What is the sum of (8a+2b-4) and (3b-5)

Find the perimeter of the following shape: Shape ABCD is shown. Point A is at 7, 5. Point B is at 6, 3. Point C is at 3, 2. Poin

Art [367]

I think it A im not sure

8 0

3 years ago

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Y = 2x x + y = −6 the answer should be like this (x,y) or ( , )

melisa1 [442]

Answer:

(-2, -4)

Step-by-step explanation:

Given:

The system of equations to solve is given as:

$y=2x\\\\x+y=-6$

In order to solve this system of linear equations, we use substitution method.

In substitution method, we substitute the value of any one variable in the other equation.

So, we substitute the value of 'y' from first equation in the second equation.

This gives,

$x+2x=-6\\\\3x=-6$

Dividing both sides by 3, we get:

$\frac{3x}{3}=\frac{-6}{3}\\\\x=-2$

Now, $y=2x=2\times -2=-4$

Therefore, the solution is (-2, -4)

6 0

3 years ago

What is the perimeter of each figure? 2x 2x X-8 x + 2 1

____ [38]

Answer:

triangle: 5x-8

square: 4x+8

Step-by-step explanation:

triangle: 2x + 2x + x -8

combine like terms

= 5x-8

square: x + 2 + x + 2 + x + 2 + x +2

combine like terms

= 4x + 8

8 0

3 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

Evaluate the range of the function f(x) = 3 -6x, when the domain is x=0

MrRissso [65]

Answer:

s

Step-by-step explanation:

7 0

3 years ago

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