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3 years ago

HELP PLZ!!! this is a very difficult question

Mathematics

1 answer:

antoniya [11.8K]3 years ago

6 0

Answer:

251.2 cm²

Step-by-step explanation:

Area of cylinder=2πrh+2πr^2

which is also=2πr(r+h)

applying formula

area=251.2 cm²

hope this helps plzz mark me brainliest

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Solve x2−4x−12=0 by completing the square. A x=−2 B x=−6 or x=2 C x=−2 or x=6 D x=−6

MrRissso [65]

Answer:

B x=-6

x=2

Step-by-step explanation:

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2 years ago

One rectangle has a base of 12 inches and an altitude of 6inches, The other has a base of 10 inches and an altitude of 5 inches.

liubo4ka [24]

Sides of first rectangle are 12 inches and 6 inches

and sides of second rectangle are 10 inches and 5 inches

So to find the ration of sides we can do

Ratio of bases = 12/10 = 6/5

Ratio of altitudes = 6/5

Area of first rectangle = base * altitude = 12 *6 = 72 square inches

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3 years ago

Which of the following graphs represents a function?

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Step-by-step explanation:

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3 years ago

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Solve for x using the given diagram

blagie [28]

Answer:

Step-by-step explanation:

Look a t the picture.

The triangles on the picture are similar.

Therefore the sides are in proportion:

$\dfrac{8x}{10}=\dfrac{4}{5x}$ <em>cross multiply</em>

$(8x)(5x)=(4)(10)$

$40x^2=40$ <em>divide both sides by 40</em>

$x^2=1\to x=\sqrt1\\\\x=1$

7 0

3 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

3 years ago