A portion of the quadratic formula proof is shown. fill in the missing reason. ax^2+bx+c=0 —— Given
ax^2+bx=-c ——subtract c from both sides of the equation
x^2+b/a x=-c/a —— divide both sides of the equation by a
x^2+ b/a x+(b/2a)^2=-c/a+(b/2a)^2 —— complete the square and add (b/2a)^2 to both sides.
x^2+ b/a x+(b/2a)^2=-c/a+b^2/4a^2 —— square (b/2a) on the right side of the equation.
x^2+ b/a x+(b/2a)^2=-4ac/4a+b^2/4a^2—— find a common denominator on the right side of the equation.
x^2+ b/a x+(b/2a)^2=b^2-4ac/4a^2 —— ?

A-multiply the fractions together on the right side of the equation
B- subtract 4ac on the right side of the equation
C- add 4ac to both sides of the equation
D- add the fractions together on the right side of the equation

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  D- add the fractions together on the right side of the equation

Step-by-step explanation:

In the previous step, the fractions are written with a common denominator. In the last step, the two fractions are combined (added together).

The best description of those provided is ...

  D- add the fractions together on the right side of the equation

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

a

step-by-step explanation: