Question

Please solve, answer choices included.

Answer 1

4. To solve this problem, we divide the two expressions step by step:

$\frac{x+2}{x-1}* \frac{x^{2}+4x-5 }{x+4}$
Here we have inverted the second term since division is just multiplying the inverse of the term.

$\frac{x+2}{x-1}* \frac{(x+5)(x-1)}{x+4}$
In this step we factor out the quadratic equation.


$\frac{x+2}{1}* \frac{(x+5)}{x+4}$
Then, we cancel out the like term which is x-1.

We then solve for the final combined expression:
$\frac{(x+2)(x+5)}{(x+4)}$

For the restrictions, we just need to prevent the denominators of the two original terms to reach zero since this would make the expression undefined:

$x-1\neq0$
$x+5\neq0$
$x+4\neq0$

Therefore, x should not be equal to 1, -5, or -4.

Comparing these to the choices, we can tell the correct answer.

ANSWER: $\frac{(x+2)(x+5)}{(x+4)}$; $x\neq1,-4,-5$

5. To get the ratio of the volume of the candle to its surface area, we simply divide the two terms with the volume on the numerator and the surface area on the denominator:

$\frac{ \frac{1}{3} \pi r^{2}h }{ \pi r^{2}+ \pi r \sqrt{ r^{2} +h^{2} } }$

We can simplify this expression by factoring out the denominator and cancelling like terms.

$\frac{ \frac{1}{3} \pi r^{2}h }{ \pi r(r+ \sqrt{ r^{2} +h^{2} } )}$
$\frac{ rh }{ 3(r+ \sqrt{ r^{2} +h^{2} } )}$
$\frac{ rh }{ 3r+ 3\sqrt{ r^{2} +h^{2} } }$

We then rationalize the denominator:

$\frac{rh}{3r+3 \sqrt{ r^{2} + h^{2} }} * \frac{3r-3 \sqrt{ r^{2} + h^{2} }}{3r-3 \sqrt{ r^{2} + h^{2} }}$
$\frac{rh(3r-3 \sqrt{ r^{2} + h^{2} })}{(3r)^{2}-(3 \sqrt{ r^{2} + h^{2} })^{2}}}=\frac{3 r^{2}h -3rh \sqrt{ r^{2} + h^{2} }}{9r^{2} -9 (r^{2} + h^{2} )}=\frac{3rh(r -\sqrt{ r^{2} + h^{2} })}{9[r^{2} -(r^{2} + h^{2} )]}=\frac{rh(r -\sqrt{ r^{2} + h^{2} })}{3[r^{2} -(r^{2} + h^{2} )]}$

Since the height is equal to the length of the radius, we can replace h with r and further simplify the expression:

$\frac{r*r(r -\sqrt{ r^{2} + r^{2} })}{3[r^{2} -(r^{2} + r^{2} )]}=\frac{ r^{2} (r -\sqrt{2 r^{2} })}{3[r^{2} -(2r^{2} )]}=\frac{ r^{2} (r -r\sqrt{2 })}{-3r^{2} }=\frac{r -r\sqrt{2 }}{-3 }=\frac{r(1 -\sqrt{2 })}{-3 }$

By examining the choices, we can see one option similar to the answer.

ANSWER: $\frac{r(1 -\sqrt{2 })}{-3 }$