Do i have the correct answer? am i supposed to be solving for a straight angle (180°)?

Suppose the clean water of a stream flows into Lake Alpha, then into Lake Beta, and then further downstream. The in and out flow

Gala2k [10]

Answer:

a) dx / dt = - x / 800

b) x = 500*e^(-0.00125*t)

c) dy/dt = x / 800 - y / 200

d) y(t) = 0.625*e^(-0.00125*t)*( 1 - e^(-4*t) )

Step-by-step explanation:

Given:

- Out-flow water after crash from Lake Alpha = 500 liters/h

- Inflow water after crash into lake beta = 500 liters/h

- Initial amount of Kool-Aid in lake Alpha is = 500 kg

- Initial amount of water in Lake Alpha is = 400,000 L

- Initial amount of water in Lake Beta is = 100,000 L

Find:

a) let x be the amount of Kool-Aid, in kilograms, in Lake Alpha t hours after the crash. find a formula for the rate of change in the amount of Kool-Aid, dx/dt, in terms of the amount of Kool-Aid in the lake x:

b) find a formula for the amount of Kook-Aid in kilograms, in Lake Alpha t hours after the crash

c) Let y be the amount of Kool-Aid, in kilograms, in Lake Beta t hours after the crash. Find a formula for the rate of change in the amount of Kool-Aid, dy/dt, in terms of the amounts x,y.

d) Find a formula for the amount of Kool-Aid in Lake Beta t hours after the crash.

Solution:

- We will investigate Lake Alpha first. The rate of flow in after crash in lake alpha is zero. The flow out can be determined:

dx / dt = concentration*flow

dx / dt = - ( x / 400,000)*( 500 L / hr )

dx / dt = - x / 800

- Now we will solve the differential Eq formed:

Separate variables:

dx / x = -dt / 800

Integrate:

Ln | x | = - t / 800 + C

- We know that at t = 0, truck crashed hence, x(0) = 500.

Ln | 500 | = - 0 / 800 + C

C = Ln | 500 |

- The solution to the differential equation is:

Ln | x | = -t/800 + Ln | 500 |

x = 500*e^(-0.00125*t)

- Now for Lake Beta. We will consider the rate of flow in which is equivalent to rate of flow out of Lake Alpha. We can set up the ODE as:

conc. Flow in = x / 800

conc. Flow out = (y / 100,000)*( 500 L / hr ) = y / 200

dy/dt = con.Flow_in - conc.Flow_out

dy/dt = x / 800 - y / 200

- Now replace x with the solution of ODE for Lake Alpha:

dy/dt = 500*e^(-0.00125*t)/ 800 - y / 200

dy/dt = 0.625*e^(-0.00125*t)- y / 200

- Express the form:

y' + P(t)*y = Q(t)

y' + 0.005*y = 0.625*e^(-0.00125*t)

- Find the integrating factor:

u(t) = e^(P(t)) = e^(0.005*t)

- Use the form:

( u(t) . y(t) )' = u(t) . Q(t)

- Plug in the terms:

e^(0.005*t) * y(t) = 0.625*e^(0.00375*t) + C

y(t) = 0.625*e^(-0.00125*t) + C*e^(-0.005*t)

- Initial conditions are: t = 0, y = 0:

0 = 0.625 + C

C = - 0.625

Hence,

y(t) = 0.625*( e^(-0.00125*t) - e^(-0.005*t) )

y(t) = 0.625*e^(-0.00125*t)*( 1 - e^(-4*t) )

6 0

3 years ago

Find all the missing sides or angles in each right triangles

astra-53 [7]

In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle △XYZ is cut by AB¯¯¯¯¯¯¯¯ where A and B are midpoints of sides XZ¯¯¯¯¯¯¯¯ and YZ¯¯¯¯¯¯¯ respectively. AB¯¯¯¯¯¯¯¯ is called a midsegment of △XYZ. Note that △XYZ has other midsegments in addition to AB¯¯¯¯¯¯¯¯. Can you see where they are in the figure above?

If we construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and construct CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ respectively, we have the following figure and see that segments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ are midsegments of △XYZ.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

parallel to the third side. half as long as the third side.

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle △XYZ. Construct the midpoint A of side XZ¯¯¯¯¯¯¯¯.

