All categories

3 years ago

A large brine tank containing a solution of salt and water is being diluted with fresh water.

Mathematics

1 answer:

kirza4 [7]3 years ago

6 0

Answer:

41.04 g/L

Step-by-step explanation:

The relationship is modeled by the following function:

S(t) = 500*e^(-0.25*t)

where t is time elapsed after the dilution begins (in hours) and S(t) the concentration of salt in the tank (in g/L)

Replacing with t = 10

S(t) = 500*e^(-0.25*10)

S(t) = 41.04 g/L

You might be interested in

Pratap Puri rowed 14 miles down a river in 2 hours, but the return trip took him 3 and one half hours. Find the rate Pratap can

SOVA2 [1]

speed of current is 1.5 mi/hr

Answer:

let the rate in still water be x and rate of the current be y.

speed down the river is:

speed=distance/time

speed=14/2=7 mi/h

speed up the river is:

speed=(14)/(3.5)=4 mi/hr

thus total speed downstream and upstream will be:

x+y=7...i

x-y=4.......ii

adding the above equations i and ii we get:

2x=11

x=5.5 mi/hr

thus

y=5-5.5=1.5 mi/r

thus the speed in still waters is 5.5 mi/hr

speed of current is 1.5 mi/hr

7 0

2 years ago

Which is the logarithmic form of 25 = 52?

goldfiish [28.3K]

Answer: 5^5=52

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Functions math, please help

kolbaska11 [484]

Answer:

Slope: 2

Y-intercept: (0,-7)

4 0

2 years ago

Read 2 more answers

6p+2+ p - 3 = 8p - 8

saul85 [17]

6p+p+2-3=8p-8
7p-1=8p-8
1=1p-8
7=1p
p=7

7 0

3 years ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago