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3 years ago

Jordan has

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

(1/3)/(5)= 1/3 * 1/5 = 1/15 of a chocolate bar each friend gets

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PLEASSSSSSSSSEEEEEEEEEEEE IS NOT A SCAMMM" I JUST NEED HELP

emmainna [20.7K]

Answer:

Step-by-step explanation:

5 0

2 years ago

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Triangle PQR has vertices P(–2, 6), Q(–8, 4), and R(1, –2). It is translated according to the rule (x, y) → (x – 2, y – 16).

Amanda [17]

Since you subtract 16 from the y value, P is (-2, 6), and the y value comes second (meaning that it's 6), we get 6-16=-10

4 0

3 years ago

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A vertical shift of 3 units down and a horizontal shift of 2 units to the right

Lubov Fominskaja [6]

Depends on what the coordinates are to get an exact answer but you go down on your y axis 3 units (so if it was at 0 it would become -3) and to the right on your x axis, which means the numbers are increasing ( so if it was 3 it would become 5.)

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3 years ago

9) Nancy starts a race at the start line and she is running 3 meters per second. Juan starts the same race 3 meters ahead of Nan

Travka [436]

Given:

Nancy is running 3 meters per second.

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second.

To find:

The equations for Nancy and Juan.

Solution:

Let x be the number of seconds.

Nancy is running 3 meters per second. So, the total distance covered by Nancy in the race is

$y=3x$

Juan starts the same race 3 meters ahead of Nancy but he is going at 2 meters per second. So, the total distance covered by Juan in the race is

$y=3+2x$

Therefore, the equations of Nancy and Juan are $y=3x$ and $y=3+2x$ respectively.

5 0

3 years ago

How many gallons of a 80% antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antif

konstantin123 [22]

440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze

Solution:

Let "x" be the gallons of 80 % antifreeze added

Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze

Final mixture is x + 80

Therefore, we can frame a equation as:

"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze

Thus, we get,

x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %

$x \times \frac{80}{100} + 80 \times \frac{15}{100} = (x+80) \times \frac{70}{100}\\\\0.8x + 80 \times 0.15 = (x+80) \times 0.7\\\\0.8x+12 = 0.7x+56\\\\0.8x-0.7x=56-12\\\\0.1x = 44\\\\x = \frac{44}{0.1}\\\\x = 440$

Thus 440 gallons of 80 % antifreeze solution must be mixed

4 0

3 years ago