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andrew11 [14]
3 years ago
14

Priya spent 1/2 of the day hiking. She was lost 5/6 ​​of the time that she was hiking. What fraction of the day was Priya lost

Mathematics
2 answers:
Vadim26 [7]3 years ago
5 0

Answer:

\frac{5}{12} fraction of the day was Priya lost

Step-by-step explanation:

As per the statement:

Priya spent 1/2 of the day hiking.

⇒She spent for hiking = 1/2 of the day.

It is also given that:

She was lost 5/6 ​​of the time that she was hiking.

⇒\text{She lost} = \frac{5}{6} \times \text{She was hiking}

Substitute the given values we have;

\text{She lost} = \frac{5}{6} \times \frac{1}{2} = \frac{5}{12} of the day

therefore, \frac{5}{12} fraction of the day was Priya lost

const2013 [10]3 years ago
5 0

5/6 of 1/2= 5/12
answer 5/12
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(Hehe...Please ignore the markings)
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Answer:

See below:

Step-by-step explanation:

Problem 1:

Multiply Equation 1 by 4, keep Equation 2 the same.

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so, the equivalent system is: 4x+4y=32 and x-y=2

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4(y+2)+4y=32---> 4y+8+4y=32--->8y=24---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 1 Answer:

Equivalent system: 4x+4y=32, x-y=2; solution: x=5, y=3

Problem 2:

Keep Equation 1 the same. Add 1 and 2.

To add an equation, add the left sides together, and then the rights.

so: x+y=8 + x-y=2 gives us: 2x=10

solve for x ---> 2x/2=10/2--->x=5

plug x into x+y=8--->5+y=8--->y=3

Problem 2 answer:

Equivalent system: x+y=8, 2x=10; solution:x=5, y=3

Problem 3:

Subtract Equation 2 from 1, and keep 2 the same.

To subtract an equation, subtract the left sides, then the rights. We are subtracting 1<em> from </em>2, so its 2-1.

x-y=2 - x+y=8 gives us: -2y=-6

Solve for y by dividing by -2-->-2y/-2=-6/-2---> y=3

Plug into x-y=2---> x-3=2---> x=5

Problem 3 answer:

Equivalent system: -2y=-6, x-y=2; solution: x=5, y=3

Problem 4:

Multiply the sum of Equation 1 and 2 by a factor of 3. Keep equation 2 the same.

First we add 1 and 2: (we did this earlier) ---> 2x=10 ---> now we multiply it all by 3---> 2x*(3)=10*(3)---> this gives us: 6x=30---> now divide by 6 to solve for x: 6x/6=30/6 gives us: x=5

Now, solve for y by plugging x into equation 2: x-y=2---> 5-y=2--->y=3

Problem 4 answer:

Equivalent system: 6x=30, x-y=2; solution: x=5, y=3

______

Quick Tip: One thing inherent of Equivalent systems is that they have the same set of solutions. Thus, we know the systems are equivalent when they have the same set of solutions for x and y. Moreover, you don't need to solve every time after you attempt to find an equivalent system, instead, just plug in the values found in problem 1 to each new set of equations to test if they are equivalent.

If we find x=5 and y=3 for x+y=8 and x-y=2, then all we have to do is plug them in to 6x=30 and -2y=-6 to see if they are equivalent.

6(5)=30 ---> true

-2(3)=-6 ---> true

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