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soldi70 [24.7K]
3 years ago
11

You created a playlist, and 100 of your friends listened to it and shared if they

Mathematics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

its yes

Step-by-step explanation:

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Evaluate the expreesion 41
irakobra [83]

Answer:

i'm pretty sure 41  is not an expression

Step-by-step explanation:

7 0
2 years ago
a certain triangle has a perimeter of 3067 mi. the shortest side measures 71 mi less than the middle side and the longest side m
UkoKoshka [18]

Answer:

The measure of the shortest side is 851 miles

Step-by-step explanation:

Let

x ----> the measure of the shortest side

y ---> the measure of the middle side

z ---> the measure of the longest side

we know that

The perimeter of triangle is equal to

P=x+y+z

P=3,067\ mi

so

3,067=x+y+z ----> equation A

the shortest side measures 71 mi less than the middle side

so

x=y-71 ----> equation B

the longest side measures 372 mi more the the middle side

so

z=y+372 ----> equation C

substitute equation B and equation C in equation A

3,067=(y-71)+y+(y+372)

solve for y

3,067=3y+301\\3y=2,766\\y=922

Find the value of x

x=y-71\\x=922-71\\x=851

therefore

The measure of the shortest side is 851 miles

6 0
3 years ago
PLSSS HELP IF YOU TURLY KNOW THISS
Novay_Z [31]

Answer:

18225

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
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
Help with 2 and 5 plzzzzz
Katyanochek1 [597]
Multiply all the numbers on 2

4 0
2 years ago
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