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
Answer: -65/6
Step-by-step explanation:
Answer:
W = 0
Z = -2
Step-by-step explanation:
While simplyfying the equation, i noticed that one is 24 and the other is -24, which meant that one value is non existent, and after i found out that w = 0 imputing the reslut made that easier
Answer:
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
Step-by-step explanation:
Given
Volume of a box = length × breadth × height= l×b×h
In this case the box have a square base. i.e l=b
Volume V = l^2 × h
The surface area of a square box
S = 2(lb+lh+bh)
S = 2(l^2 + lh + lh) since l=b
S = 2(l^2 + 2lh)
Given that the box is open top.
S = l^2 + 4lh
And Surface Area of the box is 1200cm^2
1200 = l^2 + 4lh ....1
Making h the subject of formula
h = (1200 - l^2)/4l .....2
Volume is given as
V = l^2 × h
V = l^2 ×(1200 - l^2)/4l
V = (1200l - l^3)/4
the maximum point is at dV/dl = 0
dV/dl = (1200 - 3l^2)/4
dV/dl = (1200 - 3l^2)/4 = 0
3l^2= 1200
l^2 = 1200/3 = 400
l = √400
I = 20cm
Since,
h = (1200 - l^2)/4l
h = (1200 - 20^2)/4×20
h = (800)/80
h = 10cm
The largest possible volume V is ;
V = l^2 × h
V = 20^2 × 10 = 4000cm^3
