Answer:
The rate at which the distance between the two cars is increasing is 30 mi/h
Step-by-step explanation:
Given;
speed of the first car, v₁ = 24 mi/h
speed of the second car, v₂ = 18 mi/h
Two hours later, the position of the cars is calculated as;
position of the first car, d₁ = 24 mi/h x 2 h = 48 mi
position of the second car, d₂ = 18 mi/h x 2 h = 36 mi
The displacement of the two car is calculated as;
displacement, d² = 48² + 36²
d² = 3600
d = √3600
d = 60 mi
The rate at which this displacement is changing = (60 mi) / (2h)
= 30 mi/h
Answer:
B
Step-by-step explanation:
5.98 * 10^24kg because the amount of zeros is 24 and the exponent will be 24 and you have the decimal point between the 5 and then 9
Answer:
= 50.6 miles per hour
Showing Work:
This is a fraction equal to
405 miles ÷ 8 hours
We want a unit rate where
1 is in the denominator,
so we divide top and bottom by 8
405 miles ÷ 8
8 hours ÷ 8
=
50.625 miles
1 hour
=
50.625 miles
hour
= 50.625 miles per hour
rounding to the nearest tenth:
50.6 miles per hour
Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.
An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.
Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.
Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.
Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:
Given that William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.
To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.
For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false
For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true
For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true
For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false
For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false
For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.
