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## How do you simplify (root4(x^3)*root4(x^5))^(-2)?

smendyka
Featured 2 months ago

See a solution process below:

#### Explanation:

First, we can use this rule to combine the radicals within the parenthesis:

$\sqrt[n]{\textcolor{red}{a}} \cdot \sqrt[n]{\textcolor{b l u e}{b}} = \sqrt[n]{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}}$

${\left(\sqrt[4]{\textcolor{red}{{x}^{3}}} \cdot \sqrt[4]{\textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{\textcolor{red}{{x}^{3}} \cdot \textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2$

Next, use this rule for exponents to combine the terms within the radical:

${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

${\left(\sqrt[4]{\textcolor{red}{{x}^{3}} \cdot \textcolor{b l u e}{{x}^{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{{x}^{\textcolor{red}{3} + \textcolor{b l u e}{5}}}\right)}^{-} 2 = {\left(\sqrt[4]{{x}^{8}}\right)}^{-} 2$

Then, we can use this rule to rewrite the radical into an exponent:

$\sqrt[\textcolor{red}{n}]{x} = {x}^{\frac{1}{\textcolor{red}{n}}}$

${\left(\sqrt[\textcolor{red}{4}]{{x}^{8}}\right)}^{-} 2 = {\left({\left({x}^{8}\right)}^{\frac{1}{\textcolor{red}{4}}}\right)}^{-} 2$

Next, we can use this rule to simplify the inner exponents:

${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$

${\left({\left({x}^{\textcolor{red}{8}}\right)}^{\textcolor{b l u e}{\frac{1}{4}}}\right)}^{-} 2 = {\left({x}^{\textcolor{red}{8} \times \textcolor{b l u e}{\frac{1}{4}}}\right)}^{-} 2 = {\left({x}^{2}\right)}^{-} 2$

We can use the same rule to reduce the outer exponents:

${\left({x}^{\textcolor{red}{2}}\right)}^{\textcolor{b l u e}{- 2}} = {x}^{\textcolor{red}{2} \times \textcolor{b l u e}{- 2}} = {x}^{-} 4$

We can now use this rule to eliminate the negative exponent:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

${x}^{\textcolor{red}{- 4}} = \frac{1}{x} ^ \textcolor{red}{- - 4} = \frac{1}{x} ^ 4$

## How do you solve 7z^2=70z-147?

smendyka
Featured 2 months ago

See a solution process belowL

#### Explanation:

First, write this equation in standard form by subtracting $\textcolor{red}{70 z}$ and adding $\textcolor{b l u e}{147}$ to each side of the equation:

$7 {z}^{2} - \textcolor{red}{70 z} + \textcolor{b l u e}{147} = 70 z - \textcolor{red}{70 z} - 147 + \textcolor{b l u e}{147}$

$7 {z}^{2} - 70 z + 147 = 0 - 0$

$7 {z}^{2} - 70 z + 147 = 0$

Next, divide each side of the equation by $\textcolor{red}{7}$ to reduce the coefficients:

$\frac{7 {z}^{2} - 70 z + 147}{\textcolor{red}{7}} = \frac{0}{\textcolor{red}{7}}$

$\frac{7 {z}^{2}}{\textcolor{red}{7}} - \frac{70 z}{\textcolor{red}{7}} + \frac{147}{\textcolor{red}{7}} = 0$

$1 {z}^{2} - 10 z + 21 = 0$

${z}^{2} - 10 z + 21 = 0$

Then we can factor the right side of the equation as:

$\left(z - 3\right) \left(z - 7\right) = 0$

Now, solve each term on the right side of the equation for $0$:

Solution 1:

$z - 3 = 0$

$z - 3 + \textcolor{red}{3} = 0 + \textcolor{red}{3}$

$z - 0 = 3$

$z = 3$

Solution 2:

$z - 7 = 0$

$z - 7 + \textcolor{red}{7} = 0 + \textcolor{red}{7}$

$z - 0 = 7$

$z = 7$

**The Solutions Are: $z = 3$ and $z = 7$

We could also use the quadratic formula to solve this problem. The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 10}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{21}$ for $\textcolor{g r e e n}{c}$ gives:

$z = \frac{- \textcolor{b l u e}{\left(- 10\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 10\right)}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{21}\right)}}{2 \cdot \textcolor{red}{1}}$

$z = \frac{10 \pm \sqrt{\textcolor{b l u e}{100 - 84}}}{2}$

$z = \frac{10 - \sqrt{16}}{2}$ and $z = \frac{10 + \sqrt{16}}{2}$

$z = \frac{10 - 4}{2}$ and $z = \frac{10 + 4}{2}$

$z = \frac{6}{2}$ and $z = \frac{14}{2}$

$z = 3$ and $z = 7$

## Simplify exponential expressions using factorising?

EZ as pi
Featured 1 month ago

$\frac{1}{5} ^ \left(n + 7\right)$

#### Explanation:

In order to simplify exponents, make sure that the bases are all the same. Change numbers to the product of their prime factors.

$\frac{{\textcolor{red}{25}}^{2 n - 4}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times 5} = \frac{{\textcolor{red}{\left({5}^{2}\right)}}^{2 n - 4}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times 5}$

$= \frac{{\textcolor{red}{5}}^{4 n - 8}}{{5}^{3 n + 1} \times {5}^{2 n - 3} \times {5}^{1}}$

Add the indices of like bases when multiplying

(5^(4n-8))/(5^(5n-1)

Method 1

Subtract the indices when dividing

$= \frac{1}{5} ^ \left(5 n - 4 n - 1 - \left(- 8\right)\right)$

$= \frac{1}{5} ^ \left(n + 7\right)$

Method 2

${x}^{m} = \frac{1}{{x}^{-} m}$

(5^(4n-8))/(5^(5n-1)

$= \frac{1}{{5}^{5 n - 1} \times {5}^{- 4 n + 8}}$

$= \frac{1}{5} ^ \left(n + 7\right)$

Give the answer without negative or zero indices.

## How do you find the value of the discriminant and state the type of solutions given -2x^2-x-1=0?

LM
Featured 4 weeks ago

no real roots; $\Delta < 0$

#### Explanation:

$- 2 {x}^{2} - x - 1 = 0$ is already in $a {x}^{2} + b x + c$ form, so the $a$,$b$ and $c$ -values can be used.

$a = - 2$
$b = - 1$
$c = - 1$

$\Delta = {b}^{2} - 4 a c$
$= {\left(- 1\right)}^{2} - \left(4 \cdot \left(- 2 \cdot - 1\right)\right)$
$= 1 - 8$
$- 7$

$- 7 < 0 \therefore \Delta < 0$

hence, $- 2 {x}^{2} - x - 1 = 0$ has no real roots.

this can also be seen by graphing $- 2 {x}^{2} - x - 1 = 0$:

this parabola does not meet the $x$-axis on a graph for real numbers, so there are no (real) roots.

## How do you simplify 3/5×(-1/4 - 1/6) ÷ (-7/3 + 5/4)?

Meave60
Featured 2 weeks ago

3/5 xx (-1/4-1/6) -:(-7/3+5/4)=color(blue)(3/13

#### Explanation:

Simplify:

$\frac{3}{5} \times \left(- \frac{1}{4} - \frac{1}{6}\right) \div \left(- \frac{7}{3} + \frac{5}{4}\right)$

Follow the order of operations: parentheses, exponents, multiplication and division left to right, addition and subtraction left to right.

Simplify the parentheses first.

$\left(- \frac{1}{4} - \frac{1}{6}\right)$ must have a common denominator. The least common denominator (LCD) can be found by listing the multiples of $4$ and $6$.

$4 :$$4 , 8 , \textcolor{red}{12} , 16. . .$

$6 :$$6 , \textcolor{red}{12.} . .$

LCD$=$$\textcolor{red}{12}$

Multiply both fractions by a fraction equal to $1$, so that each will have the same denominator. For example, $\frac{5}{5} = 1$. This way, the values don't change.

$\left(- \frac{1}{4} \times \frac{\textcolor{t e a l}{3}}{\textcolor{t e a l}{3}} - \frac{1}{6} \times \frac{\textcolor{m a \ge n t a}{2}}{\textcolor{m a \ge n t a}{2}}\right)$

Simplify.

$\left(- \frac{3}{12} - \frac{2}{12}\right) = \left(- \frac{5}{12}\right)$

Return $\left(- \frac{5}{12}\right)$ to the original expression.

$\frac{3}{5} \times \left(- \frac{5}{12}\right) \div \left(- \frac{7}{3} + \frac{5}{4}\right)$

Simplify $\left(- \frac{7}{3} + \frac{5}{4}\right)$.

Follow the same procedure as with $\left(- \frac{1}{4} - \frac{1}{6}\right)$.

LCD$=$$12$

$\left(- \frac{7}{3} \times \frac{\textcolor{g r e e n}{4}}{\textcolor{g r e e n}{4}} + \frac{5}{4} \times \frac{\textcolor{red}{3}}{\textcolor{red}{3}}\right)$

Simplify.

$\left(- \frac{28}{12} + \frac{15}{12}\right) = - \frac{13}{12}$

Return $\left(- \frac{13}{12}\right)$ to the original equation.

$\frac{3}{5} \times \left(- \frac{5}{12}\right) \div \left(- \frac{13}{12}\right)$

Multiply $\frac{3}{5}$ and $- \frac{5}{12}$.

$- \frac{15}{60} \div \left(- \frac{13}{12}\right)$

Divide $- \frac{15}{60}$ and $- \frac{13}{12}$.

When dividing by a fraction, invert the fraction and multiply.

$- \frac{15}{60} \times - \frac{12}{13}$

Simplify.

$\frac{180}{780}$

Reduce by dividing the numerator and denominator by $60$.

$\frac{180 \div 60}{780 \div 60} = \frac{3}{13}$

## How do you prove log_2 3 is irrational number ?

Cesareo R.
Featured 3 weeks ago

See below.

#### Explanation:

$y = {\log}_{2} 3 \Leftrightarrow {2}^{y} = 3$ If $y$ is rational then $y = \frac{n}{m}$ with $\left\{n , m\right\} \in \mathbb{Z}$ and $m \ne 0$ or

${2}^{\frac{n}{m}} = 3 \Rightarrow {2}^{n} = {3}^{m}$

This last equality is an absurd because ${3}^{m}$ is odd and ${2}^{n}$ is even.

Concluding $y = {\log}_{2} 3$ is not rational, being then irrational.

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