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3 years ago

Help pls................

Mathematics

2 answers:

Sergeeva-Olga [200]3 years ago

5 0

Answer:

x = - 10, x = 10

Step-by-step explanation:

Given

- 7 + $(x^2-19)^{\frac{3}{4} }$ = 20 ( add 7 to both sides )

$(x^2-19)^{\frac{3}{4} }$ = 27 , that is

$\sqrt[4]{(x^2-19)^{3} }$ = 27 ( raise both sides to the power of 4 )

(x² - 19)³ = $27^{4}$ = 531441 ( take the cube root of both sides )

x² - 19 = $\sqrt[3]{531441}$ = 81 ( add 19 to both sides )

x² = 100 ( take the square root of both sides )

x = ± $\sqrt{100}$ = ± 10

solution is x = - 10, x = 10

Dominik [7]3 years ago

4 0

Hai!!! Your answer is both x=-10 and x=10

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The answer is option D) r < 5 or r > – 1.

I'm going to graph each inequality below on a number line.

A) r > 5 or r > – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\\\ \mathsf{r>5~~or~~r>-1}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\overset{********************~~~}{\textsf{|||||}\!\!\underset{5}{\bullet}\!\!\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}} \end{array}$

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—————

B) r > 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r>5}&\qquad\mathsf{\textsf{|||}\!\!\!\underset{-1}{\circ}\!\!\!\textsf{|||||}\!\!\underset{5}{\circ}\!\!\overset{*****~~~}{\textsf{|||}}\!\!\!\footnotesize\begin{array}{l}\blacktriangleright \end{array}\qquad \mathbb{R}}\\\\ \mathsf{r5~~or~~r$

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C) r < 5 or r < – 1.

$\large\begin{array}{cl} \mathsf{r$

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D) r < 5 or r > – 1.

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