Approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter
<em><u>Solution:</u></em>
Given semicircle with a radius of 30 centimeters
<em><u>To find: approximate circumference of a semicircle</u></em>
To find circumference of semicircle we can divide the circumference of circle by 2
circumference of semicircle = circumference of circle 2
Substituting the given radius = 30 cm,
Thus approximate circumference of a semicircle with a radius of 30 centimeters is 94.2 centimeter
Answer: D. -40x^3+25x^2+45x
The value of arccos(-√3/2) is 5π/6.
In the given question,
We have to find the value of .
The given term is .
We using the quadrant rule.
The value is in negative so we check in which quadrant the value of cos negative.
As we now that in second quadrant only the value of sine and cosine is positive and all value are negative.
The second quadrant of 180 degree or we can also write 180 degree as π.
From the trigonometry table we know that at 30 degree or (π/6) the value of cos is √3/2.
So the value of cos at which -√3/2 is
-√3/2 = cos(π - π/6)
-√3/2 = cos(π×6/6 - π/6)
-√3/2 = cos(6π/6 - π/6)
-√3/2 = cos{(6π- π)/6}
-√3/2 = cos 5π/6
So we can write it as
arccos(-√3/2)=arccos(cos 5π/6)
Since there is a arccos and cos both so only left 5π/6.
arccos(-√3/2)=5π/6
Hence, the value of arccos(-√3/2) is 5π/6.
To learn more about trigonometry quadrant link is here
brainly.com/question/7196312
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The perimeter of a rectangle is twice ths sum of length and width, so the length of the garden can be described by (30 - x) in feet, where 30 = half the length of the available fence. The area is the product of length and width, so
f(x) = x(30 -x)
For the answer to the question above,
though the first two would start out as S, when you put the actual value into the equation, that is what you get. This is the equation based on your question above 3 * 2 + 2 = j