Am I right??????????????

Mathematics

2 answers:

zubka84 [21]3 years ago

7 0

No its the last one :)))))

yaroslaw [1]3 years ago

3 0

Yes, your answer is correct.
Hope this helps

You might be interested in

-1/2 + 1 1/2 + 3 - n

GREYUIT [131]

Answer:

The answer all together would be 5 minus n.

Step-by-step explanation:

a negative and a positive equals a positive so -1/2 + 1 1/2 would equal 2 and 2 + 3 equals 5. Then, the remaining numbers are 5 and n. The final answer is 5 minus n because n is unknown.

4 0

3 years ago

How do you solve this? Thank you

V125BC [204]

2)

a)

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

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

b)

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

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

3)

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

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

$\bf 5\sqrt{x^5}-2\sqrt{x^3}+\sqrt[10]{x^7}-25x^3+10x^2-5\sqrt[5]{x^6}$

6 0

3 years ago

11. Which of the following equations use the distributive property to multiply rational numbers?

Veronika [31]

Answer:

well you see there is go.ogle

8 0

3 years ago

Pls help with this please

Marina86 [1]

You have to do 2 5/8 4 times but in a picture or scrap

6 0