Answer:
Step-by-step explanation:
|4x + 3| = 9 + 2x
We see the absolute value so we have to consider 2 cases:
(1)
4x + 3 = -9 - 2x
6x = -12
x = -2
(2)
4x + 3 = 9 + 2x
2x = 6
x = 3
Let's plug in both results.
|4(-2) + 3| = 9 + 2(-2)
|-5| = 9 - 4
5 = 5
Therefore x = -2 is a solution
|4(3) + 3| = 9 + 2(3)
|15| = 9 + 6
15 = 15
Therefore x = 3 is also a solution.
Solution: The commutative property of addition states that when two numbers are being added, their order can be changed without affecting the sum.
The circumference is (2 pi) (radius) . The ratio of radius to circumference is (1)/(2 pi).
Answer:
(E) 0.71
Step-by-step explanation:
Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.
So, the probability that the student has a GPA lower than 3.5 and is not taking any AP classes is calculated as:
P(A'∩B') = 1 - P(A∪B)
it means that the students that have a GPA lower than 3.5 and are not taking any AP classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.
Therefore, P(A∪B) is equal to:
P(A∪B) = P(A) + P(B) - P(A∩B)
Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled in at least one AP class is 0.12
So, P(A∪B) is equal to:
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = 0.25 + 0.16 - 0.12
P(A∪B) = 0.29
Finally, P(A'∩B') is equal to:
P(A'∩B') = 1 - P(A∪B)
P(A'∩B') = 1 - 0.29
P(A'∩B') = 0.71
Let x be the distance (in feet) along the road that the car has traveled and h be the distance (in feet) between the car
and the observer.
(a) Before the car passes the observer, we have dh/dt < 0; after it passes, we have dh/dt > 0. So at the instant it passes the observer we have
dh/dt = 0, given that dh/dt varies continuously since the car travels at a constant velocity.
