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allsm [11]
3 years ago
15

(a2)(2a3)(a2 – 8a + 9)

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
8 0

Answer:

-72a^3 +108a^2

Step-by-step explanation:

(2a)(6a)(2a-8a+9)= (2a)(6a)(-6a+9)=12a^2(-6a+9)

= -72a^3 +108a^2

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There are 12 boys and 15 girls in your class for new students three boys and one girl join your class what is the ratio of the c
Finger [1]
Before the new kids the ratio of girls to boys is 15:12 because there are 15 girls and 12 boys now you can simplify by dividing 15 by 3 and 12 by 3 to get 5:4 as your simplified ratio. As far as after the new kids. Since there is 1 new girl you do 15+1 to get 16 and there are 3 new boys so you do 12+3 to get 15 now 16:15 is already in simplest form so your answer is:

Before: 15:12 simplified as 5:4
After 16:15
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3 years ago
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Answer:

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Can show me how to do this ?
elena-s [515]
You have to multiply your 20cm by 6cm and your final answer being 120 because 0 times 6 is zero and 2 times 6 is 12, then that getting you 120
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3 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)

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A rectangular prism had a volume of 540 cubic centimeters. the height of this prism was changed from 10 centimeters to 5 centime
levacccp [35]
The volume of the new rectangular prism would be 270 cubic centimeters.  The reason for this is that when only one dimension is changed, the overall volume is only changed by that same relationship.  Because only the height was cut in half, this would be only one of the dimensions.  
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