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Black_prince [1.1K]
3 years ago
14

When x = 2, y = 50 and when x = 4, y = 100. Which direct variation equation can be used to model this function?

Mathematics
1 answer:
Karolina [17]3 years ago
3 0
Well, let's first write these as points.

( 2 , 50 )

( 4 , 100 )

We can see that when "x" is reduced by 2, "y" is reduced by 50. This means that if we reduce "x" by 1, "y" will be reduced by 25. Thus, we can say that if "x" is 1, "y" will be 25.

( 1 , 25 )

What we know that 25 * 1 = 25, and that 2 * 25 = 50. We can see that multiplying "x" by 25 will give us our "y". We can now write this as an equation.

y = 25x
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Which like on the graph represents the equation y+4=-2x/3?
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Answer:

1

Step-by-step explanation:

8 0
3 years ago
Solve the following inequality for y.<br> 18x + 6y &gt; 12
Murljashka [212]

Answer:

12

Step-by-step explanation:

Simplifying

18x + -6y = 12

Solving

18x + -6y = 12

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '6y' to each side of the equation.

18x + -6y + 6y = 12 + 6y

Combine like terms: -6y + 6y = 0

18x + 0 = 12 + 6y

18x = 12 + 6y

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3 0
3 years ago
51% of the tickets sold at a school carnival were early-admission tickets. If the school sold 100 tickets in all, how many early
saul85 [17]

Answer:

51 were the total number of the tickets sold at a school carnival were early-admission tickets.

Step-by-step explanation:

Total number of tickets sold by  = 100

Let x be the early-admission tickets sold by the school.

As 51% of the tickets sold at a school carnival were early-admission tickets.

so

x=\frac{51}{100}\times 100

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Therefore, 51 were the total number of the tickets sold at a school carnival were early-admission tickets.

5 0
3 years ago
part 3. Find the coordinates of the vertices of the triangle after a reflection across the y-axis and then across the line y= 2.
garik1379 [7]

Answer:

  c.  S'(3, 1), T'(1, -1), U'(0, 1)

Step-by-step explanation:

Reflection across the y-axis negates the x-coordinate, so is equivalent to the transformation ...

  (x, y) ⇒ (-x, y)

Reflection across the horizontal line y=c is equivalent to the transformation ...

  (x, y) ⇒ (x, 2c-y)

So, the combined reflections are equivalent to the transformation ...

  (x, y) ⇒ (-x, 4 -y)

Then we have ...

  S(-3, 3) ⇒ S'(-(-3), 4-3) = S'(3, 1)

  T(-1, 5) ⇒ T'(-(-1), 4-5) = T'(1, -1)

  U(0, 3) ⇒ U'(-(0), 4-3) = U'(0, 1) . . . . matches choice C

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3 years ago
MARKIN BRAINIEST !!!
Angelina_Jolie [31]

Answer:

B-1, C-2, A-3

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