Answer:
This would be the last option I think, sorry if it's wrong.
Step-by-step explanation:
Which of the following smallest values
Answer:
-3+2sqrt7
-3-2sqrt7
Step-by-step explanation:
x^2+6x-9=10
x^2+6x-9-10=0
x^2+6x-19=0
ax^2+bx+c=0
a=1 b=6 c=-19
As cannot be solved by completing square we will use quadratic equation
x= (-b+sqrt(b^2-4ac))/2a and x= (-b-sqrt(b^2-4ac))/2a
x= (-6+sqrt(6^2-4*-19))/2 and x= (-6-sqrt(6^2-4*-19))/2
x=(-6+sqrt(36+76))/2 and x=(-6-sqrt(36+76))/2
x=(-6+4sqrt7)/2 and x=(-6-4sqrt7)/2
x=(-3+2sqrt7) and x=(-3-2sqrt7)
x=2.29 and x= -8.29
-π/2 < arctan(x) < π/2
So cos(π/2) < cos(arctan(x)) < cos(0)
0 < cos(arctan(x)) < 1
-16w-26+9w=-61
-16w+9w=-61+26
-7w=-35
w=5
