Answer:
1/1
Step-by-step explanation:
try that if that doesnt work try 1 the question was confusing im sorry if its wrong...
Answer: 5/4
Step-by-step explanation:
<span>ow far does the first car go in the 2 hours head start it gets?
Now, at t = 2 hours, both cars are moving. How much faster is the second car than the first car? How long will it take to recover the head start? You can determine this by dividing the head start by the difference in the two speeds. If car 1 has a 20 mile head start, and car 2 is 5 mph faster, then it will take 20/5 = 4 hours to catch up.
</span>You could also write two equations, one for each car, showing how far they have gone in a variable amount of time. Set the two equations equal to each other and solve for the value of the time. Note that the second car's equation will use (t-2) for the time, because it doesn't start driving until t = 2.
Answer:
X = 1 , -2
Step-by-step explanation:
Given in question as,
(log4(x² + x))² = 0.25
log4(x² + x) = 0.5
Applying log base property if log(a)X = b, then X =a^b
Or, x² + x = 4^0.5
x² + x = 2
x² + x - 2 = 0
x²+2x-x-2 = 0
(x + 2) (x - 1) = 0
So, x = 1 and -2 Answer
Answer:
1/2
Step-by-step explanation:
