There seems to be a problem with your question - AEC does not create a triangle while ADE does.
As for finding x
Draw an imaginary line from C to D. 3^2+4^2=5^2, so take the square root to find CD=5. Now we know that CA is also 5. To find x, x^2+3^2=5^2, or x^2+9=25. Subtract 9 from both sides to get 16, and take the square root of both sides to get x=4.
Answer:
D. (5, 0) and (4, 0)
Step-by-step explanation:
we know that
The x-intercepts are the values of x when the value of the function is equal to zero
Looking at the graph
The x-intercepts are the points (4,0) and (5,0)
We have to determine, The tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each system of equations to its graph.
The groups of equation are as follows;
The y-intercept of equation = 1
The x-intercept of equation = -1/2
And The slope of equation = 2
Then,
At x = 0, y = 1
And at y = 0, x = -1/2
And at x = 2, y = 3
The first and second equation Corresponding with the third graph.
The y-intercept of equation = 0
The x-intercept of equation = 0
The slope of equation = 3
So at x = 0, y = 0 and at x = 2, y = 6
The equation Corresponds with the first graph
The y-intercept of equation = -2
The x-intercept of equation = 2
The slope of equation = 2
So at x = 0, y = -2 and at x = 2, y = 0
The equation Corresponds with the first graph.
The y-intercept of equation = 3
The x-intercept of equation = -2
The slope of equation = 2
So at x = 0, y = 3 and at x = -5, y = 0
The equation Corresponds with the fourth graph.
The y-intercept of equation = 2
The x-intercept of equation = -1/2
The slope of equation = 4
So at x = 0 y = 2 and at y = 0 x = -1/2 and at x = 0.5 y = 4
The equation Corresponding with the second graph.
