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blagie [28]
2 years ago
9

Find the length of a segment in the coordinate plane with endpoints (-1 , 3) and (4 , -5).

Mathematics
2 answers:
user100 [1]2 years ago
4 0
5 , 8 that should be the answer
olya-2409 [2.1K]2 years ago
3 0

Answer:

5 , 8

Step-by-step explanation:

This is the length

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A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo
SpyIntel [72]

Answer:

R_{in}=0.2\dfrac{mL}{min}

C(t)=\dfrac{A(t)}{30000}

R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

A(t)=300+2700e^{-\dfrac{t}{1500}},$  A(0)=3000

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

R_{in}=(concentration of chlorine in inflow)(input rate of the water)

=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}

(d) Rate at which the chlorine is leaving the pool

R_{out}=(concentration of chlorine in outflow)(output rate of the water)

= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}

Taking the integral of both sides

\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}

Recall that when t=0, A(t)=3000 (our initial condition)

3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}

3 0
3 years ago
What is the product3(x)^5(2x^2+4x+1)
ryzh [129]

Answer: 6x^7+ 12x^6+ 3x^5

Hope that helped

4 0
3 years ago
A carpenter is building a rectangular shed with a fixed perimeter of 54 ft. What are the dimensions of the largest shed that can
lutik1710 [3]

Answer:

the Largest shed dimension is 13.5 ft by 13.5 ft

Largest Area is 182.25 ft²

Step-by-step explanation:

Given that;

Perimeter = 54 ft

P = 2( L + B ) = 54ft

L + B = 54/2

L + B = 27 ft

B = 27 - L ------------Let this be equation 1

Area A = L × B

from equ 1, B = 27 - L

Area A = L × ( 27 - L)

A = 27L - L²

for Maxima or Minima

dA/dL = 0

27 - 2L = 0

27 = 2L

L = 13.5 ft

Now, d²A/dL² = -2 < 0

That is, area is maximum at L = 13.5 using second derivative test

B = 27 - L

we substitute vale of L

B = 27 - 13.5 = 13.5 ft

Therefore the Largest shed dimension = 13.5 ft by 13.5 ft

Largest Area = 13.5 × 13.5 = 182.25 ft²

7 0
2 years ago
With a rectangle length of 309 cm.and width of 249cmdoes it have the perimeter of 1500 cm?
horsena [70]
1116 is the perimeter it has. The formula for perimeter is 2l+2w=p
so when you plug the numbers in you get 2(309)+2(249)=1116. So it does not have a perimeter of 1500 cm.
7 0
3 years ago
Consider this population data set: 4,6,7, 11, 12, 18, 26, 23, 14, 31, 22, and 12. The values 11, 31, 22, and 12 constitute a ran
Andrej [43]

Answer:

The population mean is the sum of the 12 data set values divided by 12:

4 + 6 + 7 + 11 + 12 + 18 + 26 + 23 + 14 + 31 + 22 + 12 = 186.

Therefore the population mean = 186/12 = 15.5.

The sample mean is the sum of the 4 sample values divided by 4:

11 + 31 + 22 + 12 = 76.

Therefore the sample mean = 76/4 = 19.

The sample mean is more than the population mean by 19 - 15.5 = 3.5.

hope it helped

Step-by-step explanation:

6 0
3 years ago
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