To add these radicals together their radicands must be the same:

6\( \sqrt{8} \) + 2\( \sqrt{2} \)
6\( \sqrt{4 \times 2} \) + 2\( \sqrt{2} \)
6\( \sqrt{2^2 \times 2} \) + 2\( \sqrt{2} \)
(6)(2)\( \sqrt{2} \) + 2\( \sqrt{2} \)
12\( \sqrt{2} \) + 2\( \sqrt{2} \)

Now that the radicands are identical, you can add them together:

First, solve for \( \left|x + 8\right| \):

\( \left|x + 8\right| \) + 5 = -4
\( \left|x + 8\right| \) = -4 - 5
\( \left|x + 8\right| \) = -9

The value inside the absolute value brackets can be either positive or negative so (x + 8) must equal - 9 or --9 for \( \left|x + 8\right| \) to equal -9:

So, x = 1 or x = -17.

The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 6 meters so the equation becomes: 2w + 2h = 6.

Putting these two equations together and solving for width (w):

2w + 2h = 6
w + h = \( \frac{6}{2} \)
w + h = 3
w = 3 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 3 - 2w
3w = 3
w = \( \frac{3}{3} \)
w = 1

Since h = 2w that makes h = (2 x 1) = 2 and the area = h x w = 1 x 2 = 2 m²

To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:

\( \sqrt{\frac{4}{4}} \)
\( \frac{\sqrt{4}}{\sqrt{4}} \)
\( \frac{\sqrt{2^2}}{\sqrt{2^2}} \)
1

To subtract these radicals together their radicands must be the same:

9\( \sqrt{175} \) - 3\( \sqrt{7} \)
9\( \sqrt{25 \times 7} \) - 3\( \sqrt{7} \)
9\( \sqrt{5^2 \times 7} \) - 3\( \sqrt{7} \)
(9)(5)\( \sqrt{7} \) - 3\( \sqrt{7} \)
45\( \sqrt{7} \) - 3\( \sqrt{7} \)

Now that the radicands are identical, you can subtract them: