Answer: i don’t know but good luck, you got this!Have a nice day!
Step-by-step explanation:
Remark
The proof is only true if m and n are equal. Make it more general.
m = 2k
n = 2v
m + n = 2k + 2v = 2(k + v).
k and v can be equal but many times they are not. From that simple equation you cannot do anything for sure but divide by 2.
There are 4 combinations
m is divisible by 4 and n is not. The result will not be divisible by 4.
m is not divisible by 4 but n is. The result will not be divisible by 4.
But are divisible by 4 then the sum will be as well. Here's the really odd result
If both are even and not divisible by 4 then their sum is divisible by 4
Answer:
A.
Step-by-step explanation:
Given the points:
(x, y) ==> (-2, 4)
slope, m = 3
To write the slope intercept form, use the form below:
y = mx + b
Where m = slope and b = y-intercept.
Now, substitute -2 for x, 4 for y, and m for 3 in the equation above to solve for b.
We have:
y = mx + b
4 = 3(-2) + b
4 = -6 + b
Add 6 to both sides:
4 + 6 = -6 + 6 + b
10 = b
b = 10
Therefore, the slope intercept form of the equation is:
y = 3x + 10
ANSWER:
y = 3x + 10
Answer:
- 12 ft parallel to the river
- 6 ft perpendicular to the river
Step-by-step explanation:
The least fence is used when half the total fence is parallel to the river. That is, the shape of the rectangle is twice as long as it is wide.
72 = W(2W)
36 = W²
6 = W . . . . . . the width perpendicular to the river
12 = 2W . . . . the length parallel to the river
_____
<em>Development of this relation</em>
Let T represent the total length of the fence for some area A. Then if x is the length along the river, the width is y=(T-x)/2, and the area is ...
A = xy = x(T -x)/2
Note that the equation for area is that of a parabola with zeros at x=0 and at x=T. That is, for some fence length T, the area will be a maximum at the vertex of this parabola. That vertex is located halfway between the zeros, at ...
x = (0 +T)/2 = T/2
The corresponding area width (y) is ...
y = (T -T/2)/2 = T/4
Equivalently, the fence length T will be a minimum for some area A when x=T/2 and y=T/4. This is the result we used above.
