3(r-6)+2=4(r+2)-21
3r-18+2=4r+8-21
3r-16=4r-13
-3=r
r=-3
46.78
Step-by-step explanation:
you divide 16 by the cos of 70 and it turns out to be 46.78
In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
We have that discriminant = 3
If
b^2 - 4ac > 0, then the roots are real.
If
b^2 - 4ac < 0 then the roots are imaginary
<span>In
this problem b^2 - 4ac > 0 3 > 0 </span>
then
the two roots must be real
Answer:
school is a problem
Step-by-step explanation:
Have a great day!!!
