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Zina [86]
3 years ago
7

For which pair of points can you use this number line to find the distance?

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

Answer:

(2,0) and (2,3)

Step-by-step explanation:

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Helpppppppp please!!!!!!!!!!!
fenix001 [56]

Answer:

a. (4, 13/6)

b. (4, -5/6)

c. (4,  5/6)

Step-by-step explanation:

sorry if it is not right.

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If the area of circle b is 1/5 the area circle a What is the area of circle b
WARRIOR [948]

Answer:

(Area of circle a)/5

Step-by-step explanation:

(Area of circle b) = (1/5)* (Area of circle a)

What is the area of circle a?  There is not enough given information

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3 years ago
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At 9:00 on Saturday morning, two bicyclists heading in opposite directions pass each other on a bicycle path.The bicyclist headi
Oksana_A [137]
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How do i solve that question?
yawa3891 [41]

a) The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }.

b) The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}.

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}.

<h3>How to solve ordinary differential equations</h3>

a) In this case we need to separate each variable (y, t) in each side of the identity:

6\cdot \frac{dy}{dt} = y^{4}\cdot \sin^{4} t (1)

6\int {\frac{dy}{y^{4}} } = \int {\sin^{4}t} \, dt + C

Where C is the integration constant.

By table of integrals we find the solution for each integral:

-\frac{2}{y^{3}} = \frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32} + C

If we know that x = 0 and y = 1<em>, </em>then the integration constant is C = -2.

The solution of this <em>ordinary</em> differential equation is y =\sqrt[3]{-\frac{2}{\frac{3\cdot t}{8}-\frac{\sin 2t}{4}+\frac{\sin 4t}{32}-2   } }. \blacksquare

b) In this case we need to solve a first order ordinary differential equation of the following form:

\frac{dy}{dx} + p(x) \cdot y = q(x) (2)

Where:

  • p(x) - Integrating factor
  • q(x) - Particular function

Hence, the ordinary differential equation is equivalent to this form:

\frac{dy}{dx} -\frac{1}{x}\cdot y = x^{2}+\frac{1}{x} (3)

The integrating factor for the <em>ordinary</em> differential equation is -\frac{1}{x}. \blacksquare

The solution for (2) is presented below:

y = e^{-\int {p(x)} \, dx }\cdot \int {e^{\int {p(x)} \, dx }}\cdot q(x) \, dx + C (4)

Where C is the integration constant.

If we know that p(x) = -\frac{1}{x} and q(x) = x^{2} + \frac{1}{x}, then the solution of the ordinary differential equation is:

y = x \int {x^{-1}\cdot \left(x^{2}+\frac{1}{x} \right)} \, dx + C

y = x\int {x} \, dx + x\int\, dx + C

y = \frac{x^{3}}{2}+x^{2}+C

If we know that x = 1 and y = -1, then the particular solution is:

y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}

The <em>particular</em> solution of the <em>ordinary</em> differential equation is y = \frac{x^{3}}{2}+x^{2}-\frac{5}{2}. \blacksquare

To learn more on ordinary differential equations, we kindly invite to check this verified question: brainly.com/question/25731911

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2 years ago
ANSWER IT FOR BRIANLESTT AND POINTS
kvasek [131]

Answer:

Top right square. It's mirroring the image on the other side of the y axis. So Quadrant 2

4 0
2 years ago
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