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chubhunter [2.5K]
2 years ago
8

Anna wants to take fitness classes. She compares two gyms to determine which would be the best deal for her. Fit Fast charges a

set fee per class. Stepping Up charges a monthly fee, plus an additional fee per class. The system of equations models the total costs for each.
y = 7.5x

y = 5.5x + 10

1. Substitute: 7.5x = 5.5x + 10

How many classes could Anna take so that the total cost for the month would be the same?

classes

What is the total monthly cost when it is the same for both gyms?
Mathematics
1 answer:
lutik1710 [3]2 years ago
7 0

The number of classes Anna can take so the total cost for the month will be the same is 5.

The monthly cost would be  $37.50.

<h3>When would the total cost be the same?</h3>

When the monthly cost is equal, both equations would be equal: 7.5x = 5.5x + 10

In order to determine the value of x, take the following steps:

  • Combine similar terms: 7.5x - 5.5x = 10
  • Add similar terms: 2x = 10
  • Divide both sides by 2 : 10 /2 = 5

Monthly cost when the cost is the same: 7.5 x 5 = $37.50

To learn more about cost, please check: brainly.com/question/25711114

#SPJ1

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3 years ago
Graph and label the image of the figure below after a dilation by a factor of 1/2.
neonofarm [45]
Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\  \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\  \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\  \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

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