Please help I dont know the answer

jonny [76]

A zero in front of the x would make this a difference of two perfect squares. All of the other option would interfere by adding another term.

7 0

3 years ago

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Find The Volume And Surface Area!

KATRIN_1 [288]

Answer:

I'm sorry the pic isn't loading for meee but I wish u all the best

8 0

2 years ago

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

Urgent answer please <br> Ax+ay+xa+ya

marin [14]

Answer:

Possible derivation:

d/dx(a x + a y(x) + x a + y(x) a)

Rewrite the expression: a x + a y(x) + x a + y(x) a = 2 a x + 2 a y(x):

= d/dx(2 a x + 2 a y(x))

Differentiate the sum term by term and factor out constants:

= 2 a (d/dx(x)) + 2 a (d/dx(y(x)))

The derivative of x is 1:

= 2 a (d/dx(y(x))) + 1 2 a

Using the chain rule, d/dx(y(x)) = (dy(u))/(du) (du)/(dx), where u = x and d/(du)(y(u)) = y'(u):

= 2 a + d/dx(x) y'(x) 2 a

The derivative of x is 1:

= 2 a + 1 2 a y'(x)

Simplify the expression:

= 2 a + 2 a y'(x)

Simplify the expression:

Answer: = 2 a

Step-by-step explanation:

8 0

3 years ago

The sum of two numbers is 56. One number is 3 times as large as the other. What are the numbers?

MissTica

42 and 14

explanation:

x+y = 56
x = 3y

Substitute for x in the first equation:
x + y = 56
x = 3y
3y + y = 56
Combine like terms:
4y = 56
Divide both sides by 4:
y = 14

So if y = 14, and x = 3y, then x = 42.

Now substitute for both x and y in both of the original equations to prove they are the correct values.

x + y = 56
x = 42
y = 4
42 + 14 = 56
56 = 56

x = 3y
x = 42
y = 14
42 = 3*14
42 = 42

So the two numbers, x and y, are 42 and 14.

8 0

2 years ago

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Please help I dont know the answer​

Please help I dont know the answer