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Sindrei [870]
2 years ago
11

What are the x-coordinates for the maximum points in the function f(x) = 4 cos(2x − π) from x = 0 to x = 2π? (1 point) a x = 3 p

i over 2 , x = π b x =
Mathematics
1 answer:
Ratling [72]2 years ago
4 0

x-coordinates for the maximum points in any  function  f(x) by f'(x) =0 would be  x = π/2 and x= 3π/2.

<h3>How to obtain the maximum value of a function?</h3>

To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.

we want to find x-coordinates for the maximum points in any  function  f(x) by f'(x) =0

Given f(x)= 4cos(2x -π)

f'(x) = 0\\- 4sin(2x -\pi ) =0\\\\sin (2x -\pi ) =0 \\2x -\pi  = k\pi ...  k in Z

In general  x=(k+1)\pi /2

from x = 0 to x = 2π :

when k =0  then  x = π/2

when k =1  then x= π

when k =2  then x= 3π/2

when k =3  then x=2π

Thus, X-coordinates of maximum points are  x = π/2 and x= 3π/2

Learn more about maximum of a function here:

brainly.com/question/13333267

#SPJ4

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At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a
morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = \dfrac{15}{120}=0.125

As we know the poisson process, we get that

P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda

So, for exactly one car would be

P(n=1) is given by

=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

0.2599\times 18=4.6798

We will find the traffic flow q such that

P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%

Hence, a) 4.6798, and b) 19.8%.

7 0
4 years ago
The engine of a car has a displacement of 305 cubic inches. what is the displacement in cubic feet? round your answer to 2 decim
ipn [44]
 solution
 1 inch = 2.54 centimeters
Therefore, to convert to cubic measurements
 (1 in)³ = (2.54 cm)³
 Thus;
1 in³ = 16.387 cm³
Hence 305 cubic inches will be equivalent to
  305 × 16.387
 = 4998.035 cm³

5 0
4 years ago
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Evaluate a + b for a = 12 and b = 6.
den301095 [7]

Answer:

Here,

a= 12

b = 6

Then,

a+b

= 12 + 6

= 18

.°. 18 is the solution

3 0
3 years ago
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I just need #21 and #23 answered please.
kykrilka [37]
#21: Even
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7 0
3 years ago
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Kate can mow lawns at a constant rate of 4/5 lawns per hour. How many lawns can Kate mow in 20 hours?
weqwewe [10]

Answer:

16 lawns

Step-by-step explanation:

set up a proportion:

(4/5 ÷ 1) = (x÷20)

cross-multiply:

5x = 80

x = 16

8 0
3 years ago
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