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2 years ago

Which combined inequality has the same solution set shown in the graph of x?

Mathematics

1 answer:

alexandr402 [8]2 years ago

4 0

Step-by-step explanation:

an open (empty) point means the point itself is not included.

a closed (filled) point means the point itself is included.

the x values going from -3 (not included) to -1 (included) means

-3 < x <= -1

so, the second answer is correct.

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Solve<br> 6(x-3) + 10 = 2(4x-5)

Firdavs [7]

6x-8=8x-10 then 2=2x which means x=1

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3 years ago

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What is an equation of the line that passes through the point (-2 1) and is parallel to the line whose equation is 4x-2y=8 ?

monitta

4x-2y=8 ==> y=2x-4 (to better show you what I mean)

2 lines are parallel if they have the same slope (m the coefficient of x).

So y=mx + b // y =2x-4 means m=2 & y=mx + b, becomes y=2x + b.
Moreover they are telling us that this function passes by (-2,1), where -2 represents x & 1, represents y. To calculate b , replace x & y by their values:
y=2x+b ==> 1 = 2(- 2) + b==> 1 = -4 + b ==> b= 5. Finally the equation is
y=-2x+5

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3 years ago

Whats file hosting//////////////////

julsineya [31]

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Step-by-step explanation: there u go

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3 years ago

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When the expression 3(n + 7) is evaluated for a given value of n, the result is 33. What is

trapecia [35]

Answer:4

Step-by-step explanation:What +7=11 bause the () make a multiplication and 11x3=33

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3 years ago

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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

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

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

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3 years ago