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yaroslaw [1]
3 years ago
11

Sketch the situation if necessary and use related rates to solve. Two airplanes are flying in the air at the same height. Airpla

ne A is flying east at 250 mph and airplane B is flying north at 300 mph. If they are both heading to the same airport, located 30 miles east of airplane A and 40 miles north of airplane B, at what rate (in mph) is the distance between the airplanes changing?
Mathematics
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

  -390 mph

Step-by-step explanation:

Let a and b represent, respectively, the distances of A and B from the airport. The distance d between the planes is then given by the Pythagorean theorem as ...

  d² = a² + b²

Differentiating with respect to time, we have ...

  2d·d' = 2a·a' +2b·b'

Solving for d', we get ...

  d' = (a/d)a' +(b/d)b'

The value of d at the time of interest is ...

  d = √(a² +b²) = √(30² +40²) = √2500 = 50

Then the rate of change of separation is ...

  d' = (30/50)(-250 mph) +(40/50)(-300 mph) = (-150 -240) mph

  d' = -390 mph

The distance between planes is decreasing at 390 miles per hour.

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Answer:

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Step-by-step explanation:

Given the expression

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Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

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M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

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8 0
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Solve:<br> 2/3x+15=17<br><br> 3x+8-x=7<br><br> 4(2x-6)=2
Volgvan

Answer:

x = 3

x = (-1)/2

x = 13/4

Step-by-step explanation:

Solve for x:

(2 x)/3 + 15 = 17

Put each term in (2 x)/3 + 15 over the common denominator 3: (2 x)/3 + 15 = (2 x)/3 + 45/3:

(2 x)/3 + 45/3 = 17

(2 x)/3 + 45/3 = (2 x + 45)/3:

1/3 (2 x + 45) = 17

Multiply both sides of (2 x + 45)/3 = 17 by 3:

(3 (2 x + 45))/3 = 3×17

(3 (2 x + 45))/3 = 3/3×(2 x + 45) = 2 x + 45:

2 x + 45 = 3×17

3×17 = 51:

2 x + 45 = 51

Subtract 45 from both sides:

2 x + (45 - 45) = 51 - 45

45 - 45 = 0:

2 x = 51 - 45

51 - 45 = 6:

2 x = 6

Divide both sides of 2 x = 6 by 2:

(2 x)/2 = 6/2

2/2 = 1:

x = 6/2

The gcd of 6 and 2 is 2, so 6/2 = (2×3)/(2×1) = 2/2×3 = 3:

Answer:  x = 3

______________________________________________________

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3 x - x + 8 = 7

Grouping like terms, 3 x - x + 8 = (3 x - x) + 8:

(3 x - x) + 8 = 7

3 x - x = 2 x:

2 x + 8 = 7

Subtract 8 from both sides:

2 x + (8 - 8) = 7 - 8

8 - 8 = 0:

2 x = 7 - 8

7 - 8 = -1:

2 x = -1

Divide both sides of 2 x = -1 by 2:

(2 x)/2 = (-1)/2

2/2 = 1:

Answer: x = (-1)/2

_______________________________________

Solve for x:

4 (2 x - 6) = 2

Divide both sides of 4 (2 x - 6) = 2 by 4:

(4 (2 x - 6))/4 = 2/4

4/4 = 1:

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The gcd of 2 and 4 is 2, so 2/4 = (2×1)/(2×2) = 2/2×1/2 = 1/2:

2 x - 6 = 1/2

Add 6 to both sides:

2 x + (6 - 6) = 1/2 + 6

6 - 6 = 0:

2 x = 1/2 + 6

Put 1/2 + 6 over the common denominator 2. 1/2 + 6 = 1/2 + (2×6)/2:

2 x = 1/2 + (2×6)/2

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2 x = 1/2 + 12/2

1/2 + 12/2 = (1 + 12)/2:

2 x = (1 + 12)/2

1 + 12 = 13:

2 x = 13/2

Divide both sides by 2:

x = (13/2)/2

2×2 = 4:

Answer: x = 13/4

8 0
3 years ago
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