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3 years ago

Bruno claims that the solution to the linear equation −3(2x+6)+25=1 is x=1. He shows the steps below to justify his solution.

Mathematics

1 answer:

jonny [76]3 years ago

6 0

Answer:

Bruno is correct x=1

Step-by-step explanation:

-3(2x+6)+25=1

-6x-18+25=1

-6x+7=1

-6x=-6

1=x

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PLEASE HELP! 50 Points!

gulaghasi [49]

The bearing of the plane is approximately 178.037°. $\blacksquare$

<h3>Procedure - Determination of the bearing of the plane</h3><h3 />

Let suppose that <em>bearing</em> angles are in the following <em>standard</em> position, whose vector formula is:

$\vec r = r\cdot (\sin \theta, \cos \theta)$ (1)

Where:

$r$ - Magnitude of the vector, in miles per hour.
$\theta$ - Direction of the vector, in degrees.

That is, the line of reference is the $+y$ semiaxis.

The <em>resulting</em> vector ( $\vec v$ ), in miles per hour, is the sum of airspeed of the airplane ( $\vec v_{A}$ ), in miles per hour, and the speed of the wind ( $\vec v_{W}$ ), in miles per hour, that is:

$\vec v = \vec v_{A} + \vec v_{W}$ (2)

If we know that $v_{A} = 239\,\frac{mi}{h}$ , $\theta_{A} = 180^{\circ}$ , $v_{W} = 10\,\frac{m}{s}$ and $\theta_{W} = 53^{\circ}$ , then the resulting vector is:

$\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})$

$\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]$

Now we determine the bearing of the plane ( $\theta$ ), in degrees, by the following <em>trigonometric</em> expression:

$\theta = \tan^{-1}\left(\frac{v_{x}}{v_{y}} \right)$ (3)

$\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)$

$\theta \approx 178.037^{\circ}$

The bearing of the plane is approximately 178.037°. $\blacksquare$

To learn more on bearing, we kindly invite to check this verified question: brainly.com/question/10649078

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3 years ago

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If x is

Brilliant_brown [7]

The solution to the system of equations is (x, y) = (10, 14), where x is the number of 3-point questions and y is the number of 5-point questions. The appropriate choice is ...
The test contains 10 three-point questions and 14 five-point questions. (2nd selection)

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4 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

3 years ago

I rlly suck at math so plz help lol

Jlenok [28]

Answer:

A. 81

Step-by-step explanation:

Combine fractions, that equals 2/5 in your exponent. Three to the 2/5 power is 81. So answer A.

6 0

3 years ago

Read 2 more answers

Suppose you are standing in a parking lot near a building, and the winter air temperature is 0 degrees Celsius. At that temperat

Naily [24]

The building is 231.7 m far away.Distance can refer to a physical length or an estimate based on other factors in physics or common use.

<h3>What is distance?</h3>

Distance is a numerical representation of the space between two objects or locations. |AB| is a symbol for the distance between two points A and B.

Given data;

Speed of sound = 331 meters per second

Time = 0.7 sec

The distance from the building is;

d=v×t

d=331 m/sec × 0.7 sec

d=231.7 m

Hence the building is 231.7 m far away.

To learn more about the distance refer to the link;

brainly.com/question/26711747

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2 years ago