1.  Solve for $h$: $S=2\pi r h+2\pi r^2$

Solution:

The variable $h$ is in the first term on the right, so isolate the first term by subtracting the second term:

$S-2\pi r^2=2\pi rh$

Now, divide $2\pi r$ on both sides:

Swicthing the sides,

$\dfrac{S-2\pi r^2}{2\pi r}=h$

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2.  Solve for $y$: $|y+2|-1=10$

First isolate the absolute value, the first term on the left hand side:

$|y+2|=10\color{red}+1$

$|y+2|=11$

$y+2=11$  or $y+2=-11$

Solving,

$y=11-2$ or $y=-11-2$

$y=9$ or $y=-13$

The solutions are,

$\{9, -13\}$  or  $\{-13, 9\}$

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3.  Solve for $f$: $|7f-2|=-9$

Solution:

Left hand side is the absolute value, $|7f-2|$ that is equal to a negative number $-9$.

Since an absolute cannot be a negative number, the equation has no solution.

We write the no solution as $\varnothing$.

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4.  Solve for $h$: $|h+3|+3\le 7$

Solution:

Isolate the absolute value:

$|h+3|=\le 7-3$

$|h+3|=\le 4$

$-4\le h+3 \le 4$

Subtracting $3$:

$-7 \le h \le 1$

Writing this in interval notation:

$[-7, 1]$

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5.  Solve for $y$: $|2y-1|+1\gt -7$

Solution:

First isolate the absolute value:$

$|2y-1|\gt -7-1$

$|2y-1|>0$

$2y-1\gt 0$ or $2y-1\lt 0$

Solving these two equations:

$y\gt \dfrac{1}{2}$ or  $y\lt \dfrac{1}{2}

Absolute value is greater than a negative number, means, it is greater than zero.

So, all the real numbers are the solution to the equation. This in interval notation is

$(-\infty, \infty)$

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10.  Determine whether the relation defines y as a function of x and give the domain.

$y=\sqrt{6x-4}$

$The given function is a square root function with a linear expression inside the square root. This is a one-to-one function:

To find the domain, set the expression inside the square root $\ge 0$

$6x-4\ge 0$

Now, solve for $x$,

$x\ge\dfrac{4}{6}$

Dividing out the common factor $2$, you get

$x\ge \dfrac{2}{3}$

Writing this inequality in interval notation:

$[\dfrac{2}{3}, \infty)$