For
ax+by=c
slope=-a/b
we have
-6x+1y=-4
slope=-6/-1=6
slope is 6
yint is where x=0
-6(0)+y=-4
0+y=-4
y=-4
yint is at y=-4 or (0,-4)
Answer:
9r^2 + 7r - 5
Step-by-step explanation:
Answer:
A) Dimensions;
Length = 20 m and width = 10 m
B) A_max = 200 m²
Step-by-step explanation:
Let x and y represent width and length respectively.
He has 40 metres to use and he wants to enclose 3 sides.
Thus;
2x + y = 40 - - - - (eq 1)
Area of a rectangle = length x width
Thus;
A = xy - - - (eq 2)
From equation 1;
Y = 40 - 2x
Plugging this for y in eq 2;
A = x(40 - 2x)
A = 40x - 2x²
The parabola opens downwards and so the x-value of the maximum point is;
x = -b/2a
Thus;
x = -40/2(-2)
x = 10 m
Put 10 for x in eq 1 to get;
2(10) + y = 40
20 + y = 40
y = 40 - 20
y = 20m
Thus, maximum area is;
A_max = 10 × 20
A_max = 200 m²
Answer:

Step-by-step explanation:
Given
![\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20-%5Cfrac%7B1%7D%7B64%7De%5E%7B-4x%7D%5BAx%5E2%20%2B%20Bx%20%2B%20E%5DC)
Required
Find 
We have:
![\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20-%5Cfrac%7B1%7D%7B64%7De%5E%7B-4x%7D%5BAx%5E2%20%2B%20Bx%20%2B%20E%5DC)
Using integration by parts

Where
and 
Solve for du (differentiate u)

Solve for v (integrate dv)

So, we have:




-----------------------------------------------------------------------
Solving

Integration by parts
---- 
---------- 
So:



So, we have:

![\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20-%5Cfrac%7Bx%5E2%7D%7B4%7De%5E%7B-4x%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B%20-%5Cfrac%7Bx%7D%7B4%7De%5E%7B-4x%7D%20%20-%5Cfrac%7B1%7D%7B4%7De%5E%7B-4x%7D%5D)
Open bracket

Factor out 
![\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20%5B-%5Cfrac%7Bx%5E2%7D%7B4%7D%20-%5Cfrac%7Bx%7D%7B8%7D%20-%5Cfrac%7B1%7D%7B8%7D%5De%5E%7B-4x%7D)
Rewrite as:
![\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20%5B-%5Cfrac%7B1%7D%7B4%7Dx%5E2%20-%5Cfrac%7B1%7D%7B8%7Dx%20-%5Cfrac%7B1%7D%7B8%7D%5De%5E%7B-4x%7D)
Recall that:
![\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20-%5Cfrac%7B1%7D%7B64%7De%5E%7B-4x%7D%5BAx%5E2%20%2B%20Bx%20%2B%20E%5DC)
![\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}](https://tex.z-dn.net/?f=%5Cint%5Climits%20%7Bx%5E2%5Ccdot%20e%5E%7B-4x%7D%7D%20%5C%2C%20dx%20%20%3D%20%5B-%5Cfrac%7B1%7D%7B64%7DAx%5E2%20-%5Cfrac%7B1%7D%7B64%7D%20Bx%20-%5Cfrac%7B1%7D%7B64%7D%20E%5DCe%5E%7B-4x%7D)
By comparison:



Solve A, B and C

Divide by 

Multiply by 64



Divide by 

Multiply by 64



Multiply by -64


So:

