Find the LCM of 6 and 15. A) 6 B) 15 C) 30 D) 90
2 answers:
You might be interested in
Using a graph tool
case 1)
2x-y=-13
y=x+9
the solution is the point (-4,5)
see the attached figure N 1
case 2)
y=3x-7
y=2x-5
the solution is the point (2,-1)
see the attached figure N 2
case 3)
3x+2y=10
6x-y=10
the solution is the point (2,2)
see the attached figure N 3
case 4)
y=6
x=-5
the solution is the point (-5,6)
see the attached figure N 4
case 5)
4x-3y=5
3x+2y=-9
the solution is the point (-1,-3)
case 6)
x+y=7
x-y=-1
the solution is the point (3,4)
the answer in the attached figure
case 1)
2x-y=-13
y=x+9
the solution is the point (-4,5)
see the attached figure N 1
case 2)
y=3x-7
y=2x-5
the solution is the point (2,-1)
see the attached figure N 2
case 3)
3x+2y=10
6x-y=10
the solution is the point (2,2)
see the attached figure N 3
case 4)
y=6
x=-5
the solution is the point (-5,6)
see the attached figure N 4
case 5)
4x-3y=5
3x+2y=-9
the solution is the point (-1,-3)
case 6)
x+y=7
x-y=-1
the solution is the point (3,4)
the answer in the attached figure
Answer:
The horizontal distance the ball travels before it hits the ground is 22. feet
Step-by-step explanation:
The given parameter are;
The function modelling the path of the ball tossed by Nate y = -14·x² + 335·x
x = The horizontal distance the ball travels from Nate in feet
y = The height of the ball in feet
The line equation modelling the hill is y = 15·x
The point where the ball hits the ground is given by the point the graph of the equation for the path of the ball and the path of the model of the line of the hill meet as follows;
Ball path is y = -14·x² + 335·x
Hill path is y = 15·x
The point both paths meet and the ball hits the ground is -14·x² + 335·x = 15·x
Which gives;
-14·x² + 335·x - 15·x = 0
-14·x² + 320·x = 0
320·x - 14·x² = 0
x × (320 - 14·x) = 0
x = 0, or x = 320/14 = 22 6/7 = 22. feet
Therefore;
The horizontal distance the ball travels before it hits the ground = x = 22. feet.