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Vlada [557]
3 years ago
11

One week you tell 3 of your friends a secret. The next week, each of them tells 3 people the secret and the

Mathematics
1 answer:
yawa3891 [41]3 years ago
5 0
By the 7th week, 2,187 people will have found out the secret.
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Help this is due in 10 minutes please help
Assoli18 [71]

Answer: acute angle, straight line

Step-by-step explanation:

6 0
2 years ago
Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] < y0 < [infinity](a) Sketch the graph
kotegsom [21]

Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of  f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0  and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane  is shown on the fourth uploaded image

Step-by-step explanation:

Step One: Sketch The Graph of  f(y) versus y

Looking at the given differential equation

       \frac{dy}{dt} = e^{y} - 1 for -∞ < y_{o} < ∞

 We can say let \frac{dy}{dt} = f(y) =e^{y} - 1

Now the dependent value is f(y) and the independent value is y so to sketch is graph we can assume a scale in this case i cm on the graph is equal to 2 unit for both f(y) and y and the match the coordinates and after that join the point to form the graph as shown on the uploaded image.

Step Two : Determine the critical point

   To fin the critical point we have to set   \frac{dy}{dt} = 0

       This means e^{y} - 1 = 0

                          For this to be possible e^{y} = 1

                          which means that  e^{y} = e^{0}

                          which implies that y = 0

Hence the critical point occurs at y = 0

meaning that the equilibrium solution is y = 0

As t → ∞, our curve is going to move away from y = 0  hence it is asymptotically unstable.

Step Three : Draw the Phase lines

A phase line can be defined as an image that shows or represents the way an ODE(ordinary differential equation ) that does not explicitly depend on the independent variable behaves in a single variable. To draw this phase line , draw the y-axis as a vertical line and mark on it the equilibrium, i.e. where  f(y) = 0.

In each of the intervals bounded  by the equilibrium draw an upward

pointing arrow if f(y) > 0 and a downward pointing arrow if f(y) < 0.

      This phase line would solely depend on y does not matter what t is

On the positive x axis it would get steeper very quickly as you move up (looking at the part A graph).

For  below the x-axis which stable (looking at the part a graph) we are still going to have negative slope but they are going to be close to 0 and they would take a little bit longer to get steeper  

Step Four : Draw a Solution Curve

A solution curve is a curve that shows the solution of a DE (deferential equation)

Here the solution curve would be drawn on the ty-plane

So the t-axis(x-axis) is its the equilibrium  that is it is the solution

If we are above the x-axis it is going to increase faster and if we are below it is going to decrease but it would be slower (looking at part A graph)

5 0
2 years ago
Answer to number 6 please
Alenkasestr [34]

Answer:

Step-by-step explanation:

C

6 0
2 years ago
if a taxi charges $3.50 for the first on-fifth of a mile, then $0.55 cents for each additional one-fifth mile, how far can one t
34kurt
Let's forget for the tie being the cost ($3.5)of the 1st fifth of a mile:

Taxi charge excluding $3.5, is $13.95 - $3.5 = $10.45.

If each of the 5th (1/5) of a mile costs $0.55, then with $10.45 we can drive:
10.45/0.55 = 19 fifth of a mile. Add to that the 1st fifth = 20 fifth of a mile
 20 fifth = 20 x 1/5 = 4 miles (answer)

7 0
3 years ago
Calculate the size of a cube​
Varvara68 [4.7K]

Answer:

<em>A</em><em> </em><em>cube</em><em> </em><em>size</em><em> </em><em>is</em><em> </em><em>150mm</em>

Step-by-step explanation:

I think it will be

4 0
3 years ago
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