Can you figure out a problem? The quotient of a number and 9 times its reciporical is 1. Find the number(s)?

Answer 1

The quotient of a and b is a/b. In your case, this is n and 9 multiplied by the reciporical of n.

The reciporical of n is 1/n, and 9 times that is 9/n

So you want to solve n(9/n) = 1

n/(9/n) = 1
Multiply top and bottom by n
n²/9 = 1

9/9 = 1
therefore
n² = 9

n = ±3

Answer 2

Let the number be x. The quotient is the result of a division problem.

The reciprocal of a number is 1 over that number; so the reciprocal of x is 1/x. 9 times that reciprocal is 9/x.

The quotient of x and 9/x is x/(9/x)

x/(9/x)=1
x/x * x/(9/x) = 1
x^2/9 = 1
9 * x^2/9 = 1 * 9
x^2 = 9
x = 3 or -3.

Answer 3

Let x be the number.

Then x / [9(1/x)] = 1

Multiply both sides by 9(1/x),

x = 9(1/x)

Multiply both sides by x.

x^2 = 9

Solve.

x^2 - 9 = 0
(x - 3)(x + 3) = 0

x = {-3, 3}

So you have two solutions: -3 and 3

Answer 4

if a number be a, it's reciprocal is 1/a

if a/(-1/a)(9)=1

-a^2/9=1

-a^2=9

a^2=-9

a=+-3

the result does not satisfy the equation, no such number exists, but i believe you are supposed to multiply the reciprocal by -9 then the equation satisfies

Answer 5

Check your question

let the number be x
let quotient be q
let the divisor be d
then

q+9/(qd)=1

find x

if we take after "of" together it will be
x/(9/x)=1
x^2=9
x1=-3
x2=3

Answer 6

x/(9/x)=1.....x=9/x.......x^2=9......x=plus or minus 3