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Tatiana [17]
3 years ago
6

Plssss helppp!!!!Find the value of x.

Mathematics
2 answers:
alina1380 [7]3 years ago
6 0

Answer:

See how X looks similar to a obtuse angle? The angle looks an awful lot like it's around 40 degrees though, it seems to be an acute angle, they are less than 90* so anything near 90 should make sense. I'd say exactly 45.

Step-by-step explanation:

CaHeK987 [17]3 years ago
3 0

Answer: I got the concluding answer as "x = 1" but I'm not sure if that's correct. I apologize if it's not! Good Luck!

Step-by-step explanation:

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52% = 0.52

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So your answer is 43.68.

Hope this answer helps! feel free to ask any additional questions :)
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4 years ago
The length of a rectangle is six times it’s width.If the area of the rectangle is 486 inches squared find its perimeter.
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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

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
How many solutions are there for the system of equations of shown on the graph?
bearhunter [10]

Answer:

one solution

(second option listed)

Step-by-step explanation:

We can that these two lines, each representing one equation/function, only meet at one specific value.

In a system of equations, we are essentially looking for a solution that works for both equations.

So, if both lines share a point/value (meaning they intersect), that point is a solution to the system of equations.

Because these lines only overlap at one point, this system of equations has one solution.

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