Lol
trouble varies directly as distance
lets say t=trouble and d=distance
t=kd
k is constant
given
when t=20, and d=400
find k
20=400k
divide by 400 both sides
1/20=k
t=(1/20)d
given, d=60
find t
t=(1/20)60
t=60/20
t=3
3 troubles
Answer:
12.33
Step-by-step explanation:
6x - 2 + 9x - 3 = 180° (linear pair)
6x + 9x - 2 - 3 = 180°
15x - 5 = 180°
15x = 180 + 5
15x = 185
x = 185/15
x = 12.33
hope this helps you!
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:
C≈59.19
Step-by-step explanation:
I hope this helps! ^^
☁️☁️☁️☁️
If you plot the coordinates on a piece of graph paper your answer would come out too (1,-8)
