Copy question and paste it and it will show up trust
Answer:
<u>1759</u><u>. </u><u> </u> or 27 58/63
63
hope it helps
Answer:
See answers below for explanations (answers are in bold)
Step-by-step explanation:
<u>Problem 1 (Top left):</u>
Assuming that the shape is an ellipse, we will use the formula A=πab where a and b are both half the major and minor axis respectively. Since a=9 and b=6, then we have A = π(9)(6) = 54π = 169.646 units^2 as our final answer
<u>Problem 2 (Top right):</u>
Again, we use the same equation as in Problem 1 but this time, a=12 and b=8, so we then have A = π(8)(12) = 96π = 301.593 units^2 as our final answer
<u>Problem 3 (Bottom left):</u>
We can make this irregular shape into a half-ellipse and a rectangle. We would find their separate areas and add them up. The area of the half-ellipse would be A=1/2πab. The value of "a" would be 13.8/2=6.9 since that's the length of the rectangle. The value of "b" would be 3.5, so that means the area of the half-ellipse would be A = 1/2(π)(6.9)(3.5) = 12.075π = 37.935 units^2. The width and length of the rectangle are 5.4 and 13.8 respectively, so this means the area of the rectangle is 5.4*13.8 = 74.52 units^2. Combining both areas, we get 37.935+74.52 = 112.455 units^2 as our final answer
<u>Problem 4 (Bottom right):</u>
This irregular shape is an ellipse, but a circle is missing from its center. Thus, we must find the area of the circle and ellipse and then subtract the circle's area from the ellipse's area. For the area of the ellipse, a=(15.5/2)=7.75, and b=(7/2)=3.5, making its area A = π(7.75)(3.5) = 27.125π = 85.216 units^2. The area of a circle is A=πr² where "r" is the radius. The radius would be 7/2=3.5, making the circle's area A = π(3.5²) = 12.25π = 38.485 units^2. This means that the shaded area is 85.216-38.435 = 46.781 units^2
Answer:
[see below]
Step-by-step explanation:
A function's inputs do not repeat. This means that any point with the x-value not repeated with the other points can be added to ensure that it continues as a function.
In this scenario:
{x| x ≠ -7, 4, 0, -2}
A point that does not have the x-value of -7, 0, 4, and -2 could be added to the relation to ensure it continues to be a function.
Hope this helps.
You can use protractors I think
