We need to determine the radius and diameter of the circle. If the area of the circle is 10 pi in^2, then, according to the formula for the area of a circle,
A = 10 pi in^2 = pi*r^2. Thus, 10 in^2 = r^2, and r = radius of circle = sqrt(10) in.
Thus, the diam. of the circle is 2sqrt(10) in. This diam. has the same length as does the hypotenuse of one of the triangles making up the square.
Thus, [ 2*sqrt(10) ]^2 = x^2 + x^2, where x represents the length of one side of the square. So, 4(10) in^2 = 2x^2. Then:
40 in^2 = 2x^2, or 20 in^2 = x^2, and so the length x of one side of the square is sqrt(20). The area of the square is the square of this result:
Area of the square = x^2 = [ sqrt(20) ]^2 = 20 in^2 (answer). Compare that to the 10 pi sq in area of the circle (31.42 in^2).
Answer:
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Step-by-step explanation:
I don’t understand
Answer:
28/5= 5 3/5 15/7= 2 1/7 21/4=5 1/4
Step-by-step explanation:
Answer: see below
<u>Step-by-step explanation:</u>
Write each equation in y = mx + b format where m is the slope and b is the y-intercept.
Left side: -4 ≤ x < -1
If you continue the line through the y-axis it will pass through (0, 4) --> b = 4
The rise over run is -1 over 1 --> m = -1
y = (-1)x + (4) --> y = -x + 4
Right side: -1 < x < 4
The line passes through (0,0) --> b = 0.
The rise over run is -1 over 1 --> m = -1
y = (-1)x + (0) --> y = -x
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I feel it is unfair to kids of today to require work in bad weather. Especially in harsh and even dangerous weather conditions. For me what if power goes out should I be required to do work?
