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3 years ago

Please help I don’t know the answer :(

Mathematics

2 answers:

gogolik [260]3 years ago

6 0

Answer:

When x=8, y=-6. When x=10, y=0

Step-by-step explanation:

$y=3x-30\\x^2+y^2=100$

You can plug the given y value into the second equation to get:

$x^2+(3x-30)^2=100\\x^2+9x^2-180x+900=100\\10x^2-180x+800=0\\10(x^2-18x+80)=0\\10(x-8)(x+10)=0\\x= 8, 10\\y= -6, 0$

Hope this helps!

GarryVolchara [31]3 years ago

4 0

-6,0 is the answer for this

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I'm confused- how do I form the equation with 2 pts?

klio [65]

Answer:

the equation of the line is y = -3x - 6

Step-by-step explanation:

Note that since (0, -6) is the y-intercept, we can write the slope-intercept equation of the line as y = mx - 6. The other given point is (-2, 0) (which happens to be the x-intercept also). Starting with y = mx - 6, replace y with 0 and x with -2:

0 = m(-2) - 6. We now solve this for the slope, m: 0 = -2m - 6 becomes

2m = -6, or m = -3.

With m = -3 and b = -6, the equation of the line is y = -3x - 6

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3 years ago

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WINSTONCH [101]

Answer:

a) $y(t) = y_{0}e^{4t} + 2$ . It does not have a steady state

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Step-by-step explanation:

a) $y' -4y = -8$

The first step is finding $y_{n}(t)$ . So:

$y' - 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r - 4 = 0$

$r = 4$

So:

$y_{n}(t) = y_{0}e^{4t}$

Since this differential equation has a positive eigenvalue, it does not have a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' -4(y_{p}) = -8$

$(C)' - 4C = -8$

C is a constant, so (C)' = 0.

$-4C = -8$

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{4t} + 2$

b) $y' +4y = 8$

The first step is finding $y_{n}(t)$ . So:

$y' + 4y = 0$

We have to find the eigenvalues of this differential equation, which are the roots of this equation:

$r + 4 =$

$r = -4$

So:

$y_{n}(t) = y_{0}e^{-4t}$

Since this differential equation does not have a positive eigenvalue, it has a steady state.

Now as for the particular solution.

Since the differential equation is equaled to a constant, the particular solution is going to have the following format:

$y_{p}(t) = C$

$(y_{p})' +4(y_{p}) = 8$

$(C)' + 4C = 8$

C is a constant, so (C)' = 0.

$4C = 8$

$C = 2$

The solution in the form is

$y(t) = y_{n}(t) + y_{p}(t)$

$y(t) = y_{0}e^{-4t} + 2$

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Answer:

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Answer:

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Step-by-step explanation:

The geometric mean of x and y is √xy.

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Answer:

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Step-by-step explanation:

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