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2 years ago

Can anyone can tell me that why 2 by 2 is 4

Mathematics

1 answer:

V125BC [204]2 years ago

7 0

I mean cause 2 x 2 is 4

So 2+2 is 4 you’re just putting 2 into 4 twice

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Step-by-step explanation:

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2 years ago

Alonzo's suitcase can weigh no more than 20 kilograms if he wants to take it on the airplane without paying a fee. His suitcase

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He will have to take out 20 hectograms in order to meet the weight limit.

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3 years ago

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A given line has the equation 10x 2y=−2 . what is the equation, in slope-intercept form, of the line that is parallel to the giv

aivan3 [116]

The equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 1

<h3>Equation of a line</h3>

A line is the distance between two points

Given the equation of a line expressed as 10x + 2y = -2. Determine the slope

2y = -10x -2

y = -5x - 1

Slope of the line is -5

The equation of a line in point-slope form is y - y1 = m(x-x1)

Substitute the point and the slope of the parallel line

y - 12 = -5(x - 0)
y - 12 = -5x
y = -5x + 12

Hence the equation, in slope-intercept form, of the line that is parallel to the given line and passes through the point (0, 12) is y = -5x + 12

Learn more on equation of a line here: brainly.com/question/13763238

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3 0

1 year ago

Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Charra [1.4K]

Separating variables, we have

$\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx$

Integrate both sides.

$\displaystyle \int y\,dy = \int x\,dx$

$\dfrac12 y^2 = \dfrac12 x^2 + C$

Given that $y(0)=-1$ , we find

$\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12$

Then the particular solution is

$\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12$

$y^2 = x^2 + 1$

$y = \pm\sqrt{x^2 + 1}$

and because $y(0)=-1$ , we take the negative solution to accommodate this initial value.

$\boxed{y(x) = -\sqrt{x^2+1}}$

7 0

1 year ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

2 years ago