51) 5x-1=0 and x+6=0
5x=1, x=1/5 and x=-6
2 solutions x=1/5 and x=-6
52)x(2x+3)=0
2x=0 and 2x+3 =0
x=0 and x=-3/2=-1.5
53) x(5x-1)=0
x=0 and 5x-1=0
x=0 and x=1/5
54) z(4z+7) =0
z=0 and 4z+7=0
z=0 and z= - 7/4 = -1.75
z=0 and z=-1.75
If you would like to find a step that can be used to find the solution to the set of equations, you can do this using the following steps:
c = 2d + 1
c = 3d + 5
__________
c = c
2d + 1 = 3d + 5
2d - 3d = 5 - 1
- d = 4
d = -4
The correct result would be <span>2d + 1 = 3d + 5.</span>
Use Pythagorean Theorem:
a² + b² = c²
"a" is one side of the right triangle.
Its length is: radius of larger circle minus radius of smaller circle: (11 - 3 = 8)
The distance between the center of the circles creates the hypotenuse (17)
8² + b² = 17²
b² = 225
b = 15
Answer: the length of the common external tangent is 15.
The range is the highest number minus the lowest number. So in this case it would be:
16 - 5 = 11
Hope this helps! :)
