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2 years ago

Convert 149g/cm to kilograms per meter

1 answer:

5 0

Set it up kind of like a picket fence. Put 149 g over one cm and multiply that by 1kg over 1000g then multiply that by 1 cm over .01 m. The answer is 14.9 kg/m

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Classify the polynomial:

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Answer:

A. Cubic binomial.

Step-by-step explanation:

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3 years ago

Suppose the number of prisoners held in a certain type of prison, measured in thousands, can be described by the polynomial - 2.

horsena [70]

Insufficient dvdsuicnfbf

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2 years ago

Please helpp

Kisachek [45]

Answer:

13,784

Step-by-step explanation:

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3 years ago

Given a term in the geometric sequence and the common ratio find the first five terms the explicit form and the recursive formul

Pani-rosa [81]

A1 = 4
a2 = 5a1 = 5 x 4 = 20
a3 = 5a2 = 5 x 20 = 100
a4 = 5a3 = 5 x 100 = 500
a5 = 5a4 = 5 x 500 = 2,500

Tn = ar^(n-1); where a = 4, r = 5
Tn = 4(5)^(n-1) = 4/5 (5)^n
Explicit formular is Tn = 4/5 (5)^n

Recursive formular is $a_{n+1} = 5a_{n-1}$

5 0

2 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

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

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

5 0

2 years ago