-3(x+3)= -3(x+1)-5
-3x+9=-3x+3-5
so you would have just distributed then you combine like terms and get
-3x+9=-3x-2
then add two to both sides and get
-3x+9=-3x-2
-3x+2=-3x+2
-3x+11=-3x
+3x+11=+3x
0x+11=0x
so 11+0x
divide both by zero and you get x+0
Answer:
$3.16
Step-by-step explanation:
Divide
add
Answer:
1. a) Square
2. c) Rectangle
Step-by-step explanation:
1. If you cut this rectangular prism in way to be PERPENDICULAR to the base, that means you're cutting it straight down, from the top to the bottom, in a vertical line. The cross-section obtained will be just like an end of the prism, which in this case is a square. Because it's a regular rectangular prism, no matter where you cut it, as long as it's vertical, perpendicular to the base, you'll get a square due to this particular form. Technically, the answer could also be a rectangle, since a square is a rectangle.
2. you cut this rectangular prism in way to be PARALLEL to the base, that means you're cutting it straight, from the front to the back, in an horizontal line. The cross-section obtained will be just like an top of the prism, which in this case is a rectangle. Because it's a regular rectangular prism, no matter where you cut it, as long as it's an horizontal line, parallel to the base, you'll get a rectangle due to this particular form.
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.
