-intercept of a line that passes through the point (-5, 1) with that slope, you will have to plug in the x and y values of that point into what you know of the equation of the line.
y = -2/5x + b
1 = -2/5(-5) + b
1 = 2 + b
-1 = b
From this, you can construct an equation.
y = -2/5x - 1
However, this is not an answer choice.
It cannot be A because this line does not have a slope of -1.
2x + 5y = 15
5y = -2x - 5
y = -2/5x - 1
It can be B because this line is the same as the equation we came up with.
It cannot be C because this line does not have a slope of -1 or a y-intercept of -3.
2x + 5y = -15
5y = -2x - 15
y = -2/5x - 3
It cannot be D because this line does not have a y-intercept of -3.
y - 1 = -2/5(x + 5)
y - 1 = -2/5x - 2
y = -2/5x - 1
It can be E because this equation matches the one we came up with.
Answer:
Uh can you please be more specific?
