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Artyom0805 [142]
3 years ago
5

Which equation would define a graph that has its vertex on (2, 0) and opens to the right?

Mathematics
1 answer:
Fed [463]3 years ago
8 0
I guess the equation would be y^2 = 4(x-2)

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Her change is $7.55 :)
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Which are the ways to finance your college education?
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Answer:

All of the above

Step-by-step explanation:

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Fitness Your goal is to take at least 10,000 steps per day. According to your pedometer, you have walked 527 4 steps. Write and
Archy [21]

The inequality is x + 5274 ≥ 10,000.

The total number of steps I have to take to reach my goal is 4726.

<h3>How many more steps do I have to take?</h3>

Here are inequality signs and what they mean:

  • > means greater than
  • < means less than
  • ≥ means greater than or equal to
  • ≤ less than or equal to

Number of steps I have to take + number of steps I have taken ≥Total steps

X + 5274 + x ≥ 10,000

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x ≥ 4726

To learn more about inequality, please check: brainly.com/question/5031619

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2 years ago
Long-Term Mehlum
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If the right side of the equation dy dx = f(x, y) can be expressed as a function of the ratio y/x only, then the equation is sai
melomori [17]

Answer:

 y = x*sqrt(Cx - 1)

Step-by-step explanation:

Given:

                                  dy / dx = (x^2 + 5y^2) / 2xy

Find:

Solve the given ODE by using appropriate substitution.

Solution:

- Rewrite the given ODE:

                               dy/dx = 0.5(x/y) + 2.5(y/x)

- use substitution y = x*v(x)

                               dy/dx = v + x*dv/dx

- Combine the two equations:

                                v + x*dv/dx = 0.5*(1/v) + 2.5*v

                                x*dv/dx = 0.5*(1/v) + 1.5*v

                                x*dv/dx = (v^2 + 1) / 2v

-Separate variables:

                                 (2v.dv / (v^2 + 1) = dx / x  

- Integrate both sides:

                                 Ln (v^2 + 1) = Ln(x) + C

                                 v^2 + 1 = Cx

                                 v = sqrt(Cx - 1)

- Back substitution:

                                (y/x) = sqrt(Cx - 1)

                               y = x*sqrt(Cx - 1)

                         

3 0
3 years ago
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