Please help me with this, idk what to do

The coordinates of the vertices of a polygon are (−2,−2), (3,−3), (4,−6), (1,−6), and (−2,−4).

krek1111 [17]

The perimeter of the polygon is 17.5

5 0

3 years ago

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

4 years ago

For each question, draw a figure. Label the information on the figure. Write an equation that solves the

Julli [10]

Answer:

See attachment for the figures drawn for each question.

10. x = 3; AB = 11

11. x = 2; AB = 6

Step-by-step explanation:

10. AC = 10x + 2

AB = 2x + 5

BC = 4x + 9

AB + BC = AC (segment addition postulate)

(2x + 5) + (4x + 9) = 10x + 2 (substitution)

The equation above is what we would use to solve the problem as follows:

2x + 5 + 4x + 9 = 10x + 2

Combine like terms

6x + 14 = 10x + 2

6x - 10x = -14 + 2

-4x = -12

Divide both sides by -4

x = 3

AB = 2x + 5

Plug in the value of x

AB = 2(3) + 5 = 6 + 5

AB = 11

11. AB = 3x

BC = x - 6

AC = 8x - 14

AB + BC = AC (segment addition theorem)

3x + (x - 6) = 8x - 14 (substitution)

Use the equation to solve for x as follows:

3x + x - 6 = 8x - 14

Collect like terms

4x - 6 = 8x - 14

4x - 8x = 6 - 14

-4x = -8

x = 2 (dividing both sides by -4)

AB = 3x

AB = 3(2)

AB = 6

5 0

4 years ago

Classify the triangle by its sides.

otez555 [7]

Answer:

It would be C

Step-by-step explanation:

This triangle has two equal sides and one side that is not equal to the other two, so it is an isosceles triangle.

8 0

3 years ago

Read 2 more answers

Very faint but the number in the root thing is 114 :) help please!!

JulsSmile [24]

dear I think the figure under the root is 144

14²- √144

(14 ×14) -12

the square root of 144 is 12

196 -12

184

3 0

3 years ago

Read 2 more answers