Answer:
5.14 km
Step-by-step explanation:
A semicircle is 1/2 of a circle so the perimeter is just 1/2 of the circumference
C = 2 * pi r for a circle so for a semicircle
1/2* 2* pi *r
pi*r
3.14 * 1
3.14
If we want to include the piece that closes the semicircle, we need to add the diameter
d = 2r = 2(1) = 2
2+3.14 = 5.14
Answer:
108 ft²
Step-by-step explanation:
For the polygon on the left, we are given the length of one side (3 ft) and the total area (27 ft²). With this information, we can determine the length of the other side - 9 ft. I found this by dividing 27 ft² by 3 ft, giving me the final result of 9 ft.
Next, we have to find the other side of the polygon on the <em>right</em> so we can ultimately determine its area. It looks like there is a scale factor of 2 between the two polygons, since 3 × 2 = 6. We know that the bottom side of the left polygon is 9, so multiplying 9 by 2 should give us the bottom side of the polygon on the right. 9 × 2 = 18.
Now, we have the side lengths for the polygon on the right and can determine its area. What is 6 ft × 18 ft? Well, the answer is 107 ft², and this is the answer to the question.
Hopefully that's helpful! :)
Answer:
what's the question? I don't know
Answer:
x= number of tickets bought
y= total
44x+12=y
Answer:
Step-by-step explanation:
![\sf 3(t+3)+5(3+2t)-76 = 0\\\\Expanding \ Parenthesis\\\\3t + 9 + 15 + 10t -76 = 0\\\\13t + 24 -76 = 0\\\\13t - 52 = 0\\\\Add \ 52 \ to \ both \ sides\\\\13t = 52\\\\Dividing\ both\ sides\ by\ 13\\\\t = 52/13\\\\t = 4\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%203%28t%2B3%29%2B5%283%2B2t%29-76%20%3D%200%5C%5C%5C%5CExpanding%20%5C%20Parenthesis%5C%5C%5C%5C3t%20%2B%209%20%2B%2015%20%2B%2010t%20-76%20%3D%200%5C%5C%5C%5C13t%20%2B%2024%20-76%20%3D%200%5C%5C%5C%5C13t%20-%2052%20%3D%200%5C%5C%5C%5CAdd%20%5C%2052%20%5C%20to%20%5C%20both%20%5C%20sides%5C%5C%5C%5C13t%20%3D%2052%5C%5C%5C%5CDividing%5C%20both%5C%20sides%5C%20by%5C%2013%5C%5C%5C%5Ct%20%3D%2052%2F13%5C%5C%5C%5Ct%20%3D%204%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
