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## Evaluate the following?

Mia
Featured 2 months ago

$\frac{7}{11}$

#### Explanation:

Computing the rational number with fractional exponent ${\left(\frac{u}{v}\right)}^{\frac{1}{6}} = {u}^{\frac{1}{6}} / {v}^{\frac{1}{6}}$ is determined by prime factorizing the numerator and denominator .

example of: #sqrtx = x^(1/2#:

$\sqrt{\left(\frac{4}{9}\right)} = {\left(\frac{4}{9}\right)}^{\textcolor{b l u e}{\frac{1}{2}}} = {\left({2}^{2}\right)}^{\textcolor{b l u e}{\frac{1}{2}}} / {\left({3}^{2}\right)}^{\textcolor{b l u e}{\frac{1}{2}}} = \frac{2}{3}$

The rational number above ${\left(\frac{117649}{1771561}\right)}^{\frac{1}{6}} = {\left(117649\right)}^{\frac{1}{6}} / {\left(1771561\right)}^{\frac{1}{6}}$

Prime Factorization :
$\textcolor{b l u e}{117649 = 7 \times 7 \times 7 \times 7 \times 7 \times 7 = {7}^{6}}$
$\textcolor{b r o w n}{1771561 = 11 \times 11 \times 11 \times 11 \times 11 \times 11 = {11}^{6}}$

${\left(\frac{117649}{1771561}\right)}^{\frac{1}{6}} = {\left(117649\right)}^{\frac{1}{6}} / {\left(1771561\right)}^{\frac{1}{6}} = {\left(\textcolor{b l u e}{{7}^{6}}\right)}^{\frac{1}{6}} / {\left(\textcolor{b r o w n}{{11}^{6}}\right)}^{\frac{1}{6}}$

Then we apply the power of a power with base $a$ :
${\left({a}^{m}\right)}^{\frac{1}{n}} = {a}^{\frac{m}{n}}$

${\left(\frac{117649}{1771561}\right)}^{\frac{1}{6}} = \left({7}^{\frac{6}{6}} / {11}^{\frac{6}{6}}\right) = \frac{7}{11}$

## The length of a rectangle is 7 feet larger than the width. The perimeter of the rectangle is 26 ft. How do you write an equation to represent the perimeter in terms of its width (w). What is the length?

smendyka
Featured 2 months ago

An equation to represent the perimeter in terms of its width is: $p = 4 w + 14$ and the length of the rectangle is $10$ ft.

#### Explanation:

Let the width of the rectangle be $w$.

Let the length of the rectangle be $l$.

If the length ($l$) is 7 feet longer than the width, then the length can be written in terms of the width as:

$l = w + 7$

The formula for perimeter of a rectangle is:

$p = 2 l + 2 w$ where $p$ is the perimeter, $l$ is the length and $w$ is the width.

Substituting $w + 7$ for $l$ gives an equation to represent the perimeter in terms of its width:

$p = 2 \left(w + 7\right) + 2 w$

$p = 2 w + 14 + 2 w$

$p = 4 w + 14$

Substituting $26$ for $p$ allows us to solve for $w$.

$26 = 4 w + 14$

$26 - 14 = 4 w + 14 - 14$

$12 = 4 w$

$\frac{12}{4} = 4 \frac{w}{4}$

$w = 3$

Sustituting $3$ for $w$ in the equation above, $l = w + 7$ allows us to determine the length:

$l = 3 + 7$

$l = 10$

## The hypotenuse of a right triangle is 39 inches, and the length of one leg is 6 inches longer than twice the other leg. How do you find the length of each leg?

George C.
Featured 2 months ago

The legs are of length $15$ and $36$

#### Explanation:

Method 1 - Familiar triangles

The first few right angled triangles with an odd length side are:

$3 , 4 , 5$

$5 , 12 , 13$

$7 , 24 , 25$

Notice that $39 = 3 \cdot 13$, so will a triangle with the following sides work:

$15 , 36 , 39$

i.e. $3$ times larger than a $5 , 12 , 13$ triangle ?

Twice $15$ is $30$, plus $6$ is $36$ - Yes.

