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yKpoI14uk [10]
3 years ago
10

What is... 55.

Mathematics
1 answer:
sukhopar [10]3 years ago
3 0

Answer:

one more would be 56

one less would be 54

ten more would 65

ten less would be 45

Step-by-step explanation:

add one more to 55 to get 56

subtract one from 55 to get 54

add ten more to 55 to get 65

subtract ten from 55 to get 45

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The angle of the elevation of the sun is 34 degrees. Find the length, l, of a shadow cast by a tree that is 53 feet tall. Round
gulaghasi [49]
\tan{34^o}= \frac{53}{x} 
\\
\\x= \frac{53}{\tan{34^o}} 
\\
\\x \approx 78.58

8 0
3 years ago
What are the answers​
LUCKY_DIMON [66]

Answer:

Step-by-step explanation:

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7 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
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Julli [10]

Answer:

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Step-by-step explanation:

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UkoKoshka [18]
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Alternately, you can write [</span><span><span>1 / (tan(2x) - cot(2x))] + [cos(8x) / (tan(2x) - cot(2x))].

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