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3 years ago

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Mathematics

2 answers:

Mariulka [41]3 years ago

6 0

The true statements about the function y = 3x are I, II, IV ⇒ 4th

Step-by-step explanation:

The direct variation is:

A mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other
The constant is called the constant of variation
The direct variation is represented graphically by a line passes through the origin
If y varies directly with x, then the equation of variation is y = kx, where k is the constant of variation

∵ y = 3x

- One variable is equal to the product of a constant and the

other variable

∵ The equation is in the form of the direct variation

∴ The function is a direct variation ⇒ I

∵ The direct variation represented by a line passes through the origin

∴ y = 3x is represented graphically by a line passes through the origin

∴ The function goes through the origin ⇒ II

∵ y = 3x is a direct variation

∴ 3 represents the constant of variation

∴ The constant of variation is not 1/3 ⇒ III

∵ y = 3x

∵ x = 2

- Substitute x by 2 to find y

∴ y = 3(2)

∴ y = 6

∴ y is 6 when x is 2 ⇒ IV

The true statements about the function y = 3x are I, II, IV

Learn more:

You can learn more about the variation in brainly.com/question/10708697

#LearnwithBrainly

-Dominant- [34]3 years ago

3 0

I do Monarch/Lifepac and the answer is I, II, IV. Hope this helps!

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3 years ago

I need help with this equation please!<br><br> 15- w/4 = 28

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Answer:

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Step-by-step explanation:

We simplify the equation to the form, which is simple to understand

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

$\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy$

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

$\bf G(x,y) = (x, y, 3-x^2-y^2)$

with 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)$

we also have

$\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)$

and so

$\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2$

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

$\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy$