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Bogdan [553]
3 years ago
6

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!

You might be interested in
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
Please help me I'm stuck​
AnnZ [28]
I don’t see any typing ?
6 0
3 years ago
196 divided by 3.5 show your work
Ilya [14]

Answer:

56

Step-by-step explanation:

Sorry for not showing my work im a mental worker

196/3.5= 56

5 0
3 years ago
You are renting a car that charges a $30 fee plus 40 cents a mile. The rate of change
Dmitry [639]

Answer:

true

is the answer

7 0
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

So

\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
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