Which measurements could NOTrepresent the side lengths of a right triangle?

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

3 years ago

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years. Five years after Brian's initial inve

pentagon [3]

Answer:

Brian has $776 more account in his account than Chris.

Step-by-step explanation:

Compound interest Formula:

$A=P(1+r)^t$

$I$ = A-P

A= Amount after t years

P= Initial amount

r= Rate of interest

t= Time in year

Given that,

Brian invests $10,000 in an account earning 4% interest, compounded annually for 10 years.

Here P = $10,000 , r= 4%=0.04, t=10 years

The amount in his account after 10 years is

$A=10000(1+0.04)^{10}$

=$14802.44

≈$14802

Five years after Brian's investment,Chris invests $10,000 in an account earning 7% interest, compounded annually for 5 years.

Here P = $10,000 , r= 7%=0.07, t=5 years

The amount in his account after 5 years is

$A=10000(1+0.07)^{5}$

=$14025.51

≈$14026

From the it is cleared that Brian has $(14802-14026)=$776 more account in his account than Chris.

6 0

2 years ago

A girl ran to the store at a speed of 6mph, and walked back at 4mp. If her round trip took 15 minutes, how far away is the store

MariettaO [177]

Answer:

0.6 miles

Step-by-step explanation:

Let's say the store is x miles away, she walks to the store in a hours, and she walks back in b hours.

First, her total time = time walked there + time walked back = a + b = 15 minutes = 1/4 of an hour = 0.25 hours

a + b = 0.25 (because a and b are in hours)

distance = speed * time

x = 6 miles/hour * a hours = 6a miles

x = 4 miles/hour * b hours = 4b miles

If we can find a or b, we can find x.

4b = x = 6a

4b = 6a

a + b = 0.25

We now have two equations with two variables. Using substitution, we can solve for one of them

a + b = 0.25

subtract a from both sides to find b

0.25 - a = b

plug that into the other equation, 4b = 6a

4(0.25-a) = 6a

1 - 4a = 6a

add 4a to both sides to isolate a and its coefficient

1 = 10a

divide both sides by 10 to solve for s

a = 1/10 of an hour = 0.1 of an hour = 60 minutes / 10 = 6 minutes

6a = x

6(0.1) = x

0.6 = x

the store is 0.6 miles away

6 0

2 years ago

Order the decimals from least to greatest...

PtichkaEL [24]

Answer:

-1.5

-0.75

0.25

2

2.250

Hope this helps!!

6 0

3 years ago

Read 2 more answers

Use a separate sheet of paper to solve the system of inequalities by graphing. Use the graph to decide if the point (1, –1) is i

Lena [83]

Answer:

The answer is no though, the point (1,-1) is below both.

Step-by-step explanation:

So the two inequalities are y≤-3x+3 and 3x+8y≥3

With inequalities you have to know a few things. First, when you graph you may have to use non solid lines. for instance, if you had x<5, 5 would not be part of the answer because it is only answers less than 5. so you represent this by making a dashed line that says "you cannot include values on this line". Second, with algebra, if you multiply both sides by a negative it flips the inequality. for instance -y > 54 you would multiply (or divide) by -1 and it would change to y < -54. Third, when graphing an inequality, treat it like an equation and graph the line as if it were said equation(with solid or dashed lines as needed) then, if it is y > f(x) you shade in the section above the line and y < f(x) shades in below. The shaded part is your answer basically. Now lets get into yours.

y≤-3x+3

It is already in the form y = f(x) so grapht he line 3x + 3 with a solid line since it is y is less than or equal to, then shade below that line.

3x+8y≥3

First get it so y is by itself

3x+8y≥3

8y≥3 - 3x

y ≥ 3/8 - 3/8 x

Now grapht his line with a solid line and shade the area above it.

Now, the shaded area below between y ≥ 3/8 - 3/8 x and y≤-3x+3 and the lines are your answer, so you need to check if the point (1, -1) is in this shared shaded area or on one of the lines.

The answer is no though, the point (1,-1) is below both.

6 0

3 years ago