The formula for the area of a rhombus is s^2=A. Plug in and solve:
s^2=25
<em>*Take the square root of both sides*</em>
s=5
Hope this helps!!
Answer:
are you from k-12? bc if so im also stuck on this question
Step-by-step explanation:
The right answer to this is (0,-5). If you substitute the points into the equation, it will give the answer 10.
![5x - 2y = 10 \\ let x = 0 \\ let y = -5 \\ \\ 5(0) - 2(-5) =10 \\ 10 = 10](https://tex.z-dn.net/?f=5x%20-%202y%20%3D%2010%20%5C%5C%20let%20x%20%3D%200%20%5C%5C%20let%20y%20%3D%20-5%20%5C%5C%20%20%5C%5C%205%280%29%20-%202%28-5%29%20%3D10%20%5C%5C%2010%20%3D%2010)
The points (0,-5) satisfy the equation because both sides of the equation is equal to 10.
Answer:
second choice
Step-by-step explanation:
![(f \circ g)(x) =](https://tex.z-dn.net/?f=%20%28f%20%5Ccirc%20g%29%28x%29%20%3D%20)
![= f(g(x))](https://tex.z-dn.net/?f=%20%3D%20f%28g%28x%29%29%20)
![= f(-7x + 7)](https://tex.z-dn.net/?f=%20%3D%20f%28-7x%20%2B%207%29%20)
![= \sqrt{-7x + 7 + 3}](https://tex.z-dn.net/?f=%20%3D%20%5Csqrt%7B-7x%20%2B%207%20%2B%203%7D%20)
![= \sqrt{-7x + 10}](https://tex.z-dn.net/?f=%20%3D%20%5Csqrt%7B-7x%20%2B%2010%7D%20)
Let x = 5:
![(f \circ g)(5) = \sqrt{-7(5) + 10}](https://tex.z-dn.net/?f=%20%28f%20%5Ccirc%20g%29%285%29%20%3D%20%5Csqrt%7B-7%285%29%20%2B%2010%7D%20)
![(f \circ g)(5) = \sqrt{-35 + 10}](https://tex.z-dn.net/?f=%20%28f%20%5Ccirc%20g%29%285%29%20%3D%20%5Csqrt%7B-35%20%2B%2010%7D%20)
![(f \circ g)(5) = \sqrt{-25}](https://tex.z-dn.net/?f=%20%28f%20%5Ccirc%20g%29%285%29%20%3D%20%5Csqrt%7B-25%7D%20)
Since using x = 5 gives the square root of a negative number, 5 is not in the domain of the composition of the functions.
Answer: second choice
Wherever the line hits on the horizontal line on a graph.