Answer:
12 = f(40)
Step-by-step explanation:
From the given information:
We are being told that:
G = tons per week
p = people in thousands
However, the relation existing between the amount of garbage that is produced by a city with population p can be expressed as:
G = f(p)
Similarly, they said there exists a total population of 40000 persons in the town of Tola. i.e. 40 thousand and also 12 tons of garbage is produced by week i.e. that will be the value for G.
Then, we have:
12 = f(40)
The Q30X program and the Intensity program are both effective because the average participant 1-mile run time was reduced for each program, but the Intensity program is the more effective program because it yielded a greater decrease in the average participant 1-mile run time.
Answer:
Step-by-step explanation:
Given the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Divide through by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The range of the solution is
0<θ<2π I.e 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n =5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 is out of range of θ
Then, the solution is from n =0 to n=9
So the equation have 10 solutions in the range 0<θ<2π
When putting data into a class for this case, 0.350 - 0.359, they should be within the class boundaries. The class boundaries are 0.3495 and 0.3595. So, the data that will go into that class are 0.356 and 0.358. The record 0.349 is not included since it is below 0.3495. Therefore, there are only two values in the class.
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We will use u-substitute:
Then for substitution:dx=u du. and integral becomes:
=u-ln(u+1)=
. Now we will change the values of limits:
=4-ln(5)-2+ln(3)=
2+ln(0,6)=2-0.51=1.49
