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puteri [66]
3 years ago
12

one family spends $134 on 2 adult tickets and 3 youth tickets at an amusement park. Another family spends $146 on 3 adult ticket

s and 2 youth tickets at the same park. What is the price of a youth ticket?
Mathematics
2 answers:
mojhsa [17]3 years ago
7 0

Step-by-step explanation:

let

adult ticket = x

youth ticket = y

1). 2x+3y = 134

2). 3x+2y = 146

*then multiply line 1). by the first number of line 2).

*then multiply line 2). by the first number of line 1).

1). (2x+3y = 134 ) x 3 = (6x + 9y = 402)

2). (3x+2y = 146) x 2 = (6x + 4y = 292)

*subtract new line 2). from new line 1).

5y = 110

y = 110/5 = $22

This is the value for youth ticket.

If another question asks for adult you do as follows:

*put this value for y in either original equation, I'll chose line 1).

2x+3y = 134

2x + (3x22) = 134

2x + 66 = 134

2x = 68

X = 68/2 = $34.

Hope the explanation helps, let me know if you need any further explanation

Eddi Din [679]3 years ago
3 0

Answer:

Let x be the price of an adult ticket and y be the price of a youth ticket.

One family spends $134 on 2 adult tickets and 3 youth tickets so we can write: 2x+3y=134

Another family spends $146 on 3 adult tickets and 2 youth tickets so we can write: 3x+2y=146

We need to find y so we eliminate x. Multiply (1) by 3 and (2) by 2 to obtain (3) and (4): 6x+9y=402

6x+4y=292

Subtract (3) and (4) and solve for y: 5y=110

y=22→$22

Step-by-step explanation:

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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.
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Answer:

The first three nonzero terms in the Maclaurin series is

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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} ) }

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

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