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kozerog [31]
2 years ago
5

Which has a less value than 0

Mathematics
2 answers:
Anna35 [415]2 years ago
8 0
Where is the graph/chart/picture/etc ?
Mumz [18]2 years ago
8 0
All negative numbers have less value than 0

Example:-9,-1,-29
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I really think it’s B ...Help?
kvv77 [185]
The correct answer is c y-5=4(x-7)
7 0
2 years ago
Solve for x in the given interval.<br><br> sec x= -2√3/3, for π/2 ≤x≤π
drek231 [11]

Answer:

b. x=\frac{5\pi}{6}

Step-by-step explanation:

The given function is

\sec x=-\frac{2\sqrt{3} }{3},\:\:for\:\:\frac{\pi}{2}\le x \le \pi

Recall that the reciprocal of the cosine ratio is the secant ratio.

This implies that;

\frac{1}{\cos x}=-\frac{2\sqrt{3} }{3}

\Rightarrow \cos x=-\frac{3}{2\sqrt{3} }

\Rightarrow \cos x=-\frac{\sqrt{3}}{2}

We take the inverse cosine of both sides to obtain;

x=\cos^{-1}(-\frac{\sqrt{3}}{2})

x=\frac{5\pi}{6}

4 0
3 years ago
How many zeros are in one million?
Julli [10]

Answer:

six

Step-by-step explanation:

1,000,000

5 0
3 years ago
Read 2 more answers
Can some solve me this one question
iogann1982 [59]
What is the question?
7 0
3 years ago
Consider the following equation of the form dy/dt = f(y)dy/dt = ey − 1, −[infinity] &lt; y0 &lt; [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
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