Amara wants to draw isosceles right triangles on a coordinate plane that have two vertices at (-2, 4) and (5, 4). If the coordinates of the third vertex are both integers, how many isosceles right triangles can she draw?

4  isosceles right triangles

Step-by-step explanation:

As the side joining (-2 , 4) and (5 , 4) is horizontal (same y-coordinates)

Its length = 5 - (-2) = 7

∴ The side from the one of the two given vertices to the third vertex must be vertical so it's parallel to the y-axis

The x-coordinate can be -2 or 5

7 is the distance between the y-coordinates of the third vertex and one of the given point

Either: 4 - y = 7 ⇒ y = 4 - 7 = -3

OR: y - 4 = 7 ⇒ y = 4 + 7 = 11

The y-coordinate of the third vertex can be -3 or 11

Therefore, the third vertex is (-2 , -3) , (-2 , 11) , (5 , -3) , (5 , 11)

 Amara can draw 4 isosceles right triangles

4  isosceles right triangles

Step-by-step explanation:

As the side joining (-2 , 4) and (5 , 4) is horizontal (same y-coordinates)

Its length = 5 - (-2) = 7

∴ The side from the one of the two given vertices to the third vertex must be vertical so it's parallel to the y-axis

The x-coordinate can be -2 or 5

7 is the distance between the y-coordinates of the third vertex and one of the given point

Either: 4 - y = 7 ⇒ y = 4 - 7 = -3

OR: y - 4 = 7 ⇒ y = 4 + 7 = 11

The y-coordinate of the third vertex can be -3 or 11

Therefore, the third vertex is (-2 , -3) , (-2 , 11) , (5 , -3) , (5 , 11)

 Amara can draw 4 isosceles right triangles

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer: options

step-by-step explanation: