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2 years ago

Lionel is planning a one-day outing.

Mathematics

2 answers:

pychu [463]2 years ago

6 0

Answer:

Option B - $y=5x+40$ and $y=3x+60$

Step-by-step explanation:

Given : The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride, x. The Splash water park charges an entry fee of $60 and an additional $3 per ride, x.

To find : Which system of equations could be used to determine the solution where the cost per ride of the two amusement parks, y, is the same?

Solution :

Let x be the number of rides and

y be the cost per ride.

According to question,

The Thrill amusement park charges an entry fee of $40 and an additional $5 per ride.

The equation form is $y=40+5x$

The Splash water park charges an entry fee of $60 and an additional $3 per ride.

The equation form is $y=60+3x$

Therefore, The required system of equations form are

$y=5x+40$ and $y=3x+60$

So,Option B is correct.

Lelechka [254]2 years ago

4 0

Answer:

$y= 40+5x$

$y= 60+3x$

Step-by-step explanation:

Thrill amusement park

Entry fee = $40

Cost of 1 ride = $5

Let x be teh no. of rides

Cost of x rides = 5x

So, Total cost = 40+5x

Let the total cost be y

So,Total cost = $y= 40+5x$

Splash water park

Entry fee = $60

Cost of 1 ride = $3

Let x be the no. of rides

Cost of x rides = 3x

So, Total cost = 60+3x

Let the total cost be y

So,Total cost = $y= 60+3x$

So, system of equations could be used to determine the solution where the cost per ride of the two amusement parks, y, is the same is :

$y= 40+5x$

$y= 60+3x$

Hence Option B is true .

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2 years ago

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This is an incomplete question, here is a complete question.

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As we are given:

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Area = A = 7.00 m × 12.0 m

Formula used :

$\Delta P=\frac{1}{2}\times \rho \times v^2$

Now put all the given values in this formula, we get:

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2 years ago

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astraxan [27]

ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
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${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

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