By the Parallel Postulate, there is exactly one line though A that is parallel to side XY¯¯¯¯¯¯¯¯. Let’s say that it intersects side YZ¯¯¯¯¯¯¯ at point B. We will show that B must be the midpoint of XY¯¯¯¯¯¯¯¯ and then we can conclude that AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is parallel to XY¯¯¯¯¯¯¯¯.

We must show that the line through A and parallel to side XY¯¯¯¯¯¯¯¯ will intersect side YZ¯¯¯¯¯¯¯ at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side YZ¯¯¯¯¯¯¯.

Since XA¯¯¯¯¯¯¯¯≅AZ¯¯¯¯¯¯¯, we have BZ¯¯¯¯¯¯¯≅BY¯¯¯¯¯¯¯¯. Hence, by the definition of midpoint, point B is the midpoint of side YZ¯¯¯¯¯¯¯. AB¯¯¯¯¯¯¯¯ is a midsegment of the triangle and is also parallel to XY¯¯¯¯¯¯¯¯.

Proof of 2. We must show that AB=12XY.

In △XYZ, construct the midpoint of side XY¯¯¯¯¯¯¯¯ at point C and midsegments CA¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯ as follows:

First note that CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ by part one of the theorem. Since CB¯¯¯¯¯¯¯¯∥XZ¯¯¯¯¯¯¯¯ and AB¯¯¯¯¯¯¯¯∥XY¯¯¯¯¯¯¯¯, then ∠XAC≅∠BCA and ∠CAB≅∠ACX since alternate interior angles are congruent. In addition, AC¯¯¯¯¯¯¯¯≅CA¯¯¯¯¯¯¯¯.

Hence, △AXC≅△CBA by The ASA Congruence Postulate. AB¯¯¯¯¯¯¯¯≅XC¯¯¯¯¯¯¯¯ since corresponding parts of congruent triangles are congruent. Since C is the midpoint of XY¯¯¯¯¯¯¯¯, we have XC=CY and XY=XC+CY=XC+XC=2AB by segment addition and substitution.

So, 2AB=XY and AB=12XY. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

A. MN. B. The perimeter of the triangle △XYZ. A. Since O is a midpoint, we have XO=5 and XY=10. By the theorem, we must have MN=5. B. By the Midsegment Theorem, OM=3 implies that ZY=6; similarly, XZ=8, and XY=10. Hence, the perimeter is 6+8+10=24.

We can also examine triangles where one or more of the sides are unknown.

Example 2

Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment XY¯¯¯¯¯¯¯¯.

By the Midsegment Theorem we have 2x−6=12(18). Solving for x, we have x=152.

Lesson Summary

8 0

3 years ago

A bowl contains 7 marbles of various colors consisting of green, brown, and red.

ZanzabumX [31]

Answer:

3/7

Step-by-step explanation:

7-4=3

7 0

2 years ago

Divided 80 by my number, and then added 13. The result was 93. What was my number ?

Elena L [17]

Answer:

6400

Step-by-step explanation:

6400/80=80

80+13=93

Hope this Helps

Pls Mark as Brainliest

4 0

3 years ago

Read 2 more answers

Help! Algebra! please answer if you only know! and make it undertanding!

snow_lady [41]

1. a. Pretty much, you just have to rearrange it so that the highest power is in the front. So, here's your answer:
$-6x^4+4x^2+2$
b. It's a 4th-degree polynomial. A degree means that "what's the highest power?"
c. It's a trinomial. It has 3 terms, hence it's a trinomial.

2. a. Since it's an odd power and a negative coefficient, it will be:
x→∞, f(x)→-∞
x→-∞, f(x)→∞
b. The degree is even and the coefficient is negative, so it will be:
x→∞, f(x)→-∞
x→-∞, f(x)→-∞

3. a. This basically means that if you solve for x, you should get -2, 1, and 2. So, to do this, you can just write it in factored form and multiply inwards using any method of your choice (remember that in the parentheses, you should get the above value if you solve for x):
$f(x) = (x-2)(x+2)(x-1)$
If you multiply it out, you get (also your answer):
$x^3-x^2-4x+4$

4. The zeros are at x = 3, 2 and -7. Multiplicity of 3 is 1, for 2 it's 2, and for -7 it's 3.

Hope this helps!

6 0

3 years ago

Read 2 more answers