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Xelga [282]
3 years ago
7

Simplify −2xy + 3x − 2xy + 3x and can you explain the answer?

Mathematics
2 answers:
irina1246 [14]3 years ago
8 0

Answer:

 -4xy+6x

Step-by-step explanation:

Given : Expression =  -2xy+3x-2xy+3x

Simplifying the expression :

Step 1. Write the expression =   -2xy+3x-2xy+3x

Step 2.  Add the similar elements:  3x+3x=6x

          =  -2xy-2xy+6x

Step 3. Add the similar elements:  -2xy-2xy=-4xy

           = -4xy+6x

Therefore, -4xy+6x is the simplified form of  -2xy+3x-2xy+3x

Karolina [17]3 years ago
6 0
First you must combine like terms, the two terms I this expression are x, and xy. So by adding -2xy to -2xy you get -4xy. Next you must add 3x to 3x. This gives you the simplified expression -4xy+6x. Hope this helps!
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x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

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r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
What is the area of the triangle in the coordinate plane?
AnnZ [28]
The answer to this is C.54 units² .

Area=1/2 base*height
base=12
height=9
12*9=108
108/2=54
Area=54
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Step-by-step explanation:

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