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lara31 [8.8K]
3 years ago
7

3x + 5 = 2x + 9 What do I do???

Mathematics
2 answers:
Yakvenalex [24]3 years ago
7 0

3+5=2+9

3x+5=2x+93x+5=2x+9

Solve

1

Subtract  

5

55

from both sides of the equation

3+5=2+9

3+5−5=2+9−5

2

Simplify

Subtract the numbers

Subtract the numbers

3=2+4

3

Subtract  

2

2x2x

from both sides of the equation

3=2+4

3−2=2+4−2

4

Simplify

Combine like terms

Multiply by 1

Combine like terms

=4

Solution

=4

maxonik [38]3 years ago
3 0
3x - 2x = 9 - 5
x = 4
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I need help please...
Jobisdone [24]

Answer:1

Step-by-step explanation:

7 0
3 years ago
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
A researcher wishes to estimate the proportion of X-ray machines that malfunction. A random 275 sample of machines is taken, and
Dominik [7]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

95\% Confidence Interval for both the percentage of all x-ray machines

p = the machinery's share is not working:

= \frac{228}{275}\\\\ = 0.829

\text{Margin of Error} = Z_{(\frac{\alpha}{2})} \times \sqrt{( p \times (1-p)}{n})

                         = 1.96 \times \sqrt{(0.829 \times \frac{0.171}{275})} \\\\= 1.96 \times 0.023 \\\\= 0.045

Lower 95\% Confidence interval = p - error margin = 0.829 - 0.045 = 0.784

Upper 95\% Confidence Interval = p + error margin= 0.829 + 0.045 = 0.874

So, 95\% Confidence Interval = ( 0.78 , 0.87 )

4 0
2 years ago
Dan's gas tank is 3/8
Neporo4naja [7]

Answer:

16 gallons

Step-by-step explanation:

8 0
2 years ago
Divid 15 into two parts such that the sum of their reciprocal is 3/10
emmasim [6.3K]
Let x and y be the 2 parts of 15 ==> x + y=15 (given)

Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)


Let's solve  1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)

 ==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy

But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50

Now we have the sum S of the 2 parts that is S = 15 and 
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
4 0
3 years ago
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