The dimensions for the rectangle are 15 by 2.
15 times 2 is 30, which is the area.
15+15+2+2= 34 which is the perimeter
Step-by-step explanation:
The graph of the function y=|x| is as shown in the attached diagram. The only possible translation of this graph right and up is option A. (In option B graph is translated right and down, in option C- left and up and in option D - left and down). Answer: correct choice is A.
Answer: I think it is -x/8+2y/3+10
Step-by-step explanation:
Answer:
![\text{Average rate}=\frac{\text{40 laps}}{\text{ hour}}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D%5Cfrac%7B%5Ctext%7B40%20laps%7D%7D%7B%5Ctext%7B%20hour%7D%7D)
Step-by-step explanation:
Please consider the complete question.
Sebastian swan laps every day in the community swimming pool. He swam 45 minutes each day, 5 days each week, for 12 weeks. In that time, he swam 1800 laps. What was his average rate in laps per hour?
Let us convert time taken into hours as:
![\text{Days}=5\times 12=60](https://tex.z-dn.net/?f=%5Ctext%7BDays%7D%3D5%5Ctimes%2012%3D60)
![\text{Minutes}=60\times 45](https://tex.z-dn.net/?f=%5Ctext%7BMinutes%7D%3D60%5Ctimes%2045)
1 hour = 60 minutes. To convert minutes into hours, we will divide total minutes by 60.
![\text{Hours}=\frac{60\times 45}{60}=45](https://tex.z-dn.net/?f=%5Ctext%7BHours%7D%3D%5Cfrac%7B60%5Ctimes%2045%7D%7B60%7D%3D45)
Now, we will divide 1800 by 45 to find average rate in laps per hour as:
![\text{Average rate}=\frac{\text{1800 laps}}{\text{45 hours}}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D%5Cfrac%7B%5Ctext%7B1800%20laps%7D%7D%7B%5Ctext%7B45%20hours%7D%7D)
![\text{Average rate}=\frac{\text{40 laps}}{\text{ hour}}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%7D%3D%5Cfrac%7B%5Ctext%7B40%20laps%7D%7D%7B%5Ctext%7B%20hour%7D%7D)
Therefore, Sebastian's average rate is 40 laps per hour.
Answer:
The second time when Luiza reaches a height of 1.2 m = 2 08 s
Step-by-step explanation:
Complete Question
Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.
H(t) = -0.6 cos (2pi/2.5)t + 1.5.
What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.
Solution
Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as
H(t) = -0.6cos(2π/2.5)t + 1.5
What is t when H = 1.2 m
1.2 = -0.6cos(2π/2.5)t + 1.5
0.6cos(2π/2.5)t = 1.2 - 1.5 = -0.3
Cos (2π/2.5)t = (0.3/0.6) = 0.5
Note that in radians,
Cos (π/3) = 0.5
This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,
Cos (5π/3) = 0.5
So,
Cos (2π/2.5)t = Cos (5π/3)
(2π/2.5)t = (5π/3)
(2/2.5) × t = (5/3)
t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.
Hope this Helps!!!