To check if distances can be a triangle add the smaller two distances together. If the sum is larger (not equal to) than the longest distance then it is a possible side. For example, 2,2,4 or 1,2,5cannot be a triangle, 2,3, 4 can be a triangle because 2+3 is greater than 4.
Answer:
e
Step-by-step explanation:
e
(-1, 5) because if you plug in -1 to both equations, you should get 5 from both equations. They also cross at that point, making that the solution.
Answer:
6 x 4 x 4 Hope I get u right<3
Step-by-step explanation:
Answer:
y = -7x
Explanation:
Take two points from table: (-4, 28), (-3, 21)
Find slope:
![\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points](https://tex.z-dn.net/?f=%5Csf%20slope%3A%20%5Cdfrac%7By_2%20-%20y_1%7D%7Bx_2-%20x_1%7D%20%5C%20%5C%20%20where%20%5C%20%28x_1%20%2C%20%5C%20y_1%29%2C%20%28%20x_2%20%2C%20%5C%20y_2%29%20%5C%20are%20%5C%20points)
![\rightarrow \sf slope\ (m): \dfrac{21-28}{-3-(-4)} = -7](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20slope%5C%20%28m%29%3A%20%20%20%5Cdfrac%7B21-28%7D%7B-3-%28-4%29%7D%20%20%3D%20-7)
Now find equation:
![\sf y - y_1 = m (x - x_1)](https://tex.z-dn.net/?f=%5Csf%20y%20-%20y_1%20%3D%20m%20%28x%20-%20x_1%29)
![\rightarrow \sf y - 21 = -7(x - (-3))](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20y%20-%2021%20%3D%20-7%28x%20-%20%28-3%29%29)
![\rightarrow \sf y = -7(x +3) + 21](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20y%20%20%3D%20-7%28x%20%2B3%29%20%2B%2021)
![\rightarrow \sf y = -7x -21 + 21](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20y%20%20%3D%20-7x%20-21%20%2B%2021)
![\rightarrow \sf y = -7x](https://tex.z-dn.net/?f=%5Crightarrow%20%5Csf%20y%20%20%3D%20-7x)