135% as a fraction or a mixed number fully simplified please

Mathematics

1 answer:

solong [7]3 years ago

6 0

Answer:

135% = 27/20 = 17/20 as a fraction

Step-by-step explanation:

Step 1: Write down the percent divided by 100 like this:

135% = 135/100

Step 2: Multiply both top and bottom by 10 for every number after the decimal point:

As 135 is an integer, we don't have numbers after the decimal point. So, we just go to step 3.

Step 3: Simplify (or reduce) the above fraction by dividing both numerator and denominator by the GCD (Greatest Common Divisor) between them. In this case, GCD(135,100) = 5. So,

(135÷5)/(100÷5) = 27/20 when reduced to the simplest form.

As the numerator is greater than the denominator, we have an IMPROPER fraction, so we can also express it as a MIXED NUMBER, thus 135/100 is also equal to 1 7/20 when expressed as a mixed number.

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An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the ref

Mazyrski [523]

Answer:

7.85 km

Step-by-step explanation:

Let x represent the distance from the refinery to point P. Then the distance under the river from point P to the storage tanks is ...

√(2² +(9 -x)²) = √(x² -18x +85)

In units of $200,000, the cost of the pipeline will be ...

c = x + 2√(x² -18x +85)

The derivative with respect to x is ...

dc/dx = 1 +(2x -18)/√(x² -18x +85)

When we set this to zero, we can get the equation ...

√(x² -18x +85) +(2x -18) = 0

And this can be rewritten as ...

3x^2 -54x +239 = 0

which has solution

x = 9 -(2/3)√3 ≈ 7.8453

Point P should be located about 7.85 km from the refinery.

_____

It is interesting to note that the angle the pipe makes with the riverbank is given by arccos(1/2), where the 1/2 is the ratio of overland to under-river costs. That angle is 60°, so the distance along the riverbank from the storage facility to point P is 2cot(60°) = (2/3)√3 ≈ 1.1547 km. You will recognize this as the value subtracted from 9 km in the solution above.

This "angle solution" is the generic solution to this sort of problem where the route costs are different and part of the route is along the edge of the higher-cost path.