A sum of money is divided in the ratio 2 : 3 : 7 .Given the largest share is $112,calculate the smallest share.

Mathematics

1 answer:

emmasim [6.3K]3 years ago

4 0

Answer:

$32

Step-by-step explanation:

A sum of money is divided in the ratio 2:3:7.

Let the share of each individual be represented by 2x,3x,7x respectively.

The largest share (7x) is given as $112.

$7x = 112$

$=>x = \frac{112}{7}$

$=>x = 16$

So the respective shares are $2\times16,3\times16,7\times16$

In other words, the shares are 32,48,112.

So the smallest share is $32

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Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

yulyashka [42]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ :

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ , then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$ , then we see that there are vertical asymptotes at $x=-10$ and $x=1$