Their are 6 equilateral triangles in a regular hexagon:
The area of one triangle = (x²√3)/4, then:
1st answer: The area of the hexagon base: is 6 TIMES the area of one equilateral triangle [or, needed for 2nd, question, TOT AREA:(6x²√3)/4] = (3x²√3)/2] unit²
2nd answer: :
Volume of pyramid: (base area)(height)/3
Volume of pyramid: (3x²√3).(3x)/3 (because height = 3x, given)
Then Volume of pyramid= 3x³√3 unit³.
3 is the answer for the second question
and 6 for the 1st
The formula of an area of a circle with radius r:

Part A.
We have diameter of the circle d = 8 in. The diameter of a circle is equal two times length of a radius.
Therefore:

Calculate the area:

Answer: C. 16π square inches.
Part 2.
Look at the picture.
Total area:

Answer: C. 36π square inches
Answer:
Volume of the prism = 140 cubic inches.
Step-by-step explanation:
The edge length of the cube is 1/2 inches.
The volume of the cube is given by

Now, it has been given that there are 1120 cube is completely filled in the prism. Hence, we have to the multiply the volume by 1120.
Volume of the prism = 1120 × 0.125
Volume of the prism = 140 cubic inches.
Answer:
The graph of f(x) is shifted k units to the right of the graph of g(x).
Step-by-step explanation:
Given the function and where k <0.
As, horizontal depends on the value of x and are when
g(x)=f(x+h) graph shifted to left.
g(x)=f(x-h) graph shifted to right.
Now, given
But here k is negative, when k is considered positive then f(x) becomes
⇒ The graph of f(x) is shifted k units to the right of the graph of g(x).
Correct option is B.
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is
As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is
. Thus, there are only 35 ways to distribute the blackboards in this case.