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3 years ago

Please can someone help me work out the surface area

Mathematics

1 answer:

alexira [117]3 years ago

4 0

Answer:

The answer to your question is Area = 620 m²

Step-by-step explanation:

Process

Find the area of the different parts of the prism and later added them.

1.- Area of the base

Area of a rectangle = 10 m x 24 m = 240 m²

2.- Area of the lateral sides (there are 2)

Area of a rectangle = 10 x 13 x 2 = 260 m²

3.- Area of the triangles (there are 2)

Calculate the height using the pythagorean theorem

13² = 12² + b²

b² = 13² - 12²

b² = 169 - 144

b² = 25

b = 5

Area = 1/2 (24 x 5) x 2 = 120 m²

4.- Total area

Area = 120 + 260 + 240

= 620 m²

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Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

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3 years ago

Which regular polygon has a minimum rotation of 45 degrees to carry the polygon onto itself?

lubasha [3.4K]

Answer:

Hence, a regular polygon that has a minimum rotation of 45 degrees to carry the polygon onto itself is:

A. octagon.

Step-by-step explanation:

We know that the for a regular polygon with 'n' sides the minimum rotation that can carry the polygon onto itself is:

$\dfrac{360\degree}{n}$

Here we are given the minimum angle of rotation such that polygon is transformed to itself as 45 degree.

i.e. we have:

$\dfrac{360}{n}=45\\\\n=\dfrac{360}{45}\\\\n=8$

Hence, the number of sides of a polygon are 8.

Hence, the polygon is:

A. octagon.

( Since a polygon with 8 sides is called a octagon)

7 0

3 years ago

Read 2 more answers

How would you do this?

sdas [7]

You can't:)
destroy the paper
(*´・ｖ・)ε/̵͇̿̿/'̿̿ ̿ ̿ ̿ ̿ ̿

7 0

4 years ago

Read 2 more answers

Please can someone help me work out the surface area​

Please can someone help me work out the surface area