Six red marbles, 4 green marbles and 2 blue marbles are placed in a jar. What is the probability of selecting a red marble and t
1 answer:
You might be interested in
The volume of the prop is calculated to be 1,875.6 cubic cm.
Step-by-step explanation:
Step 1; The prop consists of a cone and a half-sphere on top. We will have to calculate the volumes of the cone and the half-sphere separately and then add them to obtain the total volume.
Step 2; The volume of a cone is determined by multiplying with π, the square of the radius (r²) and height (h). Here we substitute π as 3.14. The radius is 8 cm and the height is 12 cm.
The volume of the cone = × 3.14 × 8 × 8 × 12 = 803.84 cubic cm.
Step 3; The area of a half-sphere is half of a full sphere. The volume of a sphere is given by multiplying with π and the cube of the radius (r³). Here the radius is 8cm. We take π as 3.14.
The volume of a full sphere = × 3.14 × r³ =
× π × 8³ = 2,143.5733 cubic cm.
The volume of a half-sphere = = 1,071.7866 cubic cm.
Step 4; The total volume = The volume of the cone + The volume of the half sphere,
The total volume = 803.84 cubic cm + 1,071.7866 cubic cm = 1,875.6266 cubic cm. By rounding this off to the nearest tenth we get 1,875.6 cubic cm.
Answer:
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)