$\textcolor{w h i t e}{}$
Method 2 - Pythagoras formula and a little algebra

If the smaller leg is of length $x$, then the larger leg is of length $2 x + 6$ and the hypotenuse is:

$39 = \sqrt{{x}^{2} + {\left(2 x + 6\right)}^{2}}$

$\textcolor{w h i t e}{39} = \sqrt{5 {x}^{2} + 24 x + 36}$

Square both ends to get:

$1521 = 5 {x}^{2} + 24 x + 36$

Subtract $1521$ from both sides to get:

$0 = 5 {x}^{2} + 24 x - 1485$

Multiply both sides by $5$ to get:

$0 = 25 {x}^{2} + 120 x - 7425$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - 144 - 7425$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - 7569$

$\textcolor{w h i t e}{0} = {\left(5 x + 12\right)}^{2} - {87}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(5 x + 12\right) - 87\right) \left(\left(5 x + 12\right) + 87\right)$

$\textcolor{w h i t e}{0} = \left(5 x - 75\right) \left(5 x + 99\right)$

$\textcolor{w h i t e}{0} = 5 \left(x - 15\right) \left(5 x + 99\right)$

Hence $x = 15$ or $x = - \frac{99}{5}$

Discard the negative solution since we are seeking the length of the side of a triangle.

Hence the smallest leg is of length $15$ and the other is $2 \cdot 15 + 6 = 36$

## How do you solve #6x + 9= 39#?

smendyka
Featured 2 months ago

$x = 5$

#### Explanation:

Take the following steps to isolate and solve for $x$ while keeping the equation balanced.

Subtract $9$ from each side of the equation:

$6 x + 9 - 9 = 39 - 9$

$6 x + 0 = 30$

$6 x = 30$

Divide each side of the equation by $6$

$\frac{6 x}{6} = \frac{30}{6}$

$1 x = 5$

$x = 5$

## How do you solve #\sqrt { 2x + 48} = x#?

HSBC244
Featured 1 month ago

$\left\{8\right\}$.

#### Explanation:

Square both sides:

${\left(\sqrt{2 x + 48}\right)}^{2} = {x}^{2}$

$2 x + 48 = {x}^{2}$

$0 = {x}^{2} - 2 x - 48$

$0 = \left(x - 8\right) \left(x + 6\right)$

$x = 8 \mathmr{and} - 6$

Check, as extraneous solutions may have been introduced in the solving process.

#sqrt(2(8) + 48) =^? 8#

#sqrt(64) = 8" "color(green)(√)#

AND

#sqrt(2(-6) + 48) =^? -6#

$\sqrt{36} \ne - 6 \text{ } \textcolor{red}{\times}$

Hopefully this helps!

## Rectangle A, (dimensions 6 by 10-x) has an area twice that of rectangle B (dimensions x by 2x+1) . What are the lengths and widths of both rectangles?

HSBC244
Featured 3 weeks ago

•Rectangle A: 6 by 7
•Rectangle B: 7 by 3

#### Explanation:

The area of a rectangle is given by $\textcolor{red}{A = l \cdot w}$.

The area of rectangle A is $6 \left(10 - x\right) = 60 - 6 x$

The area of rectangle B is $x \left(2 x + 1\right) = 2 {x}^{2} + x$

We are given that the area of rectangle A is twice the area of rectangle B. Therefore, we can write the following equation.

$60 - 6 x = 2 \left(2 {x}^{2} + x\right)$

$60 - 6 x = 4 {x}^{2} + 2 x$

$0 = 4 {x}^{2} + 8 x - 60$

$0 = 4 \left({x}^{2} + 2 x - 15\right)$

$0 = \left(x + 5\right) \left(x - 3\right)$

$x = - 5 \mathmr{and} 3$

A negative answer for $x$ is impossible, since we're talking about geometric shapes.

Therefore, the rectangles have the following measurements:

•Rectangle A: 6 by 7
•Rectangle B: 7 by 3

As you can see, rectangle A's area is twice the area of rectangle B, just as the problem indicated.

Hopefully this helps!

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