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scoray [572]
3 years ago
14

Solve the equation 12+ 0.35×=20.05

Mathematics
1 answer:
lbvjy [14]3 years ago
8 0
12 + 0.35X = 20.05
12 + 0.35X - 12 = 20.05 - 12
0.35X = 8.05 
0.35X/0.35 = 8.05/0.35
X = 23
Hope this helps!! 
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Step-by-step explanation:

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3 years ago
alex is going to share a bag of 24 jelly beans with his friend austin . alex gives some of the jelly beans to austin but , when
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4 years ago
A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the tank at 6 gallons per min
balandron [24]

Answer:

1.857 minutes

Step-by-step explanation:

From the given information:

Consider the amount of salt present in the tank as x(t) at a given time (t); &

The volume of the solution = V(t)

At t = 0 i.e. (at initial conditions)  x(0) = 30; and V(0) = 50

However, the overall increase taken place for 1 gallon per minute is:

V(t) = 50 + t

The amount of salt x(t) at any given point for time (t) is;

\dfrac{x(t)}{V(t)}= \dfrac{x(t)}{50+t}

After 5 gallons of solution exit per minute; the concentration of the salt solution changes at:

\dfrac{dx(t)}{dt}= -\dfrac{5x(t)}{50+t}

Taking the integral of what we have above, we get:

In x(t) = - 5 In(t + 50) + In (C)

In x(t) = In (t+ 50)⁻⁵ + In (C)

In x(t) = In C ( t + 50)⁻⁵

x(t) = C(t + 50)⁻⁵      (General solution)

To estimate the required solution; we apply the initial conditions x(0) = 30;

Thus;

x(0) = C(50)⁻⁵ = 30

⇒ C = 30 × 50⁵

Hence; x(t) =  30 × 50⁵ × (t + 50)⁻⁵

The above expression can be re-written as:

x(t) = 25 \implies 30 \times \bigg ( \dfrac{50}{t+50} \bigg ) ^5= 25

i.e.

\bigg ( \dfrac{50}{t+50} \bigg ) ^5= \dfrac{25}{30}

\dfrac{50}{t+50}= \bigg ( \dfrac{25}{30}\bigg) ^{\dfrac{1}{5}}

\dfrac{50}{t+50}= 0.964192504

{50}= 0.964192504(t+50)

50 = 0.964192504t + 48.2096252

50 - 48.2096252 = 0.964192504t

1.7903748 =  0.964192504t

t = 1.7903748 / 0.964192504

t ≅ 1.857 minutes

We can thereby conclude that the estimated time it will require until there are 25 pounds of salt in the tank is 1.857 minutes.

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