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3 years ago

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A cargo ship left port A and is headed across the ocean to shipping port B. After one month, the ship stopped at a refueling sta

tion along a path described by a vector with components LeftAngleBracket 14, 23 RightAngleBracket. After another month, on the same path, the ship reached port B, twice the distance from port A as the fueling station.
What are the characteristics of the vector representing the path of the ship?

components:LeftAngleBracket 7, 11.5 RightAngleBracket, magnitude: 13.46
components:LeftAngleBracket 7, 11.5 RightAngleBracket, magnitude: 53.85
components:LeftAngleBracket 28, 46 RightAngleBracket, magnitude: 13.46
components:LeftAngleBracket 28, 46 RightAngleBracket, magnitude: 53.85

2 answers:

Sergeu [11.5K]3 years ago

8 0

Answer:

D

Step-by-step explanation:

Edge 2020

Yuri [45]3 years ago

8 0

Answer:

d

Step-by-step explanation:

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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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