The dimension that would give the maximum area is 20.8569
<h3>How to solve for the maximum area</h3>
Let the shorter side be = x
Perimeter of the semi-circle is πx
Twice the Length of the longer side
![[70-(\pi )x -x]](https://tex.z-dn.net/?f=%5B70-%28%5Cpi%20%29x%20-x%5D)
Length = ![[70-(1+\pi )x]/2](https://tex.z-dn.net/?f=%5B70-%281%2B%5Cpi%20%29x%5D%2F2)
Total area =
area of rectangle + area of the semi-circle.
Total area =
![x[[70-(1+\pi )x]/2] + [(\pi )(x/2)^2]/2](https://tex.z-dn.net/?f=x%5B%5B70-%281%2B%5Cpi%20%29x%5D%2F2%5D%20%2B%20%5B%28%5Cpi%20%29%28x%2F2%29%5E2%5D%2F2)
When we square it we would have
![70x +[(\pi /4)-(1+\pi)]x^2](https://tex.z-dn.net/?f=70x%20%2B%5B%28%5Cpi%20%2F4%29-%281%2B%5Cpi%29%5Dx%5E2)
This gives
![70x - [3.3562]x^2](https://tex.z-dn.net/?f=70x%20-%20%5B3.3562%5Dx%5E2)
From here we divide by 2
The maximum side would be at
This gives us 20.8569
Read more on areas and dimensions here:
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Search the square root of ____ on Google
Hi
y = mx+b
2x - 3y + 6 = 0
3y = 2x+6
y = 2x/3+6/3
y = 2x/3+2
m = 2/3
6. The lengths of two tangent lines to a circle are equal, which can be proved using similar triangles. The perimeter is 2*(3+2+5)=10.
14. c=40 degrees, since c and the given 40 angle correspond to the same arc on the circle. b=40, since b and the given angle are alternate interior angles of two parallel lines. d=180-40-40=100 degrees. a=180-d=180-100=80 degrees.
18. a is two times the given angle according to the inscribed angle theorem. a=2*20=40. b=180-a=140. c=90 since the tangent line is perpendicular to the diameter of the circle.
19. b=85, a=95. Connect the two points on the circle near 45 degrees on the diagram. b is the sum of the two angles formed (b is the exterior angle of the small triangle), and the sum is half the central angles they correspond to, which is 360-145-45=170. So b=170/2=85, and a=180-85=95 degrees.
20. Use power of a triangle. 5*(10+5)=6(6+x), x=6.5.
Answer:
Step-by-step explanation:
You know the intercepts occur at x = 0, so set each part of the factored form equal to zero and solve for x.
The first part, plain X = 0, so X = 0 is a root, an x-intercept
The second part (X-2)^2 = 0, so X = 2 is another root. Since the function is being squared, this root will happen twice. The function crosses this root twice.
The third part (x+4) = 0, so x = -4 is another root.
Finally, (x+1) so x = -1 is another root.
So you have X = 0, X = 2, X = -4, X = -1 as your x-intercepts
