Answer:
1. B
2. A
3. A
4. C
Step-by-step explanation: Got it right on e{dg}enuity.
The density of the figure is 0.9076g/cm^3
The length of base edge is given as:
[tex]Length=2cm[/tex]
The height of the pyramid is given as:
[tex]Height =1cm[/tex]
The area of an equilateral triangle is then calculated as:
[tex]A_t = \frac{\sqrt 3}{4}l^2[/tex]
Where l represents the side length of the triangle.
So, we have:
[tex]A_t = \frac{\sqrt 3}{4} \times 2^2[/tex]
[tex]A_t = \frac{\sqrt 3}{4} \times 4[/tex]
[tex]A_t = \sqrt 3[/tex]
The area of the hexagonal base is 6 times the area of the triangle.
This is so because, an hexagon has 6 sides
So, we have:
[tex]A_h = \sqrt 3 \times 6[/tex]
[tex]A_h =6\sqrt 3[/tex]
Lastly, the volume of the pyramid is then calculated as:
[tex]V = \frac 13 \times A_h \times h[/tex]
This gives
[tex]V_p = \frac 13 \times 6\sqrt 3 \times 1[/tex]
[tex]V_p = 2\sqrt 3 [/tex]
[tex]V_p = 3.4641[/tex]
The volume of the cylindrical hole is:
[tex]V_c = \pi r^2 h[/tex]
This gives
[tex]V_c = 3.14 \times (0.4/2)^2 \times 1[/tex]
[tex]V_c = 0.1256[/tex]
The volume of the steel is the difference between the volume of the pyramid, and the hole.
So, we have:
[tex]V = V_p - V_c[/tex]
This gives
[tex]V = 3.4641 - 0.1256[/tex]
[tex]V = 3.3385[/tex]
The density of the figure is then calculated as:
[tex]Density = \frac{Mass}{Volume}[/tex]
So, we have:
[tex]Density = \frac{3.03}{3.3385}[/tex]
[tex]Density = 0.9076[/tex]
Hence, the density of the figure is 0.9076g/cm^3
Read more about volumes at:
https://brainly.com/question/16597042
