x= 3+√5/4,3-√5/4
Step-by-step explanation:
1.In general, given ax^2+bx+c=0, there exists two solutions where:
x= -b+√b^2 - 4ac/2a, -b-√b^2-4ac/2a
2.In this case, a=4, b=-6 and c=1.
x=6+√(-6)^2-4×4/2×4, 6-√(-6)^2-4×4/2×4
3.Simplify.
x=6+2√5/8, 6-√5/8
4.Simplify solutions.
x=3+√5/4, 3-√5/4
Answer:

Step-by-step explanation:
Solve for z:
-2(z + 3) = -z - 4(z + 2)
-4(z + 2) = -4z - 8:
-2(z + 3) = -z -4z - 8
Grouping like terms, -z - 4z - 8 = (-z - 4z) - 8:
-2(z + 3) = (-z - 4z) - 8
-z - 4z = -5z:
-2 (z + 3) = -5z - 8
Expand out terms of the left hand side:
-2z - 6 = -5 z - 8
Add 5z to both sides:
(5z - 2z) - 6 = (5z - 5z) - 8
5z - 5z = 0:
(5z - 2z) - 6 = -8
5z - 2z = 3z:
3z - 6 = -8
Add 6 to both sides:
3z + (6 - 6) = 6 - 8
6 - 6 = 0:
3z = 6 - 8
6 - 8 = -2:
3z = -2
Divide both sides of 3z = -2 by 3:



Answer:
(A) 9
Step-by-step explanation:
Substitute 2 for X

Answer:
<h3>
<u>Step-by-step explanation:</u></h3>
A = ( -2, 1 )
B = ( 1, -1 )


______________________________
<u>________________</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>
I think it is the graph should be redrawn with the vertical axis beginning with $0.