Answer: The gas would not have remained liquid
Step-by-step explanation:
Since the freezing point of gasoline starts at -40°F, and the temperature that day was -49°F, we can conclude that the gasoline would not have remained liquid
Answer:
x = 12
Step-by-step explanation:
Solve for x:
(-3 x)/2 - 9 = -27
Put each term in (-3 x)/2 - 9 over the common denominator 2: (-3 x)/2 - 9 = (-18)/2 - (3 x)/2:
(-18)/2 - (3 x)/2 = -27
(-18)/2 - (3 x)/2 = (-3 x - 18)/2:
(-3 x - 18)/2 = -27
Multiply both sides of (-3 x - 18)/2 = -27 by 2:
(2 (-3 x - 18))/2 = -27×2
(2 (-3 x - 18))/2 = 2/2×(-3 x - 18) = -3 x - 18:
-3 x - 18 = -27×2
2 (-27) = -54:
-3 x - 18 = -54
Add 18 to both sides:
(18 - 18) - 3 x = 18 - 54
18 - 18 = 0:
-3 x = 18 - 54
18 - 54 = -36:
-3 x = -36
Divide both sides of -3 x = -36 by -3:
(-3 x)/(-3) = (-36)/(-3)
(-3)/(-3) = 1:
x = (-36)/(-3)
The gcd of -36 and -3 is -3, so (-36)/(-3) = (-3×12)/(-3×1) = (-3)/(-3)×12 = 12:
Answer: x = 12
Answer:
x = -4 and x = 6
Step-by-step explanation:
Factor using AC method.
1 × -24 = -24
Factors of -24 that add up to -2 are +4 and -6.
P(x) = (x + 4) (x − 6)
Therefore, zeroes of P(x) are x = -4 and 6.
Answer:
a) Each individual act will last 5 minutes.
b) Each group act will last 10 minutes.
Step-by-step explanation:
I think this is right
The percentage increase from 5 laps of race on the first day to 9 laps after several weeks is 80%
<h3>How to determine the percentage increase?</h3>
The distance ran are given as:
Initial = 5 laps
Final= 9 laps
The percentage increase is calculated using:
Percentage = (Final - Initial)/Initial
So, we have:
Percentage = (9 - 5)/5
Evaluate
Percentage = 0.8
Express as percentage
Percentage = 80%
Hence, the percentage increase is 80%
Read more about percentage increase at:
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