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4 years ago

A = P (8+sd) solve for d

1 answer:

5 0

A = P (8+sd) Switch the equation P(8 + sd)=A Now divide both sides by p 8+sd=a/P Subtract 8 both sides Sd=-8+a/P Divide each term by s D= -8/s+ a/p×1/s D=-8/s+ a×1/p×s D= -8/s+a/ps

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tank contains 250 liters of fluid in which 20 grams of salt is dissolved. Pure water is then pumped into the tank at a rate of 5

eimsori [14]

Solution :

Given data :

$c_{in}$ = 1 g/L

$r_{in}$ = 5 L/min

$r_{out}$ = 5 L/min

$v_0$ = 250 L

$A_0$ = 20 g

∴ $r_{net} = r_{in}- r_{out}$

= 5 - 5

= 0

$c_{out} = \frac{A}{250} \ g/L$

Now, $\frac{dA}{dt}=(r_{in} \times c_{in}) - (r_{out} \times c_{out})$

$\frac{dA}{dt} = 5-5\left(\frac{A}{250}\right)$

$\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5$

$\frac{dA}{dt}+5 \left(\frac{A}{250}\right) = 5 \text{ with} \ A_0 = 20$

Integrating factor = exp(5 t/250)

Therefore,

$A \times \exp (5t \ /250) = \text{integral of}\ 5 \times \exp (5t / 250) + C$

Put $A_0=250+C$

C = -230

$A \times \exp(5t/250) = 250 \exp(5t/250) + (-230)$

$A(t) = 250-230 \exp(-5t/250)$

$A(t) = 250-230e^{\left(\frac{-t}{50}\right)} \ g$

5 0

3 years ago

6) A sample of water is being cooled according to the formula is given by

Vinvika [58]

Answer:

Relevant domain: [0 40]

Relevant range: [0 80]

Step-by-step explanation:

We have the following function:

$C(t) = 80 - 2t$

In which C(t) is the water temperature.

t is the time.

The colling period ends when the water freezes (0°).

So it freezes after t minutes, and t is found when C(t) = 0. So

$C(t) = 80 - 2t$

$80 - 2t = 0$

$2t = 80$

$t = \frac{80}{2}$

$t = 40$

Relevant domain:

Input value of the function, which is in minutes. So between 0(initial) and 40(when the water freezes).

Relevant range:

Initial temperature, which is C(0) = 80, until it freezes, 0. So from 0 to 80.

5 0

3 years ago

Which property justifies the fact that 5(x−2) is equivalent to 5x−10?

TiliK225 [7]

Answer:

c i believe:/

Step-by-step explanation:

hope this helps:) brainliest please!

6 0

3 years ago

Don’t get it I hate this teach need help

fomenos

7. the unit rate is 30 divide 60 by 2 and 90 by 3 to show u r work

8.the unit rate is 1.89 divide 18.90 by 10 and so on

3 0

4 years ago

Find function domain <br> f(x) = sqrt( 2sin x - 1 )

vekshin1

Answer:

{x ∈ ℝ : x ≥ π/6 +2πn and x ≤ π/6 + 2πn and n ∈ ℤ}

Step-by-step explanation:

sinx can run from -1 to +1

2sinx can run from -2 to +2

2sinx -1 can run from -3 to +1

However, the square root is imaginary when x < 0. So, the condition is

2sinx -1 ≥ 0

2sinx ≥ 1

sinx ≥ ½

x ≥ π/6 (30°)

So, in the interval [0, 2π], π/6 ≤ x ≤ 5π/6

However, the sine is a cyclic function and repeats itself every 2π.

Over all real numbers, the condition is (π/6 +2πn) ≤ x ≤ (5π/6 + 2πn).

The domain is then

{x ∈ ℝ : x ≥ π/6 +2πn and x ≤ π/6 + 2πn and n ∈ ℤ}

5 0

3 years ago