According to the law of sines:

Using the given values, we can find the angle B and find the number of possible triangles that can be formed.

The range of sin is from -1 to 1. The above expression does not yield any possible value of B, as sin of no angle can be equal to 1.45.
Therefore, we can conclude that no triangle exists with the given conditions.
Answer is 512 cu.ft.
Volume of a cube is length x width x height
So 8 x 8 x 8 = 512
Answer:
Step-by-step explanation:
Multiply the length and width, or c and b to find their area. Multiply this measurement by two to account for both sides. Add the three separate measurements together. Because surface area is the total area of all of the faces of an object, the final step is to add all of the individually calculated areas together.
Answer:
360 cm^2
Step-by-step explanation:
Find the area of each face of the triangular prism.
Imagine the triangular prism as its net, it is composed with 2 triangular faces (these are tye bases of the prism) and 3 rectangular faces.
Areas of the 2 triangular bases (they are similar triangles):
1/2 x 8 cm x 6 cm = 24 cm^2
24 x 2 = 48 cm^2
Area of the rectangular face:
8 x 13 = 104 cm^2
Area of another rectangular face:
6 cm x 13 cm = 78 cm^2
Area of another rectangular face:
13 cm x 10 cm = 130 cm^2
Add up all the areas of all faces:
48 + 104 + 78 + 130 = 360
So the SA is 360 cm^2
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.