Answer:
20
Step-by-step explanation:
Answer: 3.66 cm
Step-by-step explanation: Given a rectangular casing BCDE with segment DE = 3 cm and segment BE = 3.5 cm.
The area A of a rectangle is length multiply by width.
Where length L = 3.5 cm and
width W = 3 cm
Area A = 3.5 × 3 = 10.5 cm^2
The pipe that will fit the fiber optic line is in cylindrical shape. Where area of a cylinder = πr^2.
But area A = 10.5. Substitute the values for the area of the cylinder
10.5 = πr^2
10.5 = 3.143 × r^2
Make r^2 the subject of formula
r^2 = 10.5/3.143
r = sqrt ( 3.34225 )
r = 1.828
Diameter = 2 × radius
Diameter = 2 × 1.829
Diameter = 3.656 cm
Therefore, the smallest diameter of pipe that will fit the fiber optic line is 3.66 cm approximately.
Answer:
Step-by-step explanation
Perimeter of a rectangle =2(length +breadth)
90m = 2{(36x+2) + (2(x+2)}
Opening the inner brackets,
90 = 2{36x +2+2x +4}
90 = 2{38x +6}
Open the brackets
90 = 76x + 12
Collect like terms
90 - 12 = 76x
78=76x
Divide 78 by the coefficient of x
x = 78÷76
x =1.03m (2d.p)
Substitute the value of x into the expressions for the two sides of the rectangle
36x+2 =(36×1.03) +2= 37.08+2=39.08m
2(x+2)= 2(1.03+2)= 2(3.03) =2×3.03= 6.06m
The approximate values of the two sides of d rectangular field are 39.08m and 6.06m
