Advertisements
Advertisements
प्रश्न
Find the mean proportional of the following:
17.5, 0.007
Advertisements
उत्तर
Let x be the mean proportional between 17.5 and 0.007.
Then,
17.5 : x :: x : 0.007
⇒ 17.5, x and 0.007 are in continued proportion.
⇒ `17.5/x = x/0.007`
⇒ x × x = 17.5 × 0.007
⇒ x2 = 0.1225
⇒ x = `sqrt0.1225`
⇒ x = 0.35
APPEARS IN
संबंधित प्रश्न
If a, b, c and d are in proportion prove that `sqrt((4a^2 + 9b^2)/(4c^2 + 9d^2)) = ((xa^3 - 4yb^3)/(xc^3 - 5yd^3))^(1/3)`
Check whether the following numbers are in continued proportion.
3, 5, 8
If a, b, c are in continued proportion and a(b – c) = 2b, prove that: `a - c = (2(a + b))/a`.
Find the fourth proportional to:
x3 - y2, x4 + x2y2 + y4, x - y.
If `x/a = y/b = z/c`, show that `x^3/a^3 - y^3/b^3 = z^3/c^3 = (xyz)/(zbc).`
Choose the correct answer from the given options :
The fourth proportional to 3, 4, 5 is
Write the mean and extreme terms in the following ratios and check whether they are in proportion.
400 gm is to 50 gm and 25 rupees is to 625 rupees
Which of the following ratios are in proportion?
Determine if the following are in proportion.
15, 45, 40, 120
If `(a + b)^3/(a - b)^3 = 64/27`
- Find `(a + b)/(a - b)`
- Hence using properties of proportion, find a : b.
