Answer:
a = 1/2 (1 ±sqrt(47))
Step-by-step explanation:
a^2-a+12=0
We will complete the square
Subtract 12 from each side
a^2-a+12-12=0-12
a^2-a=-12
The coefficient of a = -1
-Divide by 2 and then square it
(-1/2) ^2 = 1/4
Add it to each side
a^2 -a +1/4=-12 +1/4
(a-1/2)^2 = -11 3/4
(a-1/2)^2= -47/4
Take the square root of each side
sqrt((a-1/2)^2) =sqrt(-47/4)
a-1/2 = ±i sqrt(1/4) sqrt(47)
a-1/2= ±i/2 sqrt(47)
Add 1/2 to each side
a-1/2+1/2 = 1/2± i/2 sqrt(47)
a = 1/2± i/2 sqrt(47)
a = 1/2 (1 ±sqrt(47))
Answer:
-8 • (6 + 3) + 390 + 6=324
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Answer:

Note: to write the domain in interval notation, you'd write [-4,5]
if you need the domain in set-builder notation, then you'd write

------------------------------------------------------------------------------
Explanation:
The domain is the set of possible x input values. Look at the left most point (-4,-1). The x coordinate here is x = -4. This is the smallest x value allowed. The largest x value allowed is x = 5 for similar reasons, but on the other side of the graph.
So that's how I got

(x is between -4 and 5; inclusive of both endpoints)
Writing [-4,5] for interval notation tells us that we have an interval from -4 to 5 and we include both endpoints. The square brackets mean "include endpoint"
Writing

is the set-builder notation way of expressing the domain. The

portion means "x is a real number"
By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)
<h3>How many more distance does the average race car travels than the average consumer car?</h3>
In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:
s' = (v' - v) · t
Where:
- v' - Speed of the average race car, in miles per hour.
- v - Speed of the average consumer car, in miles per hour.
- t - Time, in hours.
Please notice that a hour equal 3600 seconds. If we know that v' = 210 mi / h, v = 120 mi / h and t = 30 / 3600 h, then the distance surplus of the average race car is:
s' = (210 - 120) · (30 / 3600)
s' = 3 / 4 mi
The distance surplus of the average race car is equal to 3 / 4 miles.
To learn more on uniform rectilinear motion: brainly.com/question/10153269
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