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salantis [7]
3 years ago
7

Jeff wants to buy tickets for a football game for his friends. The stadium is selling tickets for $7 each. The stadium requires

a
person to buy at least 4 tickets, and at most 8 tickets. Let x represent the number of tickets, and y represent the total cost.
What is a reasonable range for the price Jeff would pay for tickets?
Group of answer choices



{11, 18, 25, 32, 39}

{28, 35, 42, 49, 56}

{4, 5, 6, 7, 8}

{28, 42, 56, 70, 84}
Mathematics
1 answer:
aniked [119]3 years ago
7 0

Your answer is D){28, 42, 56, 70, 84}

To answer this question you just have to make sure the amount he pays is a multiple of 7.If you know you times table then this should be easy

Times table of 7:

7,14,21,28,35,42,49,55,63,70,77,84.

Now you just need to find an answer choice who has these numbers.Lets check it out!

A) 11 isn't a multiple of 7

B) 56 isn't a multiple of 7

C) None of these are multiples of 7 except 7 so this is also incorrect

D) This is correct because they are all multiples of 7

Hope this makes sense.

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The answer is D

Step-by-step explanation:

Because the whole length is 29.1 so you would subtract the part you have which is 16.8 so 29.1-16.8=12.3

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Step-by-step explanation:

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Which sequence of transformations confirms congruence by mapping shape I onto shape II?
Lubov Fominskaja [6]

Answer:

  • Neither of the choices is correct.

Explanation:

<u>1. Coordinates of the vertices of the figure I (preimage):</u>

  • (10, - 5)
  • (15, -5)
  • (10, - 10)
  • (15, -10)

<u />

<u>2. Coordinates of the vertices of the figure II (image):</u>

  • (0,10)
  • (0,15)
  • (-5,10)
  • (-5, 15)

Since many different rigid transformations can map the figure I into the figue II, you will need to use trial and error.

The most important is to do it in an educated way.

I will start by eliminating some options.

The first option, a reflection across the x-axis and 15 units left, does not work, because the reflection across the x-axis would shift the figure to a lower position than what you need.

The third option, a 90º counterclokwise rotation about the origin then a translation 10 units up, would move the figure to the second quadrant, and we need it in the third quadrant.

I will try now with the second choice, a 90º clockwise rotation and then 25 units up:

First, a 90º clockwise rotation, which is the same that a 270º counterclockwise rotation, follows the rule (x, y) → (y, -x)

Then, that results in:

  • (10, - 5) → (-5, -10)
  • (15, -5) → (-5, -15)
  • (10, - 10) → (-10, -10)
  • (15, -10) → (-10, -15)

Now, you can see that shifting 25 units up will not work, because you need that two x-coordinates become 0 (zero). So, this is not the correct set of transformations either.

A 180º rotation about the origin and a translation 10 units right follow this chain of rules:

  • (x, y) → (-x, -y) → (-x + 10, -y)

That means:

  • (10, - 5) → (-10,5) → (-10 + 10, 5) = (0, 5)
  • (15, -5) → (-15, 5) → (-15 + 10, 5) = (-5, 5)
  • (10, - 10) → (-10, 10) → (-10 + 10, 10) = (0, 10)
  • (15, -10) → (-15, 10) → (-15 + 10, 10) = (-5, 10)

        These last points do not coincide either with the vertices of the figure II.

In conclusion, neither of the choices gives the correct answer to the question.

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I really need help with this question, and I would really appreciate it! I will mark brainliest!
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Answer:

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1/3n = 6
n = 6 * 3
n = 18
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