Hi I need help I'm clueless and it's due Tomorrow!!!!

Mathematics

1 answer:

gtnhenbr [62]4 years ago

7 0

To multiply fractions, multiply the numerators and denominators and reduce if possible. For problem 3, do 7 x 4 = 28. This is the numerator of the answer. The denominator is 8x5 = 40. Therefore, your unreduced fraction is 28/40. Since both 28 and 40 are multiples of 4, you can reduce it to 7/10, since 28/4=7 and 40/4=10.
Use this same process for problems 4 through 12.

For problem 13: 2/5= 4/10, so the black box = 1/2. This is because you multiply the numerators and the denominators.
Numerators: 4 x 1 = 4. Denominators: 5 x 2 = 10
So, your result fraction is 4/10, which reduces to 2/5.
Therefore, the answer is 1/2 ( above math was explanation)

14: multiply the two fractions
6/7 x 4/5 = 24/35, so the answer is 24.

15. Count the shaded squares. There are 25 shaded squares out of 40. The fraction is therefore 25/40 = 5/8.
Now count the dotted squares. There are 24 shaded squares out of 40. The fraction is 24/40= 3/5.
You could also just look at the far right row to get the same answer because it will reduce anyway, but that might confuse you more.

The answer to 15 is multiplying the two fractions, 5/8 x 3/5.

16. Since both fractions reduce to 1 (numerators are the same as denominators), the model would be 3 boxes by 4 boxes, all shaded and all with dots in the middle.

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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)