Answer:
the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
Step-by-step explanation:
From the given question.
Let p be the length of the of the printed material
Let q be the width of the of the printed material
Therefore pq = 2400 cm ²
q = 
To find the dimensions of the poster; we have:
the length of the poster to be p+30 and the width to be 
The area of the printed material can now be: 
=
Let differentiate with respect to p; we have

Also;

For the smallest area 


p² = 3600
p =√3600
p = 60
Since p = 60 ; replace p = 60 in the expression q =
to solve for q;
q =
q = 
q = 40
Thus; the printed material has the length of 60 cm and the width of 40cm
the length of the poster = p+30 = 60 +30 = 90 cm
the width of the poster =
=
= 40 + 20 = 60
Hence; the dimension of the poster = 90 cm length and 60 cm width i.e 90 cm by 60 cm.
12 - 3 = 9
3 + 9 = 12
12 - 9 = 3
Answer:29
Step-by-step explanation:
Answer: Option 'C' is correct.
Step-by-step explanation:
Since we have given that
Radius of circle = 10 inches
Central angle is given by

As we know the formula for " Length of an arc " :
Length of arc is given by

Hence, length of an arc is 20.95 inches.
Hence, Option 'C' is correct.
The numbers are 0.85 and 0.15