Answer:
208 
Step-by-step explanation:
We can find the area of the net by adding up the area of each of the 6 rectangles that make up the net. Since two of each rectangle are the same, we only have to find the area of the 3 different sized rectangles and multiply each by 2.
Rectangle pairs are:
- Left rectangle and right rectangle
- Top rectangle and the rectangle above the bottom rectangle
- Bottom rectangle and the rectangle surrounded by all for sides
Now, let's solve the question.
Left rectangle:
6 x 4 = 24, rectangle has area of 24 squared cm
Top rectangle:
6 x 8 = 48, rectangle has area of 48 squared cm
Bottom rectangle:
4 x 8 = 32, rectangle has area of 32 squared cm
Add up the areas:
(24 x 2) + (48 x 2) + (32 x 2) = 208
The rectangle has a surface area of 208 squared cm
Answer:
B
Step-by-step explanation:
f(x) = 4x - 7
What does f(x) mean? It means that whatever variable you have on the left which is (x), the same variable must be on the right.
So f(2) means that whatever variable is on the right, it must be replaced with 2 on the right. So .....
f(2) = 4(2) - 7
f(2) = 8 - 7
f(2) = 1
No because a square pyramid must hav a square for its base and a square is not a triangle
answer is not possible
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.
