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one possible answer is 1/4

another one is 25/100

and 2/8

and 3/12.............................

the list goes on

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Q: What r three fractions equivalent to 0.25?
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How do you add similar proper fraction?

Two fractions are similar if they have the same denominator.So if p/r and q/r are two such fractions, then p/r + q/r = (p+q)/r.


How do you do mixed fractions?

The mixed fraction PQ/R = (P*R + Q)/R


What r the three different fractions between 5over one and 3 over 4?

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The mixed fraction PQ/R = (P*R + Q)/R


Give the law of combination of resistance in series?

In series combination, Current(I) remains same among all the resistors, but voltage(V) changes. So: V(Equivalent) = I(Equivalent) * R(Equivalent) R(Equivalent) = V(Equivalent)/ I(Equivalent) R(Equivalent) = IR1+ IR2+...+IRn / I R(Equivalent) = I(R1+R2+...+Rn)/ I R(Equivalent) = R1+R2+...+Rn


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4/16 2/8 3/12 8/32 5/20 those r some of the fractions equivalent to 1/4


What is the area of a circle with a circumference of 410 meters?

Circumference = 2pi*r 410 = 2pi*r r = 205/pi A = pi*r^2 A = pi*(205/pi)^2 A = 205^2/pi metres A = 42 025/pi metres


What are adding fractions?

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What do you do when your adding fractions and the numerator ends up larger than the denominator?

You could leave it as an improper fraction. Alternatively, you could divide the numerator by the denominator (d) to find the number of times it goes into it (w) and the remainder (r). Then the equivalent mixed ratio is w r/d.


What is the sum of all the positive proper fractions with denominators less than or equal to 100?

Consider a denominator of r; It has proper fractions: 1/r, 2/r, ...., (r-1)/r Their sum is: (1 + 2 + ... + (r-1))/r The numerator of this sum is 1 + 2 + ... + (r-1) Which is an Arithmetic Progression (AP) with r-1 terms, and sum: sum = number_of_term(first + last)/2 = (r-1)(1 + r-1)/2 = (r-1)r/2 So the sum of the proper fractions with a denominator or r is: sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2 Now consider the sum of the proper fractions with a denominator r+1: sum{r+1} = (((r+1)-1)/2 = ((r-1)+1)/2 = (r-1)/2 + 1/2 = sum{r) + 1/2 So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2 The first denominator possible is r = 2 with sum (2-1)/2 = ½; The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½; And there are 100 - 2 + 1 = 99 terms to sum So the required sum is: sum = ½ + 1 + 1½ + ... + 49½ = 99(½ + 49½)/2 = 99 × 50/2 = 2475


When did we start using fractions?

If u r an idiot than get on this website so everyone get on.


Can you always find another fraction in between two fractions and why?

Yes. The set of rational numbers is infinitely dense.If p/q and r/s are any two fractions then (p/q + r/s)/2 is a fraction which is between the two.