Advertisements
Advertisements
प्रश्न
Find the mean proportional to (x – y) and (x3 – x2y).
Advertisements
उत्तर
Let the mean proportional to (x – y) and (x3 – x2y) be n.
`=>` (x – y), n, (x3 – x2y) are in continued proportional
`=>` (x – y) : n = n : (x3 – x2y)
`=>` n2 = x2(x – y)(x – y)
`=>` n2 = x2(x – y)2
`=>` n = x(x – y)
संबंधित प्रश्न
If a/b = c/d prove that each of the given ratio is equal to `sqrt((3a^2 - 10c^2)/(3b^2 - 10d^2))`
If a, b and c are in continued proportion, prove that: a: c = (a2 + b2) : (b2 + c2)
What least number must be added to each of the numbers 5, 11, 19 and 37, so that they are in proportion?
Show that the following numbers are in continued proportion:
48, 60, 75
If `a/c = c/d = c/f` prove that : `bd f[(a + b)/b + (c + d)/d + (c + f)/f]^3` = 27(a + b)(c + d)(e + f)
If a, b, c are in continued proportion, prove that: `(1)/a^3 + (1)/b^3 + (1)/c^3 = a/(b^2c^2) + b/(c^2a^2) + c/(a^2b^2)`
If a, b, c, d are in continued proportion, prove that: (a2 – b2) (c2 – d2) = (b2 – c2)2
Find two numbers whose mean proportional is 16 and the third proportional is 128.
Determine if the following ratios form a proportion. Also, write the middle terms and extreme terms where the ratios form a proportion.
39 litres : 65 litres and 6 bottles : 10 bottles
Unequal masses will not balance on a fulcrum if they are at equal distance from it; one side will go up and the other side will go down.
Unequal masses will balance when the following proportion is true:
`("mass"1)/("length"2) = ("mass"2)/("length"1)`
Two children can be balanced on a seesaw when
`("mass"1)/("length"2) = ("mass"2)/("length"1)`. The child on the left and child on the right are balanced. What is the mass of the child on the right?
