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pentagon [3]
4 years ago
12

What is the area of the given circle in Terms of Pi 7.4

Mathematics
2 answers:
Kryger [21]4 years ago
7 0
Square the radius and then multiply it by pi to find the area of a circle 
Sophie [7]4 years ago
5 0
If you mean that the radius is 7.4, then the area of the circle would equal to 172.03
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How to do this problem​
Mumz [18]

Answer:

y = 3/2x -14

Step-by-step explanation:

do the inverse of -2/3 what is 3/2

and now sole the equation of y+8 = 3/2(x-4)

distribute 3/2 ---> y+8 = 3/2x - 8

subtract 8 from both sided getting you y=3/2x-14

7 0
3 years ago
Use logarithmic differentiation to find dy/dx
liq [111]

Answer:

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln  (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx  =   sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx =  2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx  =   [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

7 0
3 years ago
What is the student most likely explaining?
TiliK225 [7]
Answer- embryology .....
6 0
4 years ago
Read 2 more answers
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
3 years ago
What values of x makes the expression (x-6)(x+3) positive?​
Georgia [21]

Answer:

This means that for (x-6)(x+3) to be positive, the value of x has to be greater than 6; x>6

Step-by-step explanation:

Step 1

Given the equation;

(x-6)(x+3)

Step 2

To determine the values of x that make the equation positive, the equation can be expressed as;

(x-6)(x+3)>0

x-6>0/(x+3)

x-6>0

x>0+6

x>6

This means that for (x-6)(x+3) to be positive, the value of x has to be greater than 6; x>6

4 0
3 years ago
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