What is the greatest common factor of 35 56 and 63?

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

How many numbers of solutions are there for the following equation<br> 2(4x+1) −5x +2 = 3x+4

soldier1979 [14.2K]

Answer:

infinite solutions

Step-by-step explanation:

2(4x+1) −5x +2 = 3x+4

Distribute

8x +2 -5x +2 =3x+4

Combine like terms

3x +4 = 3x+4

Subtract 3x from each side

3x-3x+4 = 3x-3x+4

4=4

This is always true, so there are infinite solutions

8 0

2 years ago

SOMEONE HELP!! 30 points!

Nostrana [21]

Answer:

AB = square root of 403

Step-by-step explanation:

use the pythagorean theorem to solve for AB.

a^2 + b^2 = c^2

9^2 + b^2 = 22^2

81 + b^2 = 484

b^2 = 484 - 81

b^2 = 403

b = 20.07

6 0

3 years ago

\sqrt{-x+26}=x-6 −x+26=x−6

schepotkina [342]

Answer:

x = 1 and 10

Step-by-step explanation:

Given the equation \sqrt{-x+26}=x-6, we are to find the value of x

Take the square of both sides

(\sqrt{-x+26})²= (x-6)²

-x+26 = (x-6)²

Expand the parenthesis

-x+26 = x²-12x+36

x²-12x+36+x-26 = 0

x²-11x+10 = 0

x²-10x-x +10 = 0

x(x-10) -1()-10) = 0

(x-1)(x-10) = 0

x - 1 =0 and x - 10 = 0

x = 1 and 10

Hence the value of x is 1 and 10

7 0

2 years ago

PLEASE HELP WILL GIVE EXTRA POINTS

Eddi Din [679]

Answer:

Part A

The p-value indicates that the largest probability of having the two remedies having a difference as observed or larger than as observed being equal to 0.3822

Part B:

Given that the p-value of 0.3822 is larger than the significance level α of 0.05, we fail to reject the null hypothesis and we conclude that there is enough statistical evidence to suggest that the salvage rates are the same for Remedy A and B

Step-by-step explanation:

The p-value

The test statistic is given as follows;

$Z=\dfrac{\hat{p}_1-\hat{p}_2}{\sqrt{\hat{p}(1-\hat{p})\left (\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}} \right )}}$

$\hat p_1$ = 35/220 = 7/44

$\hat p_2$ = 39/230

$\hat p$ = 74/220 = 37/110

We get our test statistic z = -0.2996381

Part A

The p-value indicates that the largest probability of having the two remedies having a difference as observed or larger than as observed being equal to 0.3822

Part B:

Given that the p-value of 0.3822 is larger than the significance level α of 0.05, we fail to reject the null hypothesis and we conclude that there is enough statistical evidence to suggest that the salvage rates are the same for Remedy A and B.

5 0

3 years ago