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Sedaia [141]
3 years ago
11

Ursula surveyed 50 classmates about their favorite ice cream flavors. Each classmate chose one flavor. The results are shown in

the circle graph how many more of ursula’s classmates chose chocolate than chose vanilla
Mathematics
2 answers:
Akimi4 [234]3 years ago
8 0

We need the circle graph.

dalvyx [7]3 years ago
3 0
Can you upload a picture of the circle graph so we can help answer the question for you?
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1 cm = 500km the ratio is 1 to
Murrr4er [49]

Answer:

1:500

Step-by-step explanation:

Since it has already given you the conversion factor, the ratio is 1:500, where every 1 cm is 500 km.

4 0
3 years ago
Read 2 more answers
Someone check if this is right. And stop sending z.oom codes in my questions
olya-2409 [2.1K]

Answer:

no it is not

Step-by-step explanation:

-5, and 7

8 0
3 years ago
maya wants to burs 600 calories, so far, she burned 584.3 calories. how many more calories must maya burn
MrMuchimi

Answer:

15.7

Step-by-step explanation:

600 - 584.3 = 15.7

5 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
consider these graphs: Eric thinks the equation y=(x-3)^2 will match the blue graph. Jenney thinks it will match the red graph.
Dimas [21]

Answer: Jenny

Step-by-step explanation:

By subtitue Y = 0

0=(x-3)^2

And you get 3

3 0
3 years ago
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