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 (depending if the fraction is positive or negative):
- x ≥ 18
- x ≤ -18
Step-by-step explanation:
If the fraction is positive:
- Write it out:
- Multiply each side by 6 to cancel out the 6 under x. It should now look like this: x ≥ 18
If the fraction is negative:
- Write it out:
- Multiply each side by -6 to cancel out the -6 under x. It should now look like this: x ≤ -18
I hope this helps!
Answer:
Maybe because the situation changed
Step-by-step explanation:
Answer:
<h2>a) center (2, -3)</h2><h2>b) radius r = 3</h2><h2>c) in the attachment</h2>
Step-by-step explanation:
The standard form of an equation of a circle:
(h, k) - center
r - radius
We have:
a) center (2, -3)
b) radius r = 3
c) in the attachment