Solve the equation. −10x + 1 + 7x = 37

2 answers:

8 0

You combine like terms which are -10x and 7x and 1 and 37. Then it becomes -3x=36. x would equal to 36/-3, which is -12.

ycow [4]2 years ago

6 0

−10x+1+7x=37

Simplify both sides of the equation.

−10x+1+7x=37

Simplify:

−10x+1+7x=37

(−10x+7x)+(1)=37(Combine Like Terms)

−3x+1=37

−3x+1=37

Subtract 1 from both sides.

−3x+1−1=37−1

−3x=36

Divide both sides by -3.

−3x/−3 = 36/−3

x=−12

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A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$