So 1st consider that it's a square! That's very important. So for a square, all 4 sides are equal.
And now considering that the given information is the diameter. So any angle made at the circle extended from the 2 points of diameter gives an angle of 90°
Now consider one triangle. So we already know that 2 sides of the triangle are equal (because they are 2 sides of a square) , has a side of 10 (diameter) and and angle of 90°. So remaining 2 angles are 45°
Now solve it by applying
Answer:
Hope this helps! Brainliest please
Step-by-step explanation:
1. Divide both by 5, so the correct answer would be 5/9, and then multiply both by 3, so the correct answer would be 75/135. The answer is 5/9,75/135
2. 4 and 2/7 because 7 *4=28. That leaves 2 sevenths over.
3. Divide both by 9. Your answer is 3/4
Answer:
210 in²
Step-by-step explanation:
6*2.5+6*6*2+(8+6)*2.5+10*2.5+1/2*6*8*2+6*2.5= 210 in²
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r}\\\\ -------------------------------\\\\ (x-3)^2+(y+7)^2=64\implies [x-\stackrel{h}{3}]^2+[y-(\stackrel{k}{-7})]^2=\stackrel{r}{8^2} \\\\\\ center~(3,-7)\qquad radius=8](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%7D%7B%20r%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%28x-3%29%5E2%2B%28y%2B7%29%5E2%3D64%5Cimplies%20%5Bx-%5Cstackrel%7Bh%7D%7B3%7D%5D%5E2%2B%5By-%28%5Cstackrel%7Bk%7D%7B-7%7D%29%5D%5E2%3D%5Cstackrel%7Br%7D%7B8%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Acenter~%283%2C-7%29%5Cqquad%20radius%3D8)
so, the broadcast location and range is more or less like the picture below.
Answer:c
Step-by-step explanation:none of these are good answers as they can all exist in flat Euclidean space. The best answer is c because a great circle exists in spherical (non-flat) geometry . However it’s not really correct to say you could never describe a great circle in Euclidean space and whoever wrote this question (not you, the original author of the question) deserves to bite their tongue on their dinner tonight