An odd number under 50 with least 5 factors
2 answers:
You might be interested in
If e-the initial hight of the tree, and the function imply some brackets like
f(x)=30/(1+29e-0.5x) then
for every
the rate of growth is negative, the maximum growth per unit of time would be 30ft
If
then
so the tree couldn't be 2ft tall initially
if the brackets are something like
f(x)=30:(1+29e)-0.5x then the results may differ
Refer to the figure shown below.
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²
Because the question states that the semi circles are at the ends of the rectangle, each semicircle has a radius of 30 m.
The circumference of the oval is
2π(30) + 2*120 = 60π + 240 m = 428.5 m
The area of the oval is
π(30²) + 60*120 = 900π + 7200 m² = 1.0027 x 10⁴ m²
Note:
If the semi circles are placed on the 120-m sides of the rectangle, then similar calculations yield:
Circumference = 120π + 120 m = 497 m
Area = 3600π + 7200 m² = 1.851 x 10⁴ m²
Answer:
If the semi circles are placed on the 60-m sides of the rectangle, then
Circumference = 60π + 240 m, or 428.5 m
Area = 900π + 7200 m², or 1.0027 x 10⁴ m²
if the semi circles are placed on the 120-m sides of the rectangle, then
Circumference = 120π + 120 m, or 497 m
Area = 3600π + 7500 m², or 1.851 x 10⁴ m²