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Sunny_sXe [5.5K]
3 years ago
7

*in a science experiment, farah mixed a salt solution and vinegar in the ratio 3 : 7. a. if she used 262.8 milliliters of salt s

olution, how much vinegar did she use? b. if 0.56 liter of vinegar was used, how much salt solution did she use?
Mathematics
1 answer:
Basile [38]3 years ago
7 0

n a science experiment, farah mixed a salt solution and vinegar in the ratio 3 : 7.

A.) if she used 262.8 milliliters of salt solution, the amount of vinegar is equal to:

X = 262.8 ml ( 7 ml vinegar/ 3 ml salt soln)

X =  613.2 ml vinergar

B.) y = 0.56 L vinegar ( 3 L salt soln/ 7 L vinegar)

Y = 0.24 L salt soln

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Answer:

Step-by-step explanation:

The triangle would end up back where it started.  It is hard to explain without a graph.  If you have graph paper, you might want to try drawing this out.  Say that the original points are at A (1,2) B (2,0) and C(0,0).  Now, when we reflect the points over the x axis, they will be the same distance below the x that the points original were about the x axis.  Since A was 2 units above the axis, it will now be 2 units below at (1, -2).  Points B and C will stay on the x axis and will remain in place at B(2.0) and C(0,0).  Since these points are on the line, they were not above the x axis, so they will now not be below the x axis.

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The last thing left it to rotate this final triangle 180 degrees.  Since a circle is 360 degrees and 180 is half of a circle, it does not matter if we rotate clockwise or counter-clockwise.  If you could trace our new triangle and put a plus sign at the origin (0,0).  You would put your pencil on the origin  and rotate the two turns at the plus sign.  This would put your triangle right back to the beginning.  So the original value of B would be the same.  In this case C ((2,0)

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4 0
3 years ago
Mark and peter went to an arcade where the machines took tokens. Marilk played 9 games of ping pong and 5 games of pinball, usin
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Answer:

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Let, each game of ping pong requires x number of tokens and each game of pinball requires y number of tokens.

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