S=S1+S2, where S1=12*4=48 cm², S2=0.5*(12-3-3)*(8-4)=12 cm².
S=48+12=60 cm²
Answer: 12 students
Step-by-step explanation:
Let X and Y stand for the number of students in each respective class.
We know:
X/Y = 2/5, and
Y = X+24
We want to find the number of students, x, that when transferred from Y to X, will make the classes equal in size. We can express this as:
(Y-x)/(X+x) = 1
---
We can rearrange X/Y = 2/5 to:
X = 2Y/5
The use this value of X in the second equation:
Y = X+24
Y =2Y/5+24
5Y = 2Y + 120
3Y = 120
Y = 40
Since Y = X+24
40 = X + 24
X = 16
--
Now we want x, the number of students transferring from Class Y to Class X, to be a value such that X = Y:
(Y-x)=(X+x)
(40-x)=(16+x)
24 = 2x
x = 12
12 students must transfer to the more difficult, very early morning, class.
Answer:
32x3−20x2+24xz+56x−15z−35
Step-by-step explanation:
Answer:
lateral area ≈ 668 m²
surface area ≈ 740 m²
Step-by-step explanation:
The lateral area of a triangular prism = perimeter of the base × height of the prism
lateral area = ph
where
p = perimeter of the base
h = height or altitude
Base edges are 8m , 9m and 12.04m and height is 23 m.
lateral area = ph
lateral area = (8 + 9 + 12.04) × 23
lateral area = 29.04 × 23
lateral area = 667.92 m²
lateral area ≈ 668 m²
Surface area
surface area = lateral area + 2B
where
B = area of it triangular bases
surface area = lateral area + 2B
surface area = 667.92 + 2(1/2 × base × height)
surface area = 667.92 + 2(1/2 × 8 × 9)
surface area = 667.92 + 2(1/2 × 72)
surface area = 667.92 + 2(36)
surface area = 667.92 + 72
surface area = 739.92
surface area ≈ 740 m²
