# If f(x)=5x, g(x)=1/x, and h(x)=x^3, how do you find (5h(1)-2g(3))/(3f(2))?

Jan 28, 2016

$\frac{13}{90}$, or $0.1 \overline{4}$

#### Explanation:

$\frac{5 h \left(1\right) - 2 g \left(3\right)}{3 f \left(2\right)}$

Now let's plug in the values for those functions:

$\frac{5 \cdot \left({1}^{3}\right) - 2 \cdot \left(\frac{1}{3}\right)}{3 \cdot \left(5 \cdot 2\right)}$

${1}^{3} = 1$, and since any number times $1$ is the same, we can remove it.

$\frac{5 - 2 \cdot \left(\frac{1}{3}\right)}{3 \cdot \left(5 \cdot 2\right)}$

Now we can simplify $2 \cdot \left(\frac{1}{3}\right)$ to $\frac{2}{3}$

$\frac{5 - \frac{2}{3}}{3 \cdot \left(5 \cdot 2\right)}$

And we can multiply the numbers in the denominator $3 \cdot 5 = 15$, and $15 \cdot 2 = 30$

$\frac{5 - \frac{2}{3}}{30}$

Now let's try and get rid of that ugly fraction in a fraction! We can do that by multiplying the entire fraction by $\frac{3}{3}$ which is also equal to $1$, and like we said above, we're not actually changing the value by doing that!

$\frac{5 - \frac{2}{3}}{30} \cdot \frac{3}{3}$

Multiplying through on both sides:

$\frac{5 \cdot 3 - \left(\frac{2}{3}\right) \cdot 3}{30 \cdot 3}$

Now simplify the multiplications:

$\frac{15 - 2}{90}$

Now subtract in the numerator... and...

$\frac{13}{90}$

There you have it!

Jan 28, 2016

$\frac{13}{90}$

#### Explanation:

$\frac{5 h \left(1\right) - 2 g \left(3\right)}{3 f \left(2\right)}$
=(5(1)^3-2(1/3))/(3(5*2)
$= \frac{5 - \frac{2}{3}}{3 \cdot 10}$
$= \frac{\frac{13}{3}}{30}$
$= \frac{13}{3} \cdot \frac{1}{30}$
$= \frac{13}{90}$