Let n = the number. A number exceeds 45 inequality

1 answer:

3 0

This translates to "a number" is greater than 45. All you have to do now is translate these words into an algebraic statement. Basically, you replace "a number" with the variable which it is defined for, and you use the "greater than" symbol to show that the variable is greater than the value of 45.

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Please help radical expression dividing

77julia77 [94]

$\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad 
\begin{cases}
63=3\cdot 3\cdot 7\\
6=2\cdot 3
\end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}}
\\\\\\
\cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}$

$\bf \textit{now, rationalizing the denominator}\\\\
\cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}}
\\\\\\
\cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}$

and is all you can simplify from it.

so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.

6 0

3 years ago

1.Which expressions are polynomials? Select each correct answer. Pick more then one.

madreJ [45]

If all the power of the variables are greater than or equal to zero, then the expression is a valid polynomial.

So,

$7x^{7}-x-5$