2, 6, 8) → D
each element in the domain ( - 1, 3, 5) maps to exactly one value in the range (2, 6, 8)
Answer:
Step-by-step explanation:
From figure,
In triangle 
![\Rightarrow (a+b)^2=(a-b)^2+(O' D)^2\\\Rightarrow a^2+b^2+2ab-a^2-b^2+2ab=(O' D)^2\\\Rightarrow 4ab=(O' D)^2\\\Rightarrow O'D=2\sqrt{ab} \\\Rightarrow O' D=2\sqrt{ab}=AB \quad \quad [\because O' DAB\;\; \text{is a rectangle.}]](https://tex.z-dn.net/?f=%5CRightarrow%20%28a%2Bb%29%5E2%3D%28a-b%29%5E2%2B%28O%27%20D%29%5E2%5C%5C%5CRightarrow%20a%5E2%2Bb%5E2%2B2ab-a%5E2-b%5E2%2B2ab%3D%28O%27%20D%29%5E2%5C%5C%5CRightarrow%204ab%3D%28O%27%20D%29%5E2%5C%5C%5CRightarrow%20O%27D%3D2%5Csqrt%7Bab%7D%20%5C%5C%5CRightarrow%20O%27%20D%3D2%5Csqrt%7Bab%7D%3DAB%20%5Cquad%20%5Cquad%20%5B%5Cbecause%20O%27%20DAB%5C%3B%5C%3B%20%5Ctext%7Bis%20a%20rectangle.%7D%5D)
Hence,
Number 4 and 6 are correct because you can only add variables with other variables so that is why 4 and 6 are correct
Answer:
Step-by-step explanation:
- Well I hate to break the news but 243 is not a perfect square. I'll work you through it, 243 is not a perfect square because it is not an even number. an even number must end in (0,2,4,6,8)
- Step one. Find the square root. the square root of 243 is <em>15.588. </em>
- Step two Is it a perfect square. No 243 just cant be a perfect square.
- Hope this helped :)
