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3 years ago

Find the missing number ____:7 = 12:21 A) 3 B) 14 C) 4 D) 12

2 answers:

6 0

I hope this helps and i hope wecan be great friends!!!!! and corrrect me if wrong but i think its A!!!!!!!!

~ 12 year old kakashi hatake

Maurinko [17]3 years ago

3 0

4:7
3×4:7×3
12:21

The answer is <span>C) 4</span>

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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)

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

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}}$

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3 years ago

What is the product ?

xeze [42]

Answer:

Step-by-step explanation:

2*5 3* 5 4*5 = 10 15 20

9*5 -1*5 -7*5 45 -5 -35

11*5 5*5 -3*5 55 25 -15

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2 years ago

What is the solution to the system of equations that is graphed?

konstantin123 [22]

Answer:

(-4,-3)

Step-by-step explanation:

Find the point where the 2 lines intersect. In this problem it is (-4,-3)

x=-4

y=-3

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2 years ago

jamie has $37 in his bank account on sunday. the table shows his account activity for the next four days. what was the balance i

alexgriva [62]

You need to show a picture or type the table

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3 years ago

Two students use different methods to solve this multiplication problem: 1/2 • -4 4/5

sesenic [268]

Solution:

$\frac{1}{2}\times-4 \frac{4}{5}$

As one of the fraction is a proper fraction and another one is Mixed fraction.

There are two methods of solving it.

1. $\frac{1}{2}\times-4 \frac{4}{5}=\frac{1}{2} \times \frac{-24}{5}=\frac{-12}{5}=-2\frac{2}{5}$

2. $\frac{1}{2}\times-4 \frac{4}{5}=\frac{1}{2}[-4-\frac{4}{5}]=\frac{1}{2}\times (-4)+\frac{1}{2}\times\frac{-4}{5}=-2+\frac{-2}{5}=\frac{-10-2}{5}=\frac{-12}{5}=-2\frac{2}{5}$ →→Here i have used Distributive property with respect to addition and Subtraction i.e a×(b+c)= a ×b + a×c or a×(b-c)=a×b-a×c

Now, you can fill the blanks by yourself.

5 0

3 years ago