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4 years ago

The linear equation 3x – 11y = 10 has

Mathematics

1 answer:

storchak [24]4 years ago

7 0

Answer:

Correct option: c) infinitely many solutions

Step-by-step explanation:

We have an equation with two variables, so as we have a system with less equations (just one) than variables (two), we have an infinite number of solutions.

To find some solutions, we can choose values for x and then just calculate the value of y:

x = 0:

-11y = 10 -> y = -10/11

x = 1:

3 - 11y = 10 -> y = -7/11

x = 2:

6 - 11y = 10 -> y = -4/11

And so on.

So the correct option is c): infinitely many solutions

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Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 505050 meters.

Dennis_Churaev [7]

Answer= 70.7 meters.

Step-by-step explanation:

We have been given that Elise walks diagonally from one corner of a square plaza to another. Each side of the plaza is 50 meters.

Since we know that diagonal of a square is product of side length of square and . So we will find diagonal of our given square plaza by multiplying 50 by .

Therefore, diagonal distance across the plaza is 70.7 meters.

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3 years ago

Natalie walked 3/5 mile in 1/2 hour. How many miles did she walk per hour

Marizza181 [45]

Answer:

She walked 1 1/5 miles per hour.

Step-by-step explanation:

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3 years ago

Write an equivalent fractions for 7x+4x+3-1

Diano4ka-milaya [45]

11x + 2 is combining like terms

4 0

3 years ago

Read 2 more answers

Algebra 1. Please show work.

MAVERICK [17]

Try this:
Given:
$\left \{ {{x-y=1} \ [1] \atop {x+y=3} \ [2]} \right.$
1) [1]+[2]: 2x=2 ⇒ x=2
2) [2]-[1]: 2y=1 ⇒ y=1.
3) answer: (2;1)

7 0

3 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

3 years ago