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lapo4ka [179]
2 years ago
7

Brahmagupta solved a quadratic equation of the form ax2 + bx = c using the formula x=V

Mathematics
2 answers:
Hatshy [7]2 years ago
6 0

Answer: 1st box- subtracting 8 from both sides

2nd box- -8

Step-by-step explanation:

MaRussiya [10]2 years ago
5 0

Answer:

The correct answers are

1st box:  subtracting 8 from both sides

2nd box:  -8

<h3>Explanation:</h3><h3>They' correct on edge2020</h3><h3> </h3><h3 />
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How do you solve this? Thank you
V125BC [204]
2)

a)

\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad&#10;\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\&#10;-------------------------------\\\\&#10;(4x^5\cdot x^{\frac{1}{3}})+(2x^4\cdot x^{\frac{1}{3}})-(7x^3\cdot x^{\frac{1}{3}})+(3x^2\cdot x^{\frac{1}{3}})\\\\+(9x^1\cdot x^{\frac{1}{3}})-(1\cdot x^{\frac{1}{3}})&#10;\\\\\\&#10;4x^{5+\frac{1}{3}}+2x^{4+\frac{1}{3}}-7x^{3+\frac{1}{3}}+9x^{1+\frac{1}{3}}-x^{\frac{1}{3}}

\bf 4x^{\frac{16}{3}}+2x^{\frac{13}{3}}-7x^{\frac{10}{3}}+9x^{\frac{4}{3}}-x^{\frac{1}{3}}&#10;\\\\\\&#10;4\sqrt[3]{x^{16}}+2\sqrt[3]{x^{13}}-7\sqrt[3]{x^{10}}+9\sqrt[3]{x^4}-\sqrt[3]{x}

b)

\bf \cfrac{4x^5+2x^4-7x^3+3x^2+9x-1}{x^{\frac{1}{3}}}\impliedby \textit{distributing the denominator}&#10;\\\\\\&#10;\cfrac{4x^5}{x^{\frac{1}{3}}}+\cfrac{2x^4}{x^{\frac{1}{3}}}-\cfrac{7x^3}{x^{\frac{1}{3}}}+\cfrac{3x^2}{x^{\frac{1}{3}}}+\cfrac{9x}{x^{\frac{1}{3}}}-\cfrac{1}{x^{\frac{1}{3}}}&#10;\\\\\\&#10;(4x^5\cdot x^{-\frac{1}{3}})+(2x^4\cdot x^{-\frac{1}{3}})-(7x^3\cdot x^{-\frac{1}{3}})+(3x^2\cdot x^{-\frac{1}{3}})\\\\+(9x^1\cdot x^{-\frac{1}{3}})-(1\cdot x^{-\frac{1}{3}})

\bf 4x^{5-\frac{1}{3}}+2x^{4-\frac{1}{3}}-7x^{3-\frac{1}{3}}+9x^{1-\frac{1}{3}}-x^{-\frac{1}{3}}&#10;\\\\\\&#10;4x^{\frac{14}{3}}+2x^{\frac{11}{3}}-7x^{\frac{8}{3}}+9x^{\frac{2}{3}}-x^{-\frac{1}{3}}&#10;\\\\\\&#10;4\sqrt[3]{x^{14}}+2\sqrt[3]{x^{11}}-7\sqrt[3]{x^{8}}+9\sqrt[3]{x^{2}}-\frac{1}{\sqrt[3]{x}}



3)

\bf \begin{cases}&#10;f(x)=\sqrt{x}-5x\implies &f(x)x^{\frac{1}{2}}-5x\\\\&#10;g(x)=5x^2-2x+\sqrt[5]{x}\implies &g(x)=5x^2-2x+x^{\frac{1}{5}}&#10;\end{cases}&#10;\\\\\\&#10;\textit{let's multiply the terms from f(x) by each term in g(x)}&#10;\\\\\\&#10;x^{\frac{1}{2}}(5x^2-2x+x^{\frac{1}{5}})\implies x^{\frac{1}{2}}5x^2-x^{\frac{1}{2}}2x+x^{\frac{1}{2}}x^{\frac{1}{5}}

\bf 5x^{\frac{1}{2}+2}-2x^{\frac{1}{2}+1}+x^{\frac{1}{2}+\frac{1}{5}}\implies \boxed{5x^{\frac{5}{2}}-2x^{\frac{3}{2}}+x^{\frac{7}{10}}}&#10;\\\\\\&#10;-5x(5x^2-2x+x^{\frac{1}{5}})\implies -5x5x^2-5x2x+5xx^{\frac{1}{5}}&#10;\\\\\\&#10;-25x^3+10x^2-5x^{1+\frac{1}{5}}\implies \boxed{-25x^3+10x^2-5x^{\frac{6}{5}}}

\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}
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2 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
agasfer [191]

Answer:

rectangle

Step-by-step explanation:

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Step-by-step explanation:

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J/9=5
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j = 45
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X is equal to 19. Hoped this helped
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