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iogann1982 [59]
3 years ago
12

6 1/3 - 5 4/9 what is this answer?​

Mathematics
2 answers:
svetlana [45]3 years ago
5 0

Answer:

Step-by-step explanation:

1/5

tester [92]3 years ago
3 0

Answer:

The answer is 8/9

Step-by-step explanation:

Convert the mixed numbers to improper fractions, then find the LCD and combine.

Hoped this helped!

Could I perhaps get Brainly?

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The answer is (-1,-4)
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(3x+2)+(2x^2+3x-1)<br><br> Please answer ty:)
Paul [167]
Answer = 2x^2+6x+1

3x+2+2x^2+3x-1
3x+1+2x^2+3x
Now combine like terms
3x+1+2x^2+3x
6x+1+2x^2
Rearrange terms
6x+1+2x^2
2x^2+6x+1
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3 years ago
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

So, the roots are given by:

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
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sdas [7]

Answer:

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Step-by-step explanation:

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Answer:

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