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3 years ago

Can someone plz help me with this one!!!!

Mathematics

2 answers:

AfilCa [17]3 years ago

5 0

The answer to the question is yes.

luda_lava [24]3 years ago

3 0

Answer:

Yes

Step-by-step explanation:

Plug in the x and y for the x and y coordinates to get

-95 = -1+(-94)

simplify to

-95 = -1-94

solve

-95 = -95

Hope this helps!

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Which expressions are equivalent to (5⋅x)⋅3 ? Drag and drop the equivalent expressions into the box.

marysya [2.9K]

<h2>Greetings!</h2>

Answer:

3⋅(5⋅x)

5⋅(x⋅3)

15x

Step-by-step explanation:

As the values are inside the brackets, it does not matter what side the (x3) is on, so 3⋅(5⋅x) is equivalent.

Multiplying the contents of the brackets in the third one (x * 3) by 5 gives the same value as 3 * (x * 5) so 5⋅(x⋅3) is also equivalent.

On multiplying the brackets out:

5 * x = 5x

5x * 3 = 15x

So 15x is also equivalent.

<h2>Hope this helps!</h2>

3 0

3 years ago

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AnnZ [28]

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3 years ago

196 divided by 3.5 show your work

Ilya [14]

Answer:

Step-by-step explanation:

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196/3.5= 56

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3 years ago

You are renting a car that charges a $30 fee plus 40 cents a mile. The rate of change

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Answer:

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3 years ago

Read 2 more answers

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

5 0

3 years ago