The mixed fraction PQ/R = (P*R + Q)/R
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
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.
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.
Yes. a/b / u/v / r/s = (a*v)/(b*u) / r/s = (a * v * s) / (b * u * r)
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.
The mixed fraction PQ/R = (P*R + Q)/R
4/4, 8/4, 12/4
The mixed fraction PQ/R = (P*R + Q)/R
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
4/16 2/8 3/12 8/32 5/20 those r some of the fractions equivalent to 1/4
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
adding fractions r easy math prblems. like 1/4 +1/4 = 1/2.
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.
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
If u r an idiot than get on this website so everyone get on.
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.