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3 years ago

Let p(x)=90/9+50e^−x.

Mathematics

1 answer:

murzikaleks [220]3 years ago

6 0

p(x) = 90 / (9 + 50 * e^(-x))

e = 2.718, so we can use e = 2.7 as a quick substitute

p(3) =>

90 / (9 + 50 * e^(-3)) =>

90 / (9 + 50 * 2.7^(-3)) =>

90 / (9 + 50 / 2.7^3) =>

90 / ((9 * 2.7^3 + 50) / 2.7^3) =>

90 * 2.7^3 / (9 * 2.7^3 + 50)

2.7^3 =>

(27/10)^3 =>

27^3 / 1000 =>

729 * 27 / 1000 =>

2187 * 9 / 1000 =>

6561 * 3 / 1000 =>

19683 / 1000 =>

19.683

90 * 2.7^3 / (9 * 2.7^3 + 50) =>

90 * 19.683 / (9 * 19.683 + 50)

9 * 19.683 =>

9 * (20 - 0.317) =>

180 - 9 * 0.317 =>

180 - 9 * 0.3 - 9 * 0.017 =>

180 - 2.7 - 9 * 0.01 - 9 * 0.007 =>

177.3 - 0.09 - 0.063 =>

177.3 - 0.153 =>

177.15 - 0.003 =>

177.147

90 * 19.683 / (9 * 19.683 + 50) =>

10 * 177.147 / (177.147 + 50) =>

1771.47 / 227.147 =>

1771470 / 227147 =>

1770000 / 225000 =>

1770/225 =>

4 * 1770 / 900 =>

4 * 177 / 90 =>

708/90 =>

70.8 / 9 =>

23.6 / 3 =>

(21 + 2.6) / 3 =>

7 + 0.866666.... =>

7.8666666...

Actual value: 7.8333389811057471507345888320992.....

Not bad for some estimates, is it?

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The difference of the quotient m and 9 and the quotient of n and 30

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Answer: (30m-9n)/270

First we start with m/9-n/30.

We multiply the denominators to get 270.

Then we multiply the numerators by the original denominators to get (30m-9n).

To check, we can use m=2, and n=4

2/9-4/30=4/45

(30*2-9*4)/270=24/270=4/45

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3 years ago

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A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprink

deff fn [24]

Answer:

a) see below

b) 32 minutes after turn-on

c) 52 minutes after turn-on

d) 20 minutes

e) 6856.6 ft²

Step-by-step explanation:

a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).

The lines forming the sidewalk boundaries are y=0 and y=10.

b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...

(80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk

c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...

(130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk

Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).

d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.

e) In one hour, the sprinkler travels a distance of ...

(60 min)(5/2 ft/min) = 150 ft

Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...

A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²

The area of grass watered in 1 hour is about 6856.6 ft².

5 0

4 years ago