ISASP High School Algebra. Practice it free.

The factor $m$ is multiplied by $(x - h)$, so $x - h$ is isolated by using the inverse operation (division) on both sides, $\dfrac{y - k}{\textcolor{red}{m}} = x - h$. Then, $h$ is subtracted from $x$, so $x$ is isolated by using the inverse operation (addition) on both sides, $\dfrac{y - k}{m} + \textcolor{red}{h} = x$. Simplifying gives $x = \dfrac{y - k}{m} + h$.

Divide both sides by $\pi h$: $r^{2} = \dfrac{V}{\pi h}$. Take the positive square root: $r = \sqrt{\dfrac{V}{\pi h}}$.

The equation means $x + 8 = 21$ or $x + 8 = -21$. So $x = 13$ or $x = -29$. Both solutions are $-29$ and $13\textsf{.}$

Rewrite $|2x + 10| = |2(x + 5)| = |2|\,|x + 5| = 2|x + 5|$. Since $2|x + 5| = 26$, divide by $2$ to get $|x + 5| = 13$. The expression $x + 5$ equals $13$ or $-13$; the positive value is $13\textsf{.}$

The greatest common factor of $5x^{2}$ and $10x$ is $5x\textsf{.}$ Factoring gives $5x(x + 2)$.

More than $70$ means $s > 70$. At most $100$ means $s \le 100$, so $70 < s \le 100$.

$81 - x^{2} = 9^{2} - x^{2} = (9 - x)(9 + x)$.

Here $a = 1$, $b = 0$, and $c = -25$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-25) = 100\textsf{,}$ so there are $2$ distinct real roots.