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Jobisdone [24]
2 years ago
5

last year, Nicole biked n miles. This year, she biked 524 miles. Using n, write an expression for the total number of miles she

biked.​
Mathematics
1 answer:
Fittoniya [83]2 years ago
7 0

Answer:

n+524

Step-by-step explanation:

you add both years worth of biking together!

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goldenfox [79]

Answer:

2. x = 135

Step-by-step explanation:

im not sure about question 1

8 0
3 years ago
The library is halfway between your house and your school. If your house is located at (-4,-3) and the library is at (2,-1), wha
aleksandrvk [35]
I believe it would be (-2,-2) i may be wrong though i dont have my graphing papers
6 0
3 years ago
Find the value of B - A if the graph of Ax + By = 3 passes through the point (-7,2), and is parallel to the graph of x + 3y = -5
Gnoma [55]

The required simplified value of B - A = -6.

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Since, line Ax + By = 3 and x + 3y = -5 are parallel than slope of both the line will be same,
m = -1/3 = -A/B
From above
A = B/3  - - - - -(1)


Now line Ax + By = 3 passed through the point (-7, 2). So,
-7A + 2B = 3
from equation 1
-7B/3 + 2B = 3
-B/3 = 3
B = -9
Now put B in equation 1
A = -9 / 3
A = -3

Here,
B - A = -9 + 3 = -6

Thus, the required simplified value of B - A = -6.

Learn more about simplification here: brainly.com/question/12501526

#SPJ1



7 0
1 year ago
The equation of line p is y=ax+b. Which of these could be the equation of line q? A.y=−3a(x+6)+b
Ilia_Sergeevich [38]
The slope of line q is that of line p scaled by a factor of 3 (not -3). The y-intercept of line q is 6 less than the y-intercept of line p. The appropriate choice is
  D. y = 3ax +b -6
7 0
2 years ago
Read 2 more answers
A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit
Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

7 0
3 years ago
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