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Eddi Din [679]
3 years ago
15

Can someone plz help me with this!

Mathematics
1 answer:
asambeis [7]3 years ago
3 0

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

2x+y=10\\x-3y=-2

To solve, start by isolating your first equation for y.

2x+y=10\\y=-2x+10

Now, plug this value of y into your second equation.

x-3(-2x+10)=-2\\x+6x-30=-2\\7x=28\\x=4

System #2: x=1

x+2y=5\\2x+y=4

Isolate your second equation for y.

2x+y=4\\y=-2x+4

Plug this value of y into your first equation.

x+2(-2x+4)=5\\x+(-4x)+8=5\\x-4x+8=5\\-3x=-3\\x=1

System #3: x=6

5x+y=33\\x=18-4y

Isolate your first equation for y.

5x+y=33\\y=-5x+33

Plug this value of y into your second equation.

x=18-4(-5x+33)\\x=18+20x-132\\-19x=-114\\x=6

System #4: all real numbers (not included in your diagram)

y=13-2x\\8x+4y=52

Plug your value of y into your second equation.

8x+4(13-2x)=52\\8x+52-8x=52\\0=0

<em>all real numbers are solutions</em>

System #5: x=2

x+3y=5\\6x-y=11

Isolate your second equation for y.

6x-y=11\\-y=-6x+11\\y=6x-11

Plug in your value of y to your first equation.

x+3(6x-11)=5\\x+18x-33=5\\19x=38\\x=2

System #6: no solution (not included in your diagram)

2x+y=10\\-6x-3y=-2

Isolate your first equation for y.

2x+y=10\\y=-2x+10

Plug your value of y into your second equation.

-6x-3(-2x+10)=-2\\-6x+6x-30=-2\\-30=-2

<em>no solution</em>

System #7: x=3

y=10+x\\2x+3y=45

Plug your value of y into your second equation.

2x+3(10+x)=45\\2x+30+3x=45\\5x=15\\x=3

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5x-7&lt;7x-11&lt;5x+9 <br> Give your answer in interval notation
skelet666 [1.2K]

Answer:

In pic below

Step-by-step explanation:

4 0
3 years ago
Appropriate Inverse Operation
Nimfa-mama [501]
Multiplying both sides of the equation by 3 and substituting 4 for p
5 0
3 years ago
Solve the following equation with the initial conditions. x¨ + 4 ˙x + 53x = 15 , x(0) = 8, x˙ = −19
Katen [24]

x''+4x'+53x=15

has characteristic equation

r^2+4r+53=0

with roots at r=-2\pm7i. Then the characteristic solution is

x_c=C_1e^{(-2+7i)t}+C_2e^{(-2-7i)t}=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)

For the particular solution, consider the ansatz x_p=a_0, whose first and second derivatives vanish. Substitute x_p and its derivatives into the equation:

53a_0=15\implies a_0=\dfrac{15}{53}

Then the general solution to the equation is

x=e^{-2t}\left(C_1\cos(7t)+C_2\sin(7t)\right)+\dfrac{15}{53}

With x(0)=8, we have

8=C_1+\dfrac{15}{53}\implies C_1=\dfrac{409}{53}

and with x'(0)=-19,

-19=-2C_1+7C_2\implies C_2=-\dfrac{27}{53}

Then the particular solution to the equation is

\boxed{x(t)=\dfrac1{53}e^{-2t}(409cos(7t)-27\sin(7t)+15)}

8 0
3 years ago
Help ASAP please I rlly need the help
Katena32 [7]

Answer:

2x=x+x doesn't belong

Step-by-step explanation:

7 0
3 years ago
(75 points) need help now
Igoryamba

Answer:

A is not a function

Step-by-step explanation:

In table A, the x value of 4 goes to two different y values

To be a function each x value can only go to one y value

A is not a function

4 0
3 years ago
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