Answer:
24 i think
Step-by-step explanation:
I evaluated using he given value
Answer:
31.4 inches
Step-by-step explanation:
If a circle is inscribed in a square then diameter of circle inscribed is same as side as of square.
In the given problem it is given that side of square is 10 inches.
So diameter of circle inscribed is 10 inches
we know radius of circle is half of diameter of circle
Thus, radius of circle inscribed = diameter of circle/2 = 10/2 = 5inches.
Expression to calculate circumference of circle is given by 
where r is the radius of circle.
Thus circumference of circle inscribed is

Thus, circumference of circle inscribed is 31.4 inches
Answer:






Step-by-step explanation:
Given
--- terminal side of 
Required
Determine the values of trigonometric functions of
.
For
, the trigonometry ratios are:


Where:


In 
and 
So:






<u>Solving the trigonometry functions</u>


Rationalize:






Rationalize
















Answer:
40x30= 1200, 20x15=300, 300/1200= 1/4.
Step-by-step explanation:
You can also visualize it. Imagine a field 40’ long and the boy cut the top 20’ of it but left 20’ of the bottom, he would have cut half of it at this point. Now it’s 30’ wide and he cut the left 15’ of what’s remaining but left the right 15’ uncut. So now there’s a 20’x15’ bottom right corner left uncut. There are 4 sections that are 20’x15’ and he left one out of four uncut, otherwise written as 1/4.
Answer:
360
Step-by-step explanation:
T = K(1/H)
Whereas T = seconds
H = cooker setting
K = the constant of proportionality
T = (240/6)/(1/9) = (240/6)x9 = 360