$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - 5 + - \sqrt{ {5}^{2} - 4(5) } }{2} \\ x = \frac{ - 5 + - \sqrt{25 - 20} }{2} \\ x = \frac{ - 5 + - \sqrt{5} }{2}$

Therefore, the zeroes would be:

$x1 = \frac{ - 5 + \sqrt{5} }{2} \\ x2 = \frac{ - 5 - \sqrt{5} }{2}$

8 0

3 years ago

Solve for e.<br> I need help I have to hurry and turn this in!!

yarga [219]

1st option.

T9=e^2

e=sq root of T9

9 can be sq rooted to 3

hence e=3×sqroot of T

7 0

3 years ago

What is the domain of the following function?

SashulF [63]

Answer:

third option (-3,3,5,9)

Step-by-step explanation:

domain is the starting point of the set (x coordinate )

6 0

3 years ago

Need math help please!!

Eva8 [605]

Answer:

Step-by-step explanation:

The fact that DE is perpendicular to AC means triangles AED and CED are congruent and side AD is equal in length to side CD.

5x -20 = 2x +4

3x = 24 . . . . . . . add 20-2x

x = 8 . . . . . . . . . .divide by 3

The value of x is 8.

8 0

3 years ago