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elena-14-01-66 [18.8K]

2 years ago

12

2x - 3y = 1 and x + 2y = 18

1 answer:

Yuliya22 [10]2 years ago

5 0

Answer:

x = 8 and y = 5

Step-by-step explanation:

2x - 3y = 1 --(1)

x + 2y = 18

x = 18 - 2y --(2)

sub (2) into (1):

2(18 - 2y) - 3y = 1

36 - 4y - 3y = 1

-7y = 1 -36

7y = 35

y = 5

x = 18 - 2(5) = 8

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3 years ago

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

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

7 0

3 years ago

Find the rectangular coordinates of the point with the polar coordinates ordered pair 7 comma 2 pi divided by 3.

Morgarella [4.7K]

Answer:

$\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

Step-by-step explanation:

The given point is

$\left(7,\dfrac{2\pi}{3}\right)$

We need to find the rectangular coordinates of the given point.

If a polar coordinate is $(r,\theta)$ , then

$x=r\cos theta$

$y=r\sin theta$

In the given point $\left(7,\dfrac{2\pi}{3}\right)$ ,

$r=7,\theta=\dfrac{2\pi}{3}$

Now,

$x=7\cos \dfrac{2\pi}{3}$

$x=7\cos \left(\pi-\dfrac{\pi}{3}\right)$

$x=-7\cos \left(\dfrac{\pi}{3}\right)$

$x=-7\left(\dfrac{1}{2}\right)$

$x=-\dfrac{7}{2}$

and,

$y=7\sin \dfrac{2\pi}{3}$

$y=7\sin \left(\pi-\dfrac{\pi}{3}\right)$

$y=7\sin \left(\dfrac{\pi}{3}\right)$

$y=7\left(\dfrac{\sqrt{3}}{2}\right)$

$y=\dfrac{7\sqrt{3}}{2}$

Therefore, the required point is $\left(-\dfrac{7}{2},\dfrac{7\sqrt{3}}{2}\right)$ .

3 0

2 years